A First Course in Linear Algebra: Study Guide for the Undergraduate Linear Algebra Course
AA First Course in Linear Algebra
About the Author
Mohammed Kaabar is a math tutor at the Math Learning Center (MLC) at Washington State University, Pullman, and he is interested in linear algebra, scientific computing, numerical analysis, differential equations, and several programming languages such as SQL, C
This book is available under a Creative Commons license. http://creativecommons.org/licenses/by-nc-nd/4.0/ Note: If you find this book helpful and useful, and you see an opportunity to cite it in any of your publications, I would be very happy and appreciative of the citation. Please cite this book as: Kaabar, M.K.: A First Course in Linear Algebra: Study Guide for the Undergraduate Linear Algebra Course. Printed by CreateSpace, Charleston, SC (2014)
Table of Contents
Introduction vii 1 Systems of Linear Equations 1 1.1 Row Operations Method……………….........….1 1.2 Basic Algebra of Matrix……………………….17 1.3 Linear Combinations…………………….…….19 1.4 Square Matrix……………………….………….29 1.5 Inverse Square Matrix…………………..…….33 1.6 Transpose Matrix……………..………….…….42 1.7 Determinants……….……………………..…....46 1.8 Cramer’s Rule……………………………..…….53 1.9 Adjoint Method………………………………….55 1.10 Exercises………………………………...……….60 2 Vector Spaces 62 2.1 Span and Vector Space...................................62 2.2 The Dimension of Vector Space………….......65 2.3 Linear Independence……….………………….66 2.4 Subspace and Basis…………………………….69 2.5 Exercises…………………………………….…...82 vi TABLE OF CONTENTS 3 Homogeneous Systems 84 3.1 Null Space and Rank......................................84 3.2 Linear Transformation…………………..…….91 3.3 Kernel and Range …………………...….…….100 3.4 Exercises……………………………………..…105 4 Characteristic Equation of Matrix 107 4.1 Eigenvalues and Eigenvectors……..…..……107 4.2 Diagonalizable Matrix…...............................112 4.3 Exercises………………………………………...115 5 Matrix Dot Product 116 5.1 The Dot Product in ℝ 𝑛 ..................................116 5.2 Gram-Schmidt Orthonormalization ……….118 5.3 Exercises........................................................120 Answers to Odd-Numbered Exercises 121 Bibliography 125 Introduction
In this book, I wrote five chapters: Systems of Linear Equations, Vector Spaces, Homogeneous Systems, Characteristic Equation of Matrix, and Matrix Dot Product. I also added exercises at the end of each chapter above to let students practice additional sets of problems other than examples, and they can also check their solutions to some of these exercises by looking at “Answers to Odd-Numbered Exercises” section at the end of this book. This book is very useful for college students who studied Calculus I, and other students who want to review some linear algebra concepts before studying a second course in linear algebra. According to my experience as a math tutor at Math Learning Center, I have noticed that some students have difficulty to understand some linear algebra concepts in general, and vector spaces concepts in particular. Therefore, my purpose is to provide students with an interactive method to explain the concept, and then provide different sets of examples related to that concept. If you have any comments related to the contents of this book, please email your comments to [email protected]. I wish to express my gratitude and appreciation to my father, my mother, and my brother. I would also like to give a special thanks to my mathematics professor Dr. Ayman Rateb Badawi for his brilliant efforts in revising the content of this book, and I would also like to thank all my mathematics professors who taught me math courses at Washington State University, Pullman and American University of Sharjah. Ultimately, I would appreciate to consider this book as a milestone for devolving more math books that can serve our mathematical society. vii hapter 1 Systems of Linear Equations
In this chapter, we discuss how to solve 𝑛 × 𝑚 systems of linear equations using row operations method. Then, we give an introduction to basic algebra of matrix including matrix addition, matrix subtraction and matrix multiplication. We cover in the remaining sections some important concepts of linear algebra such as linear combinations, determinants, square matrix, inverse square matrix, transpose matrix, Cramer’s Rule and Adjoint Method.
First of all, let’s start with a simple example about 𝑛 ×𝑚 systems of linear equation. Example 1.1.1 Solve for x and y for the following system of linear equations: { 3𝑥 + 2𝑦 = 5−2𝑥 + 𝑦 = −6
Solution: Let’s start analyzing this system. 1
2 M. Kaabar First of all, each variable in this system is to the power 1 which means that this system is a linear system. As given in the question itself, implies that the number of equations is 2, and the number of unknown variables is also 2. (𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧𝐬) × (𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐔𝐧𝐤𝐨𝐰𝐧 𝐕𝐚𝐫𝐢𝐚𝐛𝐥𝐞𝐬)
Then, the unknown variables in this question are x and y. To solve for x and y, we need to multiply the first equation
3𝑥 + 2𝑦 = 5 by 2, and we also need to multiply the second equation −2𝑥 + 𝑦 = −6 by 3. Hence, we obtain the following: { 6𝑥 + 4𝑦 = 10 … … … … … . .1−6𝑥 + 3𝑦 = −18 … … … … .2
By adding equations 1 and 2, we get the following:
7𝑦 = −8
Therefore, 𝑦 = −87
Now, we need to substitute the value of y in one of the two original equations. Let’s substitute y in the first equation
3𝑥 + 2𝑦 = 5 as follows:
3𝑥 + 2 ( −87 ) = 5 is equivalent to
3𝑥 = 5 − 2 ( −87 ) = 5 + ( ) =
Then, 𝑥 =
Therefore, 𝑥 = 𝑎𝑛𝑑 𝑦 = −87
Hence, we solve for x and y. xample 1.1.2 Describe the following system: {2𝑥 + 3𝑦 − 2𝑧 = 105𝑥 + 2𝑦 + 𝑧 = 13
Solution: Let’s start asking ourselves the following questions. Question 1: How many equations do we have? Question 2: How many unknown variables do we have? Question 2: What is the power of each unknown variable? If we can answer the three questions above, then we can discuss the above systems. Let’s start now answering the three questions above. Answer to Question 1: We have 2 equations:
2𝑥 + 3𝑦 −2𝑧 = 10 and
5𝑥 + 2𝑦 + 𝑧 = 13 . Answer to Question 2: We have 3 unknown variables: x, y and z. Answer to Question 3: Each unknown variable is to the power 1. After answering the above three question, we have now an idea about the above system. As we remember from example 1.1.1 that the system of equations has the following form: (𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧𝐬) × (𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐔𝐧𝐤𝐨𝐰𝐧 𝐕𝐚𝐫𝐢𝐚𝐛𝐥𝐞𝐬)
Since the number of equations in this example is 2 and the number of unknown variables is 3, then using the above form, the above system is system. Recall that each unknown variable is to the power 1. This means that it is a linear system. 3
4 M. Kaabar Solving for unknown variables in or systems of linear equations using some arithmetic operations such as additions, subtractions and multiplications of linear equations is very easy. But if we have for example system of linear equations, using some arithmetic operations in this case will be very difficult and it will take long time to solve it. Therefore, we are going to use a method called Row Operation Method to solve complicated 𝑛 × 𝑚 systems of linear equations. Now, let’s start with an example discussing each step of Row Operation Method for solving 𝑛 × 𝑚 system of linear equations. Example 1.1.3 Solve for x , x and x for the following system of linear equations: { 𝑥 + 𝑥 − 𝑥 = 22𝑥 − 𝑥 + 𝑥 = 2−𝑥 + 2𝑥 = 1 Solution: In the above system, we have 3 equations and 3 unknown variables x , x and x . To solve this system, we need to use Row Operation Method. The following steps will describe this method. Step 1: Rewrite the system as a matrix, and this matrix is called Augmented Matrix. Each row in this matrix is a linear equation. x x x
3 C (1 1 −12 −1 10 −1 2 |221) s we can see from the above augmented matrix, the first column represents the coefficients of x in the three linear equations. The second column represents the coefficients of x in the three linear equations. The third column represents the coefficients of x in the three linear equations. The fourth column is not an actual column but it just represents the constants because each linear equation equals to a constant. Step 2: Start with the first row in matrix and make the first non-zero number equals to 1. This 1 is called a leader number. x x x
3 C (1 1 −12 −1 10 −1 2 |221)
Since the leader number is already 1in this matrix, then we do not need to do anything with this number. Otherwise, we need to make this leader number equals to 1 by multiplying the whole row with a non-zero number. Step 3: Eliminate all numbers that are exactly below the leader number in other words use the leader number to eliminate all things below it. Step 4: Move to the second row and repeat what has been done in step 2. 5
6 M. Kaabar -2R +R -- R (This means that we multiply the first row by -2 and we add it to the second row and the change will be only in the second row and no change in the first row) Hence, we obtain: (1 1 −10 −3 30 −1 2 | 2−21 ) −13 R (This means that we multiply the second row by −13 and the change will be only in the second row and no change in other rows) Hence, we obtain: (1 1 −10 1 −10 −1 2 |2 R +R -- R (This means that we add the second row to the third row and the change will be only in the third row and no change in the second row) R +R -- R (This means that we multiply the second row by -1, and we add it to the first row and the change will be only in the first row and no change in the second row) Hence, we obtain: ( 1 0 00 1 −10 0 1 ||432353) Step 5: Move to the third row and do the same as we did in the first row and the second row of matrix. R +R -- R (This means that we add the third row to the second row and the change will be only in the second row and no change in the third row) Hence, we obtain: ( 1 0 00 1 00 0 1||437353) Therefore, x = , x = and x = These are one solution only. Hence, the system has a unique solution (one solution). 7 8 M. Kaabar
After this example, we will get introduced to two new definitions. Definition 1.1.1 𝑛 × 𝑚 system of linear equations is consistent if it has a solution; otherwise, it is called inconsistent. Definition 1.1.2 𝑛 × 𝑚 system of linear equations has a unique solution if each variable has exactly one value. Now, let’s apply what we have learned from example 1.1.3 for the following example 1.1.4. Example 1.1.4 Solve for x , x , x , x and x for the following system of linear equations: { 𝑥 − 𝑥 + 𝑥 − 𝑥 = 1−2𝑥 + 𝑥 − 𝑥 = 0−𝑥 + 𝑥 + 2𝑥 − 10𝑥 = 12 Solution: In the above system, we have 3 equations and 5 unknown variables x , x , x , x and x . To solve this system, we need to use Row Operation Method. First, we need to construct the augmented matrix. x x x x x
5 C ( 0 1 −1−2 0 10 −1 1 1 −10 −12 −10 | 1012)
The leader number here is 1. ( 0 1 −1−2 0 10 −1 1 1 −10 −12 −10 | 1012) y doing the same steps in example 1.1.3, we obtain the following: R +R -- R x x x x x
5 C ( 0 1 −1−2 0 10 0 0 1 −10 −13 −11 | 1013) −12 R x x x x x
5 C (0 1 −11 0 −120 0 0 1 −10 123 −11 | 1013) R x x x x x
5 C (0 1 −11 0 −1/20 0 0 1 −10 1/21 −11/3 | 10133 ) -R +R -- R x x x x x
5 C (0 1 −11 0 −1/20 0 0 0 8/30 1/21 −11/3 |−10/30 ) Hence, we obtain from the above matrix: 9
10 M. Kaabar 𝑥 − 𝑥 + 83 𝑥 = − 103𝑥 − 12 𝑥 + 12 𝑥 = 0𝑥 − 113 𝑥 = 133 Since we have 3 leader numbers (numbers equal to 1) in x , x and x columns, then x , x and x are called leading variables. All other variables are called free variables such that x , x ∈ ℝ . Since we have leading and free variables, we need to write the leading variables in terms of free variables. 𝑥 = 12 𝑥 − 12 𝑥 𝑥 = 𝑥 − 83 𝑥 − 103𝑥 = 113 𝑥 + 133 Now, let x = 0 and x = 0 (You can choose any value for x and x because both of them are free variables). Therefore, we obtain 𝑥 = 12 (0) − 12 (0) = 0𝑥 = 0 − 83 (0) − 103 = − 103𝑥 = 113 (0) + 133 = 133 Hence, 𝑥 = 0 , 𝑥 = − 𝑎𝑛𝑑 𝑥 = nother possible solution for example 1.1.4: Now, let x = 1 and x = 0 (You can choose any value for x and x because both of them are free variables). Therefore, we obtain 𝑥 = 12 (1) − 12 (0) = 12𝑥 = 1 − 83 (0) − 103 = 1 − 103𝑥 = 113 (0) + 133 = 133 = − 73 Hence, we obtain 𝑥 = 12 , 𝑥 = − 73 𝑎𝑛𝑑 𝑥 = 133 This is another solution. Summary of Row Operations Method in 𝒏 × 𝒎
Systems of Linear Equations: Suppose ∝ is a non-zero constant, and 𝑖 𝑎𝑛𝑑 𝑘 are row numbers in the augmented matrix. * ∝ R i , ∝≠ 0 (Multiply a row with a non-zero constant ∝ ). * R i ↔ R k (Interchange two rows). * ∝ R i +R k -- R k (Multiply a row with a non-zero constant ∝, and add it to another row ). Note: The change is in R k and no change is in R i . Let’s start with other examples that will give us some important results in linear algebra. 11
12 M. Kaabar Example 1.1.5 Solve for x , x and x for the following system of linear equations: { 𝑥 + 2𝑥 − 𝑥 = 22𝑥 + 4𝑥 − 2𝑥 = 6 Solution: In the above system, we have 2 equations and 3 unknown variables x , x and x . To solve this system, we need to use Row Operation Method. First, we need to construct the augmented matrix. x x x
3 C (1 2 −12 4 −2 |26) -2R +R -- R x x x
3 C (1 2 −10 0 0 |22)
We stop here and read the above matrix as follows: 𝑥 + 2𝑥 − 𝑥 = 20 = 2 Hence, we get introduced to a new result. Result 1.1.1 The system is inconsistent if and only if after reducing one of the equations, we have 0 equals to a non-zero number.
Example 1.1.6 Solve for x , x and x for the following system of linear equations: { 𝑥 + 2𝑥 − 𝑥 = 22𝑥 + 4𝑥 − 2𝑥 = 4 Solution: In the above system, we have 2 equations and 3 unknown variables x , x and x . To solve this system, we need to use Row Operation Method. First, e need to construct the augmented matrix as we did in the previous example. x x x
3 C (1 2 −12 4 −2 |24) -2R +R -- R x x x
3 C (1 2 −10 0 0 |20)
We stop here and read the above matrix as follows: 𝑥1 + 2𝑥2 − 𝑥3 = 20 = 0
The solution is x = -2 x + x + 2 x is a leading variable, and x , x ∈ ℝ free variables. Now, let x = 1 and x = 0 (You can choose any value for x and x because both of them are free variables). Therefore, we obtain 𝑥1 = −2(0) + (0) + 2 = 2 𝑎𝑛𝑑 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 0. Hence, we get introduced to new results. Result 1.1.2 The 𝑛 × 𝑚 system is consistent if and only if it has a unique solution (no free variable).
Result 1.1.3 The 𝑛 × 𝑚 system is consistent if and only if it has infinitely many solutions (free variables).
Result 1.1.4 Assume the 𝑛 × 𝑚 system is consistent, and we have more variables than equations (m>n). Then, the system has infinitely many solutions.
Example 1.1.7 Given an augmented matrix of a system: ( 2 5 3−2 4 𝑏−2 −5 −3 |−13𝑑 )
14 M. Kaabar For what values of b and d will the system be consistent? Solution: We need to multiply the leader number (2) by a non-zero constant to make it equal to 1 instead of 2 as follows: R ( 1 52 32−2 4 𝑏−2 −5 −3 |− 123𝑑 ) +R -- R +R -- R (1 52 320 9 𝑏 + 30 0 0 | − 122𝑑 − 1) Therefore, 𝑏 ∈ ℝ and d=1 When the system is consistent, then we must have infinitely many solutions. In this case, b is a free variable. Example 1.1.8 Given an augmented matrix of a system: ( 1 −1 2−1 2 −2−2 2 −4 |𝑎𝑏𝑐) For what values of a, b and c will the system be consistent? Solution: We need to use the Row Operation Method to find a, b and c as follows: R +R -- R R +R -- R (1 −1 20 1 00 0 0 | 𝑎𝑎 + 𝑏2𝑎 + 𝑐) R +R -- R (1 0 20 1 00 0 0 |2𝑎 + 𝑏𝑎 + 𝑏2𝑎 + 𝑐) We stop here and read the above matrix. Hence, we obtain: x + 2x = 2a + b x = a + b 0 = 2a + c Therefore, 2a + c = 0 which means that c = -2a Hence, the solution is 𝑎, 𝑏 ∈ ℝ (free variables), and c = -2a. Let’s now discuss some new definitions and results of linear algebra. Definition 1.1.3 Suppose we have the following system of linear equations: { 2𝑥 − 3𝑥 + 𝑥 = 0𝑥 + 𝑥 − 3𝑥 = 0−𝑥 + 3𝑥 − 𝑥 = 0 This system is called homogeneous system of linear equations because all constants on the right-side are zeros. Result 1.1.5 Every 𝑛 × 𝑚 homogeneous system is consistent.
Result 1.1.6 Since every 𝑛 × 𝑚 homogeneous system is consistent, then x =0, x =0, x =0, …, x m =0 is a solution. This solution is called Trivial Solution. 15
16 M. Kaabar Result 1.1.7 The solution is called Non-Trivial if it has at least one value that is not zero. Result 1.1.8 In a homogeneous system if number of variables is greater than number of equations, then the system has infinitely many solutions. Result 1.1.9 Given the following matrix: [0 1 01 0 30 0 0 0 1 12 5 60 0 1]
This matrix is called Reduced Matrix because all the numbers exactly below each leader number are zeros. Result 1.1.10 Given the following matrix: [0 0 11 5 00 0 0 2 02 00 1]
This matrix is called Completely-Reduced Matrix because all the numbers exactly below and above each leader number are zeros. Result 1.1.11 The matrix is called Echelon Matrix if it is a combination of Reduced Matrix and Leader Numbers that are arranged so that the leader number in i th row is to the right of the leader number in the previous row, and the rows that are entirely zeros come at the end of the matrix. [ 0 1 20 0 10 0 0 3 75 61 30 0 00 0 00 0 0 0 10 00 0] This is an Echelon Matrix. [0 1 01 0 30 0 0 0 1 12 5 60 0 1]
This is NOT Echelon Matrix.xample 1.1.9 Given the following reduced matrix: [0 1 01 0 30 0 0 0 1 12 5 60 0 1]
Covert this reduced matrix to echelon matrix. Solution: We just need to interchange two rows as follows: R -- R [1 0 30 1 00 0 0 2 5 60 1 10 0 1] Now, the matrix is an echelon matrix. Result 1.1.12 To convert Reduced Matrix to Echelon Matrix, we just need to use interchange rows.Result 1.1.13 To convert Completely-Reduced Matrix to Reduced-Echelon Matrix, we just need to use interchange rows.
In this section, we discuss an example of basic algebra of matrix including matrix addition, subtraction and multiplication. Example 1.2.1 Given the following three matrices: A = [2 1 30 1 5] , B = [1 26 13 1] , C = [−3 2 51 7 3] a) Find 3A. b)
Find A+B. c)
Find A-B. d)
Find 3A-2C. Solution: Part a: We just need to multiply 3 by matrix A as follows: 17
18 M. Kaabar 3A =
Part b: In matrix addition we can only add matrices of the same order only. The order of matrix A is . The order of matrix B is . Since matrix A and matrix B have different orders, then we cannot find A+B. Hence, there is no answer for part b. Part c: In matrix subtraction we can only subtract matrices of the same order only. The order of matrix A is . The order of matrix B is . Since matrix A and matrix B have different orders, then we cannot find A-B. Hence, there is also no answer for part c. Part d: First, we need to multiply 2 by matrix C as follows: 2C =
Since we are already have 3A from part a, we just need to find 3A-2C. As we know, in matrix subtraction we can only subtract matrices of the same order only. The order of matrix 3A is . The order of matrix 2C is . Both of them have the same order. Hence, we can find 3A-2C as follows: 3A-2C = [6 3 90 3 15] − [−6 4 102 14 6 ] = [ 12 −1 −1−2 −11 9 ] .3 Linear Combinations
In this section, we discuss the concept of linear combinations of either columns or rows of a certain matrix. Example 1.3.1 Suppose we have 5 apples, 7 oranges and 12 bananas. Represent them as a linear combination of fruits. Solution: To represent them as a linear combination, we need to do the following steps: Step 1: Put each fruit individually: Apples Oranges Bananas Step 2: Separate each one of them by addition sign. Apples + Oranges + Bananas Step 3: Put 1 box in front of each one. Apples + Oranges + Bananas Step 4: Write the number of each fruit in each box. Apples + Oranges + Bananas This representation is called a Linear Combination of Fruits. Now, we can apply what we have learned in example 1.3.1on rows and columns of a certain matrix. Example 1.3.2 Given the following matrix A: A = [1 2 30 5 12 1 2] a) Find a linear combination of the columns of matrix A. b)
Find a linear combination of the rows of matrix A. 19
52 12 7
20 M. Kaabar Solution: Part a: To represent the columns of matrix A as a linear combination, we need to do the same steps as we did in example 1.3.1. Step 1: Put each column of matrix A individually: [102] [251] [312]
Step 2: Separate each one of them by addition sign. [102] + [251] + [312] Step 3: Put 1 box in front of each one. [102] + [251] + [312] Step 4: Write random number for each column in each box because it is not mentioned any number for any column in the question. [102] + [251] + [312] This representation is called a Linear Combination of the Columns of Matrix A.
52 2 -3 art b: To represent the rows of matrix A as a linear combination, we need to do the same steps as we did in part a. Step 1: Put each row of matrix A individually: [1 2 3] [0 5 1] [2 1 2]
Step 2: Separate each one of them by addition sign. [1 2 3] + [0 5 1] + [2 1 2] Step 3: Put 1 box in front of each one. [1 2 3] + [0 5 1] + [2 1 2] Step 4: Write random number for each row in each box because it is not mentioned any number for any row in the question. [1 2 3] + [0 5 1] + [2 1 2] This representation is called a Linear Combination of the Rows of Matrix A. Definition 1.3.1 Suppose matrix A has an order (size) 𝑛 × 𝑚 , and matrix B has an order (size) 𝑚 × 𝑘 , then the number of columns of matrix A equals to the number of rows of matrix B. If we assume that the usual multiplication of matrix A by matrix B equals to a new matrix C such that A ∙ B = C, then the matrix C has an order (size) 𝑛 × 𝑘 . (i.e. If A has an order (size) 3 × 7 , and B has an order (size) 7 × 10 , then C has an order (size) 3 × 10 ). 21
4 112 -8
22 M. Kaabar Example 1.3.3 Given the following matrix A and matrix B: A = [1 23 61 1] , B= [1 1 10 2 1] a) Find AB such that each column of AB is a linear combination of the columns of A.b)
Find AB such that each row of AB is a linear combination of the rows of B.c)
Find AB using the usual matrix multiplication.d)
If AB = C, then find c .e) Find BA such that each column of BA is a linear combination of the columns of B.f)
Find BA such that each row of BA is a linear combination of the rows of A. Solution: Part a: Since A has an order (size) 3 × 2 , and B has an order (size) 2 × 3 , then AB will have an order (size) 3 × 3 according to Definition 1.3.1. Step 1: 1 st column of AB: 1 * [131] + 0 * [261] = [131] Step 2: 2 nd column of AB: 1 * [131] + 2 * [261] = [ 5153 ] tep 3: 3 rd column of AB: 1 * [131] + 1 * [261] = [392] Hence, AB = [1 5 33 15 91 3 2]
Part b: As we did in part a but the difference here is rows instead of columns. Step 1: 1 st row of AB: 1 * [1 1 1] + 2 * [0 2 1] = [1 5 3] Step 2: 2 nd row of AB: 3 * [1 1 1] + 6 * [0 2 1] = [3 15 9] Step 3: 3 rd row of AB: 1 * [1 1 1] + 1 * [0 2 1] = [1 3 2] Hence, AB = [1 5 33 15 91 3 2]
Part c: Here we use the usual matrix multiplication. AB = [1 23 61 1] ∗ [1 1 10 2 1]
24 M. Kaabar AB = [1 ∗ 1 + 2 ∗ 0 1 ∗ 1 + 2 ∗ 2 1 ∗ 1 + 2 ∗ 13 ∗ 1 + 6 ∗ 0 3 ∗ 1 + 6 ∗ 2 3 ∗ 1 + 6 ∗ 11 ∗ 1 + 1 ∗ 0 1 ∗ 1 + 1 ∗ 2 1 ∗ 1 + 1 ∗ 1]
Hence, AB = [1 5 33 15 91 3 2]
Part d: Since AB = C, then c means that we need to find the number that is located in 2 nd row and 3 rd column. C = [𝑐 𝑐 𝑐 𝑐 𝑐 𝑐 𝑐 𝑐 𝑐 ] = [1 5 33 15 91 3 2] Now, let’s explain each element of matrix C. c means that the number that is located in 1 st row and 1 st column. c = 1. c means that the number that is located in 1 st row and 2 nd column. c = 5. c means that the number that is located in 1 st row and 3 rd column. c = 3. c means that the number that is located in 2 nd row and 1 st column. c = 3. c means that the number that is located in 2 nd row and 2 nd column. c = 15. c means that the number that is located in 2 nd row and 3 rd column. c = 9. c means that the number that is located in 3 rd row and 1 st column. c = 1. c means that the number that is located in 3 rd row and 2 nd column. c = 3. c means that the number that is located in 3 rd row and 3 rd column. c = 2. ence, c = 9. Part e: As we did in part a. Since B has an order (size) 2 × 3, and A has an order (size) 3 × 2 , then BA will have an order (size) 2 × 2 according to Definition 1.3.1. Step 1: 1 st column of BA: 1 * [10] + 3 * [12] + 1 * [11] = [57] Step 2: 2 nd column of BA: 2 * [10] + 6 * [12] + 1 * [11] = [ 913] Hence, BA = [5 97 13]
Part f: As we did in part b. Step 1: 1 st row of AB: 1 * [1 2] + 1 * [3 6] + 1 * [1 1] = [5 9] Step 2: 2 nd row of AB: 0 * [1 2] + 2 * [3 6] + 1 * [1 1] = [7 13] Hence, BA = [5 97 13]
Result 1.3.1 In general, matrix multiplication is not commutative (i.e. AB is not necessarily equal to BA). 25
26 M. Kaabar Example 1.3.4 Given the following matrix A and matrix B: A = [1 00 0] , B= [0 01 0] a) Find AB. b)
Find BA. Solution: Part a: Using the usual matrix multiplication, we obtain: AB = [1 00 0] [0 01 0] = [0 00 0] Part b: Using the usual matrix multiplication, we obtain: BA = [0 01 0] [1 00 0] = [0 01 0] Result 1.3.2 It is possible that the product of two non-zero matrices is a zero matrix. However, this is not true for product of real numbers. Example 1.3.5 Given the following matrix A, matrix B and matrix D: A = [1 00 0] , B= [0 01 0] , D= [0 00 1] a) Find AB. b)
Find AD. Solution: Part a: Using the usual matrix multiplication, we obtain: AB = [1 00 0] [0 01 0] = [0 00 0] zero matrix. art b: Using the usual matrix multiplication, we obtain: AD = [1 00 0] [0 00 1] = [0 00 0] zero matrix. Result 1.3.3 In general, If AB = AD and A is not a zero matrix, then it is possible that B ≠ D. Definition 1.3.2 Suppose matrix A has an order (size) 𝑛 × 𝑚 , A is a zero matrix if each number of A is a zero number. Example 1.3.5 Given the following system of linear equations: { 𝑥 + 𝑥 + 𝑥 = 3−𝑥 − 𝑥 + 2𝑥 = 0𝑥 + 4𝑥 = 5 Write the above system in matrix-form. Solution: We write the above system in matrix-form as follows: (Coefficient Matrix) ∙ (Variable Column) = (Constant Column) CX = A where C is a Coefficient Matrix, X is a Variable Column, and A is a Constant Column. [ 1 1 1−1 −1 20 1 4] [𝑥 𝑥 𝑥 ] = [305] The above matrix-form means the following: 𝑥 ∗ [ 1−10 ] + 𝑥 ∗ [ 1−11 ] + 𝑥 ∗ [124] Hence, we obtain: [ 𝑥 + 𝑥 + 𝑥 = 3−𝑥 − 𝑥 + 2𝑥 = 0𝑥 + 4𝑥 = 5 ] = [305]
28 M. Kaabar Now, let 𝑥 = 1, 𝑥 = 1, and 𝑥 = 1. [ 1 1 1−1 −1 20 1 4] [111] = [305] Result 1.3.4 Let CX = A be a system of linear equations C [ 𝑥 𝑥 ..𝑥 𝑛 ] = [ 𝑎 𝑎 ..𝑎 𝑛 ] Then, x = b , x = b , … . , x n = b n is a solution to the system if and only if b C + b C + ⋯ + b n C m = A, C is the 1 st column of C, C is the 2 nd column of C, …. , C m is n th column of C. Example 1.3.6 Given the following system of linear equations: { 𝑥 + 𝑥 = 22𝑥 + 2𝑥 = 4 Write the above system in matrix-form. Solution: We write the above system in matrix-form CX = A as follows: C = [1 12 2] , X = [𝑥1𝑥2] , A = [24]
Hence, CX = A: [1 12 2] [𝑥 𝑥 ] = [24] 𝑥 ∗ [12] + 𝑥 ∗ [12] = [24] Now, we need to choose values for x and x such that these values satisfy that above matrix-form. First, let’s try x = 3 and x = 4. herefore, x = 3, x = 4 is not solution, and our assumption is wrong. Now, let’s try x = 0 and x = 2. Therefore, x = 0, x = 2 is solution. Result 1.3.5 Let CX = A be a system of linear equations. The constant column A can be written uniquely as a linear combination of the columns of C. In this section, we discuss some examples and facts of square matrix and identity matrix. Definition 1.4.1 In general, let A is a matrix, and n be a positive integer such that A n is defined only if number of rows of A = number of columns of A. (i.e. Let A is a matrix with a size (order) × 𝑚 , A = A ∙ A is defined only if n = m). This matrix is called a Square-Matrix. The following is some facts regarding square-matrix and set of real numbers ℝ. Fact 1.4.1 Let A is a matrix, A −n is not equal to n . Fact 1.4.2 Let A is a matrix, A −n has a specific meaning if A is a square-matrix and A is invertible (nonsingular). We will talk about it in section 1.5. Fact 1.4.3 A set of real numbers ℝ has a multiplicative identity equals to 1. (i.e. 𝑎 ∙ 1 = 1 ∙ 𝑎 = 𝑎 ). Fact 1.4.4 A set of real numbers ℝ has an additive identity equals to 0. (i.e. 𝑎 + 0 = 0 + 𝑎 = 𝑎 ). 29
30 M. Kaabar Fact 1.4.5 ℝ = M (ℝ) = set of all matrices. Now, we give some helpful notations: ℤ : Set of all integers. ℚ : Set of all rational numbers. ℝ : Set of all real number. ℕ : Set of all natural numbers. The following figure 1.4.1 represents three main sets: ℤ, ℚ and ℝ.
Figure 1.4.1: Representation of Three Sets of Numbers The above figure shows that the largest set is ℝ , and the smallest set is ℤ . In addition, we know that a rational number is an integer divided by a non-zero integer. ℚ ℝ ℤ act 1.4.6 means that 3 is an element of the set ℝ. (“ ∈ ” is a mathematical symbol that means “belong to”). Fact 1.4.7 √2 ∈ ℝ means that √2 is an element of the set ℝ. (“ ∈ ” is a mathematical symbol that means “belong to”). Fact 1.4.8 ∉ ℤ means that is not an element of the set ℤ . (“ ∉ ” is a mathematical symbol that means “does not belong to”). Fact 1.4.9 A ∈ M (ℝ) means that A is a matrix. Fact 1.4.10 ℝ 𝒏×𝒏 = M 𝑛 (ℝ) = set of all 𝑛 × 𝑛 matrices. (i.e. ℝ = M (ℝ) = set of all matrices). After learning all above facts, let’s test our knowledge by asking the following question. Question 1.4.1 Does M 𝑛 (ℝ) has a multiplicative identity? Solution: Yes; there is a matrix called I 𝑛 where I 𝑛 ∈M 𝑛 (ℝ) such that AI 𝑛 = I 𝑛 A = A for every
A ∈ M 𝑛 (ℝ) . Example 1.4.1 Given M (ℝ) = ℝ . What is I ? Solution: We need to find the multiplicative identity for all matrices. This means that AI = I A = A for every
A ∈ M (ℝ) . I = [1 00 1] and A = [𝑎 𝑎 𝑎 𝑎 ] Now, let’s check if the multiplicative identity is right: AI = [𝑎 𝑎 𝑎 𝑎 ] [1 00 1] = [𝑎 𝑎 𝑎 𝑎 ] = A I A = [1 00 1] [𝑎 𝑎 𝑎 𝑎 ] = [𝑎 𝑎 𝑎 𝑎 ] = A
32 M. Kaabar Hence, the multiplicative identity for all matrices is I = [1 00 1]. Example 1.4.2 Given M (ℝ) = ℝ . What is I ? Solution: We need to find the multiplicative identity for all matrices. This means that AI = I A = A for every
A ∈ M (ℝ) . I = [1 0 00 1 00 0 1] and A = [𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 ] Now, let’s check if the multiplicative identity is right: AI = [𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 ] [1 0 00 1 00 0 1] = [𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 ] = A I A = [1 0 00 1 00 0 1] [𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 ] = [𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 ] = A Hence, the multiplicative identity for all matrices is I = [1 0 00 1 00 0 1]. Result 1.4.1 Assume we have a square-matrix with 𝑛 ×𝑛 size and the main-diagonal is 𝑎 , 𝑎 , 𝑎 , … , 𝑎 𝑛𝑛 [ 𝑎 ⋯ 𝑎 ⋮ 𝑎 …… … ⋮ 𝑎 𝑛1 ⋯ 𝑎 𝑛𝑛 ] Then, the multiplicative identity for the above 𝑛 × 𝑛 matrix is as follows: 𝑛 = [1 ⋯ ⋮ ⋮ ⋯ All ones on the main-diagonal and zeros elsewhere.
In this section, we discuss the concept of inverse square matrix, and we give some examples of elementary matrix. Example 1.5.1Given the following matrices with the following row-operations steps:
Z = [1 20 10 1 3 4−1 21 3] , D = [−2 −40 10 1 −6 −8−1 21 3 ] [1 20 10 1 3 4−1 21 3] -2R [−2 −40 10 1 −6 −8−1 21 3 ] +R -- R -3R +R -- R Matrix B R R Matrix C a)
Find a matrix D such that DZ=W. b)
Find a matrix K such that KW=Z. c)
Find Elementary Matrices, F , F , F , …, F m such that F F …. F m Z = C. 33
34 M. Kaabar d)
Find Elementary Matrices, L , L ,…, L n such that L L …. L n B = Z. e)
Find a matrix S such that SC=B. f)
Find a matrix X such that XB=C. Solution: Part a: Since Z is matrix and we multiply from left by -2, then D must be a square matrix. Hence, D must be . We use the multiplicative identity, and we multiply it by -2 from left as follows: I = [1 0 00 1 00 0 1] -2R [−2 0 00 1 00 0 1] = D D is called Elementary Matrix of Type I. Thus, [−2 0 00 1 00 0 1] Z = W
Part b: Since W is matrix and we multiply from right by − because the inverse of -2 is − , then K must be a square matrix. Hence, K must be . We use the multiplicative identity, and we multiply it by − from left as follows: I = [1 0 00 1 00 0 1] − R [− K is also called Elementary Matrix of Type I. Thus, [− esult 1.5.1 Each row-operation corresponds to one and only one elementary matrix. Part c: According to the given four row-operations steps, we need to find four elementary matrices from Z to C which means m=4 because we want to go 4 steps forward from Z to C. From part a, we already have the following: I = [1 0 00 1 00 0 1] -2R [−2 0 00 1 00 0 1] (Step One) Hence, [−2 0 00 1 00 0 1] Z = C I = [1 0 00 1 00 0 1] +R -- R [1 0 00 1 02 0 1] (Step Two) Hence, [1 0 00 1 02 0 1] [−2 0 00 1 00 0 1] Z = C I = [1 0 00 1 00 0 1] -3R +R -- R [ 1 0 0−3 1 00 0 1] (Step Three) Hence, [ 1 0 0−3 1 00 0 1] [1 0 00 1 02 0 1] [−2 0 00 1 00 0 1] Z = C
36 M. Kaabar I = [1 0 00 1 00 0 1] R R [0 0 10 1 01 0 0] (Step Four) Hence, we obtain the following four elementary matrices: [0 0 10 1 01 0 0] [ 1 0 0−3 1 00 0 1] [1 0 00 1 02 0 1] [−2 0 00 1 00 0 1] Z = C Part d: According to the given three row-operations steps from B to Z, we need to find three elementary matrices which means n=3 because we want to go 3 steps backward from B to Z. Backward steps mean that we need to do inverse steps (i.e. The inverse of -2R is − R because it is row-multiplication step). We start from the third step as follows: The inverse of -3R +R -- R is 3R +R -- R because it is row-addition step. I = [1 0 00 1 00 0 1] +R -- R [1 0 03 1 00 0 1] (Step Three) Hence, [1 0 03 1 00 0 1] B = Z The inverse of 2R +R -- R is -2R +R -- R because it is row-addition step. I = [1 0 00 1 00 0 1] -2R +R -- R [ 1 0 00 1 0−2 0 1] (Step Two) Hence, [ 1 0 00 1 0−2 0 1] [1 0 03 1 00 0 1] B = Z The inverse of -2R is − R because it is row-multiplication step I = [1 0 00 1 00 0 1] − R [− (Step One) Hence, we obtain the following three elementary matrices: [− 12 0 00 1 00 0 1] [ 1 0 00 1 0−2 0 1] [1 0 03 1 00 0 1] B = Z Part e: According to the given last row-operations step from C to B, we need to find one elementary matrix because we want to go 1 step backward from C to B. The inverse of R R is R R which is the same as R R . I = [1 0 00 1 00 0 1] R R [0 0 10 1 01 0 0] Hence, we obtain: [0 0 10 1 01 0 0] C = B . This means that S = [0 0 10 1 01 0 0] . 37
38 M. Kaabar Part f: According to the given last row-operations step from B to C, we need to find one elementary matrix because we want to go 1 step forward from B to C. I = [1 0 00 1 00 0 1] R R [0 0 10 1 01 0 0] Hence, we obtain: [0 0 10 1 01 0 0] B = C . This means that X = [0 0 10 1 01 0 0] . Example 1.5.2 Given the following matrix:
A = [2 14 0]
Find A -1 if possible. Solution: Finding A -1 if possible means that we need to find a possible inverse matrix called A -1 such that AA -1 = I = A -1 A. To find this possible matrix A -1 , we need to do the following steps: Step 1: Write A and I in the following form: ( Matrix A | Identity Matrix I ) ( | ) Step 2: Do some Row-Operations until you get the following: ( Do you see I ? If yes, then X = A −1 | X )( Do you see I ? If no, then A has no inverse matrix | X ) ow, let’s do the above step until we get the Completely-Reduced-Echelon matrix. ( | ) − R ( |
00 1 ) ( |
00 1 ) -4R +R -- R (1 (1 R (1
01 − ) (1
01 − ) − R +R -- R (1 00 1|0 ) Since we got the Completely-Reduced-Echelon matrix which is the identity matrix I , then A has an inverse matrix which is A -1 . Hence, A -1 = [0 ] Example 1.5.3 Given the following matrix:
A = [1 0 20 1 00 −1 1]
Find A -1 if possible. 39
40 M. Kaabar Solution: Finding A -1 if possible means that we need to find a possible inverse matrix called A -1 such that AA -1 = I = A -1 A. To find this possible matrix A -1 , we need to do the following steps: Step 1: Write A and I in the following form: ( Matrix A | Identity Matrix I ) ( | ) Step 2: Do some Row-Operations until you get the following: ( Do you see I ? If yes, then X = A −1 | X )( Do you see I ? If no, then A has no inverse matrix | X ) Now, let’s do the above step until we get the Completely-Reduced-Echelon matrix. ( | ) R +R -- R ( | ) -2R +R -- R ( | ) Since we got the Completely-Reduced-Echelon matrix which is the identity matrix I , then A has an inverse matrix which is A -1 . Hence, A -1 = [1 −2 −20 1 00 1 1 ] Result 1.5.2 Given 𝑛 × 𝑚 matrix A and identity matrix I 𝑛 , and with some row-operations, we got A′ and I 𝑛 ′ as follows: ( A | I 𝑛 ) Row-Operations- ( A′ | I 𝑛 ′ ) Then, I 𝑛′ A = A′ . Result 1.5.3 Given 𝑛 × 𝑛 matrix A and identity matrix I 𝑛 , and with some row-operations, we got A −1 and I 𝑛 as follows: ( A | I 𝑛 ) Row-Operations- ( I 𝑛 | A −1 ) Then, AA -1 = I 𝑛 = A -1 A. Fact 1.5.1 Given 𝑛 × 𝑚 matrix A and identity matrix I 𝑛 , I 𝑛 is called a left identity for all 𝑛 × 𝑚 matrices such that I 𝑛 A = A. (i.e. Given matrix A, then I A = A). Fact 1.5.2 Given 𝑛 × 𝑛 matrix A and identity matrix I 𝑚 , I 𝑚 is called a right identity for all 𝑛 × 𝑛 matrices such that AI 𝑚 = A. (i.e. Given matrix A, then AI = A). Result 1.5.4 Given 𝑛 × 𝑛 matrix A, A -4 has a meaning if and only if A -1 exists. Result 1.5.5 Given 𝑛 × 𝑛 matrix A, If A -1 exists, then A -4 = A -1 × A -1 × A -1 × A -1 . 41
42 M. Kaabar
In this section, we introduce the concept of transpose matrix, and we discuss some examples of symmetric and skew-symmetric matrices. Definition 1.6.1 Given 𝑛 × 𝑚 matrix A, A T is called A transpose and it is 𝑚 × 𝑛 matrix. We obtain A T from A by making columns of A rows or by making rows of A columns. Example 1.6.1 Given the following matrix: A = [2 35 11 1 1 02 30 5]
Find A T . Solution: According to definition 1.6.1, A is matrix. Thus, A T should be . By making columns of A rows, we obtain the following: A T = [2 5 13 1 11 2 00 3 5] Definition 1.6.2 Given 𝑛 × 𝑚 matrix A and 𝑚 × 𝑛 matrix A T , then AA T is always defined, and it is 𝑛 × 𝑛 matrix. (i.e. Let AA T = D , then A 𝑛×𝑚 A 𝑚×𝑛T = D 𝑛×𝑛 ). Definition 1.6.3 Given 𝑛 × 𝑚 matrix A and 𝑚 × 𝑛 matrix A T , then A T A is always defined, and it is 𝑚 × 𝑚 matrix. (i.e. Let A T A = F , then A 𝑚×𝑛T A 𝑛×𝑚 = F 𝑚×𝑚 ). efinition 1.6.4 Given 𝑛 × 𝑛 matrix A and 𝑛 × 𝑛 matrix A T , then A is symmetric if A T = A. Definition 1.6.5 Given 𝑛 × 𝑛 matrix A and 𝑛 × 𝑛 matrix A T , then A is skew-symmetric if A T = −A. Example 1.6.2 Given the following matrix:
A = [ 1 5 105 3 710 7 10]
Show that A is symmetric. Solution: According to definition 1.6.4, A is matrix. Thus, A T should be . By making columns of A rows, we obtain the following: A T = [ 1 5 105 3 710 7 10] Since A T = A = [ 1 5 105 3 710 7 10] , then A is symmetric. Example 1.6.3 Given the following matrix: A = [ 0 2 5−2 0 3−5 −3 0]
Show that A is skew-symmetric. Solution: According to definition 1.6.5, A is matrix. Thus, A T should be . By making columns of A rows, we obtain the following: A T = [0 −2 −52 0 −35 3 0 ]
44 M. Kaabar Since A T = −A = [0 −2 −52 0 −35 3 0 ] , then A is skew-symmetric. Fact 1.6.1 Given 𝑛 × 𝑛 matrix A, if A is skew-symmetric, then all numbers on the main diagonal of A are always zeros. Fact 1.6.2 Given 𝑛 × 𝑚 matrix A, then (A T ) T = A. Fact 1.6.3 Given 𝑛 × 𝑚 matrix A and 𝑚 × 𝑘 matrix B, then (AB T ) T = B T A T . (Warning: (AB T ) T ≠ A T B T ). Fact 1.6.4 Given 𝑛 × 𝑚 matrices A and B, then (A ± B) T = A T ± B T . Result 1.6.1 Given matrix A and constant 𝛼 , if A is symmetric, then 𝛼A is symmetric such that (𝛼A) T = 𝛼 A T . Result 1.6.2 Given matrix A and constant 𝛼 , if A is skew-symmetric, then 𝛼A is skew-symmetric such that (𝛼A) T = 𝛼 A T . Result 1.6.3 Let A be 𝑛 × 𝑛 matrix, there exists a symmetric matrix B and a skew-symmetric matrix C such that A is a linear combination of B and C. This means that those should be numbers 𝛼 and 𝛼 such that A = 𝛼 B + 𝛼 C . Proof of Result 1.6.3 We will show that A is a linear combination of B and C. e assume that B is symmetric such that B = A + A T . B T = (A + A T ) T = A T + (A T ) T = A T + A = B. Now, we assume that C is skew-symmetric such that
C = A − A T . C T = (A − A T ) T = A T − (A T ) T = A T − A = −(A − A T ) = −C By using algebra, we do the following:
12 B + 12 C = 12 (A + A T ) + 12 (A − A T )= 12 A + 12 A T + 12 A − 12 A T = A. Thus, A is a linear combination of B and C. Example 1.6.4 Given the following matrix:
A = [2 1 43 0 15 6 7]
Find symmetric matrix B and skew-symmetric matrix C such that
A = 𝛼 B + 𝛼 C for some numbers 𝛼 and 𝛼 . Solution: As we did in the above proof, we do the following: B = A + A T = [2 1 43 0 15 6 7] + [2 3 51 0 64 1 7] = [4 4 94 0 79 7 14] C = A − A T = [2 1 43 0 15 6 7] − [2 3 51 0 64 1 7] = [0 −2 −12 0 −51 5 0 ] Let 𝛼 = 𝛼 = Thus,
A = B + C = [4 4 94 0 79 7 14] + [0 −2 −12 0 −51 5 0 ]
46 M. Kaabar
In this section, we introduce step by step for finding determinant of a certain matrix. In addition, we discuss some important properties such as invertible and non-invertible. In addition, we talk about the effect of row-operations on determinants. Definition 1.7.1 Determinant is a square matrix. Given M (ℝ) = ℝ = ℝ , let A ∈ M (ℝ) where A is matrix, A = [𝑎 𝑎 𝑎 𝑎 ]. The determinant of A is represented by det(A) or |A| . Hence, det(A) = |A| = 𝑎 𝑎 − 𝑎 𝑎 ∈ ℝ . (Warning: this definition works only for matrices). Example 1.7.1 Given the following matrix: A = [3 25 7]
Find the determinant of A. Solution: Using definition 1.7.1, we do the following: det(A) = |A| = (3)(7) − (2)(5) = 21 − 10 = 11.
Thus, the determinant of A is 11. Example 1.7.2 Given the following matrix:
A = [1 0 23 1 −11 2 4 ]
Find the determinant of A. olution: Since A is matrix such that
A ∈ M (ℝ) = ℝ , then we cannot use definition 1.7.1 because it is valid only for matrices. Thus, we need to use the following method to find the determinant of A. Step 1: Choose any row or any column. It is recommended to choose the one that has more zeros. In this example, we prefer to choose the second column or the first row. Let’s choose the second column as follows: A = [1 0 23 1 −11 2 4 ]𝑎 = 0, 𝑎 = 1 and 𝑎 = 2. Step 2: To find the determinant of A, we do the following: For 𝑎 , since 𝑎 is in the first row and second column, then we virtually remove the first row and second column. A = [1 0 23 1 −11 2 4 ](−1) 𝑎 det [3 −11 4 ] For 𝑎 , since 𝑎 is in the second row and second column, then we virtually remove the second row and second column. A = [1 0 23 1 −11 2 4 ](−1) 𝑎 det [1 21 4] For 𝑎 , since 𝑎 is in the third row and second column, then we virtually remove the third row and second column. 47
48 M. Kaabar
A = [1 0 23 1 −11 2 4 ](−1) 𝑎 det [1 23 −1] Step 3: Add all of them together as follows: det(A) = (−1) 𝑎 det [3 −11 4 ] + (−1) 𝑎 det [1 21 4]+ (−1) 𝑎 det [1 23 −1] det(A) = (−1) (0)det [3 −11 4 ] + (−1) (1)det [1 21 4]+ (−1) (2)det [1 23 −1] det(A) = (−1)(0)det [3 −11 4 ] + (1)(1)det [1 21 4]+ (−1)(2)det [1 23 −1] det(A) = (−1)(0)(12 − −1) + (1)(1)(4 − 2) + (−1)(2)(−1− 6) det(A) = 0 + 2 + 14 = 16. Thus, the determinant of A is 16. Result 1.7.1 Let
A ∈ M 𝑛 (ℝ) . Then, A is invertible (non-singular) if and only if det(A) ≠ 0. The above result means that if det(A) ≠ 0 , then A is invertible (non-singular), and if A is invertible (non-singular), then det(A) ≠ 0 . Example 1.7.3 Given the following matrix:
A = [2 34 6] s A invertible (non-singular)? Solution: Using result 1.7.1, we do the following: det(A) = |A| = (2)(6) − (3)(4) = 12 − 12 = 0.
Since the determinant of A is 0, then A is non-invertible (singular). Thus, the answer is No because A is non-invertible (singular). Definition 1.7.2 Given
A = [𝑎 𝑎 𝑎 𝑎 ] . Assume that det(A) ≠ 0 such that det(A) = 𝑎 𝑎 − 𝑎 𝑎 . To find A −1 (the inverse of A), we use the following format that applies only for matrices: A −1 = 1det(𝐴) [ 𝑎 −𝑎 −𝑎 𝑎 ] A −1 = 1𝑎 𝑎 − 𝑎 𝑎 [ 𝑎 −𝑎 −𝑎 𝑎 ] Example 1.7.4 Given the following matrix:
A = [ 3 2−4 5]
Is A invertible (non-singular)? If Yes, Find A −1 . Solution: Using result 1.7.1, we do the following: det(A) = |A| = (3)(5) − (2)(−4) = 15 + 8 = 23 ≠ 0. Since the determinant of A is not 0, then A is invertible (non-singular). Thus, the answer is Yes, there exists A −1 according to definition 1.7.2 as follows: A −1 = 1det(𝐴) [5 −24 3 ] = 123 [5 −24 3 ] = [ 523 − 223423 323 ]
50 M. Kaabar Result 1.7.2 Let
A ∈ M 𝑛 (ℝ) be a triangular matrix. Then, det(A) = multiplication of the numbers on the main diagonal of A. There are three types of triangular matrix: a) Upper Triangular Matrix: it has all zeros on the left side of the diagonal of 𝑛 × 𝑛 matrix. (i.e.
A = [1 7 30 2 50 0 4] is an Upper Triangular Matrix). b)
Diagonal Matrix: it has all zeros on both left and right sides of the diagonal of 𝑛 × 𝑛 matrix. (i.e.
B = [1 0 00 2 00 0 4] is a Diagonal Matrix). c)
Lower Triangular Matrix: it has all zeros on the right side of the diagonal of 𝑛 × 𝑛 matrix. (i.e.
C = [1 0 05 2 01 9 4] is a Diagonal Matrix). Fact 1.7.1 Let
A ∈ M 𝑛 (ℝ) . Then, det(A) = det (A T ) . Fact 1.7.2 Let A ∈ M 𝑛 (ℝ) . If A is an invertible (non-singular) matrix, then A T is also an invertible (non-singular) matrix. (i.e. (A T ) −1 = (A −1 ) T ). Proof of Fact 1.7.2 We will show that (A T ) −1 = (A −1 ) T . We know from previous results that AA −1 = I 𝑛 . By taking the transpose of both sides, we obtain: AA −1 ) T = (I 𝑛 ) T Then, (A −1 ) T A T = (I 𝑛 ) T Since (I 𝑛 ) T = I 𝑛 , then (A −1 ) T A T = I 𝑛 . Similarly, (A T ) −1 A T = (I 𝑛 ) T = I 𝑛 . Thus, (A T ) −1 = (A −1 ) T . The effect of Row-Operations on determinants: Suppose ∝ is a non-zero constant, and 𝑖 𝑎𝑛𝑑 𝑘 are row numbers in the augmented matrix. * ∝ R i , ∝≠ 0 (Multiply a row with a non-zero constant ∝ ). i.e. A = [1 2 30 4 12 0 1] -- [1 2 30 12 32 0 1] = B Assume that det(A) = γ where γ is known, then det(B) = 3γ . Similarly, if det(B) = β where β is known, then det(A) = β . * ∝ R i +R k -- R k (Multiply a row with a non-zero constant ∝, and add it to another row ). i.e. A = [1 2 30 4 12 0 1] ∝ R i +R k -- R k [1 2 30 12 32 0 1] = B Then, det(A) = det(B) . 51
52 M. Kaabar * R i ↔ R k (Interchange two rows). It has no effect on the determinants. In general, the effect of Column-Operations on determinants is the same as for Row-Operations. Example 1.7.5 Given the following matrix A with some Row-Operations: A -- A -- A -2R -- A If det(A) = 4 , then find det(A ) Solution: Using what we have learned from the effect of determinants on Row-Operations: det(A ) = 2 ∗ det(A) = 2 ∗ 4 = 8 because A has the first row of A multiplied by 2. det(A ) = 3 ∗ det(A ) = 3 ∗ 8 = 24 because A has the third row of A multiplied by 3. Similarly, det(A ) = −2 ∗ det(A ) = −2 ∗ 24 = −48 because A has the fourth row of A multiplied by -2. Result 1.7.3 Assume A is 𝑛 × 𝑛 matrix with a given det(A) = 𝛾 . Let 𝛼 be a number. Then, det(𝛼A) = 𝛼 𝑛 ∗ 𝛾 . Result 1.7.4 Assume A and B are 𝑛 × 𝑛 matrices.
Then: a) det(A) = det(A) ∗ det(B). b) Assume A −1 exists and B −1 exists. Then, (AB) −1 = B −1 A −1 . c) det(AB) = det(BA). d) det(A) = det(A T ). ) If A −1 exists, then det(A −1 ) = . Proof of Result 1.7.4 (b) We will show that (AB) −1 = B −1 A −1 . If we multiply ( B −1 A −1 ) by (AB), we obtain: B −1 (A −1 A)B = B −1 (I 𝑛 ) B = B −1 B = I 𝑛 . Thus, (AB) −1 = B −1 A −1 . Proof of Result 1.7.4 (e) We will show that det(A −1 ) = . Since AA −1 = I 𝑛 , then det(AA −1 ) = det(𝐼 𝑛 ) = 1. det(AA −1 ) = det(A) ∗ det(A −1 ) = 1. Thus, det(A −1 ) = . In this section, we discuss how to use Cramer’s Rule to solve systems of linear equations. Definition 1.8.1 Given 𝑛 × 𝑛 system of linear equations. Let
CX = A be the matrix form of the given system:
C [ 𝑥 𝑥 𝑥 ⋮𝑥 𝑛 ] = [ 𝑎 𝑎 𝑎 ⋮𝑎 𝑛 ] The system has a unique solution if and only if det(C) ≠ 0 . Cramer’s Rule tells us how to find 𝑥 , 𝑥 , … , 𝑥 𝑛 as follows: 53
54 M. Kaabar Let C = [1 3 41 2 17 4 3]
Then, the solutions for the system of linear equations are: 𝑥 = det [𝑎 𝑛 𝑥 = det [1 𝑎
41 ⋮ 17 𝑎 𝑛 𝑥 = det [1 3 𝑎 𝑛 ]det (C) Example 1.8.1 Use Cramer’s Rule to solve the following system of linear equations: { 2𝑥 + 7𝑥 = 13−10𝑥 + 3𝑥 = −4 Solution: First of all, we write system in the form
CX = A according to definition 1.8.1. [ 2 7−10 3] [𝑥 𝑥 ] = [ 13−4] Since C in this form is [ 2 7−10 3] , then det(C) = (2 ∗ 3) − (7 ∗ (−10)) = 6 − (−70) = 76 ≠ 0.
The solutions for this system of linear equations are: 𝑥 = det [ 13 7−4 3]det (C) = det [ 13 7−4 3]76 = 6776 𝑥 = det [ 2 13−10 −4]det (C) = det [ 2 13−10 −4]76 = 12276 Thus, the solutions are 𝑥 = and 𝑥 = . In this section, we introduce a new mathematical method to find the inverse matrix. We also give some examples of using Adjoint Method to find inverse matrix and its entry elements. Example 1.9.1 Given the following matrix:
A = [1 0 22 1 −20 0 2 ]
Find A −1 using Adjoint Method. Solution: To find the inverse matrix of A, we need to do the following steps: Step 1: Find the determinant of matrix A, and check whether A −1 exists or not. Using the previous discussed method for finding the inverse matrix, we get: det(A) = 2 ≠ 0. Therefore, A −1 exists. Step 2: Calculate the coefficient matrix of A such that A = C because A is matrix. 55
56 M. Kaabar
C = [𝑐 𝑐 𝑐 𝑐 𝑐 𝑐 𝑐 𝑐 𝑐 ] In general, to find every element of coefficient matrix A, we need to use the following form: 𝑐 𝑖𝑘 = (−1) 𝑖+𝑘 det [𝑅𝑒𝑚𝑜𝑣𝑒 𝑡ℎ𝑒 𝑖 𝑡ℎ 𝑟𝑜𝑤 𝑎𝑛𝑑 𝑘 𝑡ℎ 𝑐𝑜𝑙𝑢𝑚𝑛 𝑜𝑓 𝐴] Now, using the above form, we can find all coefficients of matrix A: 𝑐 = (−1) det [1 −20 2 ] = 2 𝑐 = (−1) det [0 −20 2 ] = 0 𝑐 = (−1) det [0 10 0] = 0 𝑐 = (−1) det [0 20 2] = 0 𝑐 = (−1) det [1 20 2] = 2 𝑐 = (−1) det [1 00 0] = 0 𝑐 = (−1) det [0 21 −2] = −2 𝑐 = (−1) det [1 20 −2] = 2 𝑐 = (−1) det [1 00 1] = 1 Hence,
C = [ 2 0 00 2 0−2 2 1] tep 3: Find the adjoint of matrix A as follows: In general, adjoint of A = adj(A) = C T . Thus, adj(A) = C T = [2 0 −20 2 20 0 1 ] This formula is always true such that
A ∗ adj(A) = det(A) I 𝑛 . If det(A) ≠ 0, then A [ ∗ adj(A)] = I 𝑛 . Hence, A −1 = ∗ adj(A) = ∗ C T . A −1 = 1det(A) ∗ C T = 12 ∗ [2 0 −20 2 20 0 1 ] = [1 0 −10 1 10 0 12 ] Example 1.9.2 Given the following matrix:
A = [ 2 0 −2−2 1 2−4 −1 3 1400 0 0 4]
Find (2,4) entry of A −1 . Solution: To find the (2,4) entry of A −1 , we need to do the following steps: Step 1: Find the determinant of matrix A, and check whether A −1 exists or not. Use the Row-Operation Method, we obtain the following: [ 2 0 −2−2 1 2−4 −1 3 1400 0 0 4] R + R → R + R → R [2 0 −20 1 00 −1 −1 1520 0 0 4]
58 M. Kaabar R + R → R [2 0 −20 1 00 0 −1 1570 0 0 4] Now, using result 1.7.2 for finding the determinant of the upper triangular matrix: Since
A ∈ M 𝑛 (ℝ) is a triangular matrix. Then, det(A) = multiplication of the numbers on the main diagonal of A. [2 0 −20 1 00 0 −1 1570 0 0 4] det(A) = 2 ∗ 1 ∗ −1 ∗ 4 = −8 ≠ 0. Therefore, A −1 exists. Step 2: Use the following general form for (i,k) entry of A −1 : (𝑖, 𝑘) − 𝑒𝑛𝑡𝑟𝑦 𝑜𝑓 A −1 = 𝑐 𝑘𝑖 det(A) = (−1) 𝑖+𝑘 det[𝑅𝑒𝑚𝑜𝑣𝑒 𝑡ℎ𝑒 𝑘 𝑡ℎ 𝑟𝑜𝑤 𝑎𝑛𝑑 𝑖 𝑡ℎ 𝑐𝑜𝑙𝑢𝑚𝑛 𝑜𝑓 𝐴]det(A) Now, using the above form, we can find (2,4)-entry of matrix A: 𝑐 det(A) = (−1) det [ 2 −2 1−2 2 4−4 3 0]det(A) = (−1) det [ 2 −2 1−2 2 4−4 3 0]−8 Step 3: Find the determinant of [ 2 −2 1−2 2 4−4 3 0] . et’s call the above matrix F such that
F =[ ] To find the determinant of F, we need to use the Row-Operation Method to reduce the matrix F.
F = [ ] R + R → R + R → R [ ] R ↔ R [ ] Let’s call the above matrix D such that
D = [ ] Now, using result 1.7.2 for finding the determinant of this upper triangular matrix:
D = [ ] det(D) = 2 ∗ −1 ∗ 5 = −10 ≠ 0. Therefore, det(F) = −det(D) = 10 ≠ 0.
Thus, The (2,4)-entry of matrix A is: 𝑐 det(A) = (−1) det [ 2 −2 1−2 2 4−4 3 0]−8 = (−1) ∗ 10−8 = 10−8 = − 54= −1.25 ∴ (2,4)-entry of matrix A = −1.25
60 M. Kaabar
1. Solve the following system of linear equations: { 2𝑥 − 𝑥 + 𝑥 = 10−𝑥 + 3𝑥 + 𝑥 = 0𝑥 + 2𝑥 + 2𝑥 − 𝑥 + 2𝑥 + 𝑥 = 12
2. Given an augmented matrix of a system: ( 1 −1 𝑐−1 1 44 −3 𝑏 | 2310) a. For what values of c and b will the system be consistent? b. If the system is consistent, when do you have a unique solution?. 3. Let
F = [−1 2 −1 3 0 1 1 1−2 0 −1 1] and
H = [ 2 1 1−4 1 01−1 1 10 3] a. Find the third row of HF. b. Find the third column of FH. c. Let HF = A. Find 𝑎 .
4. Let
A = [ 1 2 4−1 −2 −3−2 −3 −8] . If possible find A −1 .
5. Given A is matrix such that
𝐀 R → 𝐀 + R → R 𝐀 . a. Find two elementary matrices say 𝐸 , 𝐸 such that 𝐸 𝐸 𝐴 = 𝐴 . b. Find two elementary matrices say 𝐹 , 𝐹 such that 𝐹 𝐹 𝐴 = 𝐴.
6. Let
A = [ 1 2 4−1 −2 3−2 −3 −7] . Find det(A) . Is A invertible? Explain.
7. Let
A = [ 4 −2−3 2 ] . Is A invertible? If yes, find 𝐴 −1 .
8. Use Cramer’s Rule to find the solution to 𝑥 in the system: { 2𝑥 + 𝑥 − 𝑥 = 2−2𝑥 + 4𝑥 + 2𝑥 = 8−2𝑥 − 𝑥 + 8𝑥 = −2
9. Let
A = [ 2 −4 2−2 0 21 −2 12 1−14−2 4 −2 12 ] . Find (2,4) entry of A −1 . 10. Find a matrix 𝐴 such that [5 51 4] 𝐴 + 3𝐼 = 2𝐴 + [−2 23 2] . 11. Given 𝐴 −1 = [ 2 −2 −2−2 3 0−2 2 3 ] and 𝐵 = [ 1 −2 −2−4 2 20 1 −2]
Solve the system
𝐴𝑋 = [ 0−11 ]
62 M. Kaabar
Chapter 2 Vector Spaces
We start this chapter reviewing some concepts of set theory, and we discuss some important concepts of vector spaces including span and dimension. In the remaining sections we introduce the concept of linear independence. At the end of this chapter we discuss other concepts such as subspace and basis.
In this section, we review some concepts of set theory, and we give an introduction to span and vector spaces including some examples related to these concepts. Before reviewing the concepts of set theory, it is recommended to revisit section 1.4, and read the notations of numbers and the representation of the three sets of numbers in figure 1.4.1. Let’s explain some symbols and notations of set theory: means that 3 is an element of ℤ. ∉ ℤ means that is not an element of ℤ. { } means that it is a set. 5} means that 5 is a subset of ℤ , and the set consists of exactly one element which is 5. Definition 2.1.1 The span of a certain set is the set of all possible linear combinations of the subset of that set. Example 2.1.1 Find Span{1}. Solution: According to definition 2.1.1, then the span of the set {1} is the set of all possible linear combinations of the subset of {1} which is 1. Hence, Span{1} = ℝ . Example 2.1.2 Find Span{(1,2),(2,3)}. Solution: According to definition 2.1.1, then the span of the set {(1,2),(2,3)} is the set of all possible linear combinations of the subsets of {(1,2),(2,3)} which are (1,2) and (2,3). Thus, the following is some possible linear combinations: (1,2) = 1 ∗ (1,2) + 0 ∗ (2,3) (2,3) = 0 ∗ (1,2) + 1 ∗ (2,3) (5,8) = 1 ∗ (1,2) + 2 ∗ (2,3) Hence, {(1,2), (2,3), (5,8)} ∈ Span{(1,2), (2,3)} . Example 2.1.3 Find Span{0}. Solution: According to definition 2.1.1, then the span of the set {0} is the set of all possible linear combinations of the subset of {0} which is 0. 63
64 M. Kaabar Hence, Span{0} = 0. Example 2.1.4 Find Span{c} where c is a non-zero integer. Solution: Using definition 2.1.1, the span of the set {c} is the set of all possible linear combinations of the subset of {c} which is 𝑐 ≠ 0 . Thus, Span{c} = ℝ . Definition 2.1.2 ℝ 𝑛 = {(𝑎 , 𝑎 , 𝑎 , … , 𝑎 𝑛 )|𝑎 , 𝑎 , 𝑎 , … , 𝑎 𝑛 ∈ ℝ} is a set of all points where each point has exactly 𝑛 coordinates. Definition 2.1.3 (𝑉, +,∙) is a vector space if satisfies the following: a. For every 𝑣 , 𝑣 ∈ 𝑉 , 𝑣 + 𝑣 ∈ 𝑉. b. For every 𝛼 ∈ ℝ and 𝑣 ∈ 𝑉 , 𝛼𝑣 ∈ 𝑉. (i.e. Given 𝑆𝑝𝑎𝑛{𝑥, 𝑦}and 𝑠𝑒𝑡 {𝑥, 𝑦}, then √10𝑥 + 2𝑦 ∈ 𝑆𝑝𝑎𝑛{𝑥, 𝑦} . Let’s assume that 𝑣 ∈ 𝑆𝑝𝑎𝑛{𝑥, 𝑦} , then 𝑣 = 𝑐 𝑥 + 𝑐 𝑦 for some numbers 𝑐 and 𝑐 ). .2 The Dimension of Vector Space In this section, we discuss how to find the dimension of vector space, and how it is related to what we have learned in section 2.1. Definition 2.2.1 Given a vector space 𝑉 , the dimension of 𝑉 is the number of minimum elements needed in 𝑉 so that their 𝑆𝑝𝑎𝑛 is equal to 𝑉 , and it is denoted by dim (𝑉) . (i.e. dim(ℝ) = 1 and dim(ℝ ) = 2 ). Result 2.2.1 dim(ℝ 𝑛 ) = 𝑛 . Proof of Result 2.2.1 We will show that dim(ℝ 𝑛 ) = 𝑛. Claim:
𝐷 = 𝑆𝑝𝑎𝑛{(1,0), (0.1)} = ℝ 𝛼 (1,0) + 𝛼 (0,1) = (𝛼 , 𝛼 ) ∈ ℝ Thus, 𝐷 is a subset of ℝ ( 𝐷 ⊆ ℝ ). For every 𝑥 , 𝑦 ∈ ℝ , (𝑥 , 𝑦 ) ∈ ℝ . Therefore, (𝑥 , 𝑦 ) = 𝑥 (1,0) + 𝑦 (0,1) ∈ 𝐷. We prove the above claim, and hence dim(ℝ 𝑛 ) = 𝑛 . Fact 2.2.1 𝑆𝑝𝑎𝑛{(3,4)} ≠ ℝ . Proof of Fact 2.2.1 We will show that 𝑆𝑝𝑎𝑛{(3,4)} ≠ ℝ .
66 M. Kaabar Claim:
𝐹 = 𝑆𝑝𝑎𝑛{(6,5)} ≠ ℝ where (6,5) ∈ ℝ . We cannot find a number 𝛼 such that (6,5) = 𝛼(3,4) We prove the above claim, and hence
𝑆𝑝𝑎𝑛{(3,4)} ≠ ℝ . Fact 2.2.2 𝑆𝑝𝑎𝑛{(1,0), (0,1)} = ℝ . Fact 2.2.3 𝑆𝑝𝑎𝑛{(2,1), (1,0.5)} ≠ ℝ . In this section, we learn how to determine whether vector spaces are linearly independent or not. Definition 2.3.1 Given a vector space (𝑉, +,∙) , we say 𝑣 , 𝑣 , … , 𝑣 𝑛 ∈ 𝑉 are linearly independent if none of them is a linear combination of the remaining 𝑣 𝑖 ′𝑠 . (i.e. (3,4), (2,0) ∈ ℝ are linearly independent because we cannot write them as a linear combination of each other, in other words, we cannot find a number 𝛼 , 𝛼 such that (3,4) = 𝛼 (2,0) and (2,0) = 𝛼 (3,4) ). Definition 2.3.2 Given a vector space (𝑉, +,∙) , we say 𝑣 , 𝑣 , … , 𝑣 𝑛 ∈ 𝑉 are linearly dependent if at least one of 𝑣 𝑖 ′𝑠 is a linear combination of the others. Example 2.3.1 Assume 𝑣 and 𝑣 are linearly independent. Show that 𝑣 and + 𝑣 are linearly independent. olution: We will show that 𝑣 and + 𝑣 are linearly independent. Using proof by contradiction, we assume that 𝑣 and + 𝑣 are linearly dependent. For some non-zero number 𝑐 , 𝑣 = 𝑐 (3𝑣 + 𝑣 ) . Using the distribution property and algebra, we obtain: 𝑣 = 3𝑣 𝑐 + 𝑣 𝑐 𝑣 − 3𝑣 𝑐 = 𝑣 𝑐 𝑣 (1 − 3𝑐 ) = 𝑣 𝑐 (1 − 3𝑐 )𝑐 𝑣 = 𝑣 Thus, none of 𝑣 and + 𝑣 is a linear combination of the others which means that 𝑣 and + 𝑣 are linearly independent. This is a contradiction. Therefore, our assumption that 𝑣 and + 𝑣 were linearly dependent is false. Hence, 𝑣 and + 𝑣 are linearly independent. Example 2.3.2 Given the following vectors: 𝑣 = (1,0, −2) 𝑣 = (−2,2,1) 𝑣 = (−1,0,5) Are these vectors independent elements? Solution: First of all, to determine whether these vectors are independent elements or not, we need to write these vectors as a matrix. 67
68 M. Kaabar [ 1 0 −2−2 2 1−1 0 5 ]
Each point is a row-operation. We need to reduce this matrix to Semi-Reduced Matrix. Definition 2.3.3 Semi-Reduced Matrix is a reduced-matrix but the leader numbers can be any non-zero number. Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: [ 1 0 −2−2 2 1−1 0 5 ] 2𝑅 + 𝑅 → 𝑅 𝑅 + 𝑅 → 𝑅 [1 0 −20 2 −30 0 3 ] This is a Semi-Reduced Matrix. Since none of the rows in the Semi-Reduced Matrix become zero-row, then the elements are independent because we cannot write at least one of them as a linear combination of the others. Example 2.3.3 Given the following vectors: 𝑣 = (1, −2,4,6) 𝑣 = (−1,2,0,2) 𝑣 = (1, −2,8,14) Are these vectors independent elements? Solution: First of all, to determine whether these vectors are independent elements or not, we need to write these vectors as a matrix.
Each point is a row-operation. We need to reduce this matrix to Semi-Reduced Matrix. Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: [ 1 −2−1 21 −2 4 60 28 14] 𝑅 + 𝑅 → 𝑅 −𝑅 + 𝑅 → 𝑅 [1 −20 00 0 4 64 84 8] −𝑅 + 𝑅 → 𝑅 [1 −20 00 0 4 64 80 0] This is a Semi-Reduced Matrix. Since there is a zero-row in the Semi-Reduced Matrix, then the elements are dependent because we can write at least one of them as a linear combination of the others.
In this section, we discuss one of the most important concepts in linear algebra that is known as subspace. In addition, we give some examples explaining how to find the basis for subspace. Definition 2.4.1 Subspace is a vector space but we call it a subspace because it lives inside a bigger vector space. (i.e. Given vector spaces 𝑉 and 𝐷 , then according to the figure 2.4.1, 𝐷 is called a subspace of 𝑉 ). 69
70 M. Kaabar Figure 2.4.1: Subspace of 𝑉 Fact 2.4.1 Every vector space is a subspace of itself. Example 2.4.1 Given a vector space
𝐿 = {(𝑐, 3𝑐)|𝑐 ∈ ℝ} . a.
Does 𝐿 live in ℝ ? b. Does 𝐿 equal to ℝ ? c. Is 𝐿 a subspace of ℝ ? d. Does 𝐿 equal to 𝑆𝑝𝑎𝑛{(0,3)}? e. Does 𝐿 equal to 𝑆𝑝𝑎𝑛{(1,3), (2,6)}?
Solution: To answer all these questions, we need first to draw an equation from this vector space, say 𝑦 = 3𝑥 . The following figure represents the graph of the above equation, and it passes through a point (1,3) . V D
Figure 2.4.2: Graph of 𝑦 = 3𝑥
Now, we can answer the given questions as follows: Part a: Yes; 𝐿 lives in ℝ . Part b: No; 𝐿 does not equal to ℝ . To show that we prove the following claim: Claim: 𝐿 = 𝑆𝑝𝑎𝑛{(5,15)} ≠ ℝ where (5,15) ∈ ℝ . It is impossible to find a number 𝛼 = 3 such that (20,60) = 𝛼(5,15) because in this case 𝛼 = 4 where (20,60) = 4(5,15) . We prove the above claim, and
𝑆𝑝𝑎𝑛{(5,15)} ≠ ℝ . Thus, 𝐿 does not equal to ℝ Part c: Yes; 𝐿 is a subspace of ℝ because 𝐿 lives inside a bigger vector space which is ℝ . 71
72 M. Kaabar Part d: No; according to the graph in figure 2.4.2, (0,3) does not belong to 𝐿 . Part e: Yes; because we can write (1,3) and (2,6) as a linear combination of each other. 𝛼 (1,3) + 𝛼 (2,6) = {(𝛼 + 2𝛼 ), (3𝛼 + 6𝛼 )} 𝛼 (1,3) + 𝛼 (2,6) = {(𝛼 + 2𝛼 ), 3(𝛼 + 2𝛼 )} Assume 𝑐 = (𝛼 + 2𝛼 ) , then we obtain: 𝛼 (1,3) + 𝛼 (2,6) = {(𝑐, 3𝑐)|𝑐 ∈ ℝ} = 𝐿 . Thus, 𝐿 = 𝑆𝑝𝑎𝑛{(1,3), (2,6)} . Result 2.4.1 𝐿 is a subspace of ℝ if satisfies the following: a. 𝐿 lives inside ℝ . b. 𝐿 has only lines through the origin (0,0) . Example 2.4.2 Given a vector space 𝐷 = {(𝑎, 𝑏, 1)|𝑎, 𝑏 ∈ ℝ} . a.
Does 𝐷 live in ℝ ? b. Is 𝐷 a subspace of ℝ ? Solution: Since the equation of the above vector space is a three-dimensional equation, there is no need to draw it because it is difficult to draw it exactly. Thus, we can answer the above questions immediately. Part a: Yes; 𝐷 lives inside ℝ . Part b: No; since (0,0,0) ∉ 𝐷 , then 𝐷 is not a subspace of ℝ . Fact 2.4.2 Assume 𝐷 lives inside ℝ 𝑛 . If we can write 𝐷 as a 𝑆𝑝𝑎𝑛 , then it is a subspace of ℝ 𝑛 . Fact 2.4.3 Assume 𝐷 lives inside ℝ 𝑛 . If we cannot write 𝐷 as a 𝑆𝑝𝑎𝑛 , then it is not a subspace of ℝ 𝑛 . Fact 2.4.4 Assume 𝐷 lives inside ℝ 𝑛 . If (0,0,0, … ,0) is in 𝐷 , then 𝐷 is a subspace of ℝ 𝑛 . Fact 2.4.5 Assume 𝐷 lives inside ℝ 𝑛 . If (0,0,0, … ,0) is not in 𝐷 , then 𝐷 is not a subspace of ℝ 𝑛 . Now, we list the main results on ℝ 𝑛 : Result 2.4.2 Maximum number of independent points is 𝑛 . Result 2.4.3 Choosing any 𝑛 independent points in ℝ 𝑛 , say 𝑄 , 𝑄 , … , 𝑄 𝑛 , then ℝ 𝑛 = 𝑆𝑝𝑎𝑛{𝑄 , 𝑄 , … , 𝑄 𝑛 } . Result 2.4.4 dim(ℝ 𝑛 ) = 𝑛 . Results 2.4.3 and 2.4.4 tell us the following: In order to get all ℝ 𝑛 , we need exactly 𝑛 independent points. Result 2.4.5 Assume ℝ 𝑛 = 𝑆𝑝𝑎𝑛{𝑄 , 𝑄 , … , 𝑄 𝑘 } , then 𝑘 ≥𝑛 ( 𝑛 points of the 𝑄 𝑘 ′𝑠 are independents). Definition 2.4.2 Basis is the set of points that is needed to 𝑆𝑝𝑎𝑛 the vector space. 73
74 M. Kaabar Example 2.4.3 Let
𝐷 = 𝑆𝑝𝑎𝑛{(1, −1,0), (2,2,1), (0,4,1)} . a.
Find dim (𝐷) . b.
Find a basis for 𝐷 . Solution: First of all, we have infinite set of points, and 𝐷 lives inside ℝ . Let’s assume the following: 𝑣 = (1, −1,0) 𝑣 = (2,2,1) 𝑣 = (0,4,1) Part a: To find dim (𝐷) , we check whether 𝑣 , 𝑣 and 𝑣 are dependent elements or not. Using what we have learned so far from section 2.3: We need to write these vectors as a matrix. [1 −1 02 2 10 4 1] Each point is a row-operation. We need to reduce this matrix to Semi-Reduced Matrix. Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: [1 −1 02 2 10 4 1] −2𝑅 + 𝑅 → 𝑅 [1 −1 00 4 10 4 1] −𝑅 + 𝑅 → 𝑅 [1 −1 00 4 10 0 0] This is a Semi-Reduced Matrix. Since there is a zero-row in the Semi-Reduced Matrix, then these elements are dependent because we can write at least one of them as a linear combination of he others. Only two points survived in the Semi-Reduced Matrix. Thus, dim(𝐷) = 2 . Part b: 𝐷 is a plane that passes through the origin (0,0,0) . Since dim(𝐷) = 2 , then any two independent points in 𝐷 will form a basis for 𝐷 . Hence, the following are some possible bases for 𝐷 : Basis for 𝐷 is {(1, −1,0), (2,2,1)} . Another basis for 𝐷 is {(1, −1,0), (0,4,1)} . Result 2.4.6 It is always true that |𝐵𝑎𝑠𝑖𝑠| = 𝑑𝑖𝑚 (𝐷) . Example 2.4.4 Given the following: 𝑀 = 𝑆𝑝𝑎𝑛{(−1,2,0,0), (1, −2,3,0), (−2,0,3,0)} . Find a basis for 𝑀 . Solution: We have infinite set of points, and 𝑀 lives inside ℝ . Let’s assume the following: 𝑣 = (−1,2,0,0) 𝑣 = (1, −2,3,0) 𝑣 = (−2,0,3,0) We check if 𝑣 , 𝑣 and 𝑣 are dependent elements. Using what we have learned so far from section 2.3 and example 2.4.3: We need to write these vectors as a matrix. [−1 21 −2−2 0 0 03 03 0] Each point is a row-operation. We need to reduce this matrix to Semi-Reduced Matrix. 75
76 M. Kaabar Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: [−1 21 −2−2 0 0 03 03 0] 𝑅 + 𝑅 → 𝑅 −2𝑅 + 𝑅 → 𝑅 [−1 20 00 −4 0 03 03 0] −𝑅 + 𝑅 → 𝑅 [−1 20 00 −4 0 03 00 0] This is a Semi-Reduced Matrix. Since there is no zero-row in the Semi-Reduced Matrix, then these elements are independent. All the three points survived in the Semi-Reduced Matrix. Thus, dim(𝑀) = 3 . Since dim(𝑀) = 3 , then any three independent points in 𝑀 from the above matrices will form a basis for 𝑀 . Hence, the following are some possible bases for 𝑀 : Basis for 𝑀 is {(−1,2,0,0), (0,0,3,0), (0, −4,0,0)} . Another basis for 𝑀 is {(−1,2,0,0), (0,0,3,0), (0, −4,3,0)} . Another basis for 𝑀 is {(−1,2,0,0), (1, −2,3,0), (−2,0,3,0)} . Example 2.4.5 Given the following: 𝑊 = 𝑆𝑝𝑎𝑛{(𝑎, −2𝑎 + 𝑏, −𝑎)|𝑎, 𝑏 ∈ ℝ} . a.
Show that 𝑊 is a subspace of ℝ . b. Find a basis for 𝑊 . c. Rewrite 𝑊 as a 𝑆𝑝𝑎𝑛 . Solution: We have infinite set of points, and 𝑊 lives inside ℝ . Part a: We write each coordinate of 𝑊 as a linear combination of the free variables 𝑎 and 𝑏 . 𝑎 = 1 ∙ 𝑎 + 0 ∙ 𝑏 −2𝑎 + 𝑏 = −2 ∙ 𝑎 + 1 ∙ 𝑏 −𝑎 = −1 ∙ 𝑎 + 0 ∙ 𝑏 Since it is possible to write each coordinate of 𝑊 as a linear combination of the free variables 𝑎 and 𝑏 , then we conclude that 𝑊 is a subspace of ℝ . Part b: To find a basis for 𝑊 , we first need to find dim (𝑊) . To find dim (𝑊) , let’s play a game called (ON-OFF GAME) with the free variables 𝑎 and 𝑏. 𝑎 𝑏 𝑃𝑜𝑖𝑛𝑡 (1, −2, −1) (0,1,0) Now, we check for independency: We already have the Semi-Reduced Matrix: [1 −2 −10 1 0 ].
Thus, dim(𝑊) = 2 . Hence, the basis for 𝑊 is {(1, −2, −1), (0,1,0)} . Part b: Since we found the basis for 𝑊 , then it is easy to rewrite 𝑊 as a 𝑆𝑝𝑎𝑛 as follows:
𝑊 = 𝑆𝑝𝑎𝑛{(1, −2, −1), (0,1,0)}.
Fact 2.4.6 dim(𝑊) ≤ 𝑁𝑢𝑚𝑏𝑒𝑟𝑜𝑓 𝐹𝑟𝑒𝑒 − 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠.
78 M. Kaabar Example 2.4.6 Given the following:
𝐻 = 𝑆𝑝𝑎𝑛{(𝑎 , 3𝑏 + 𝑎, −2𝑐, 𝑎 + 𝑏 + 𝑐)|𝑎, 𝑏, 𝑐 ∈ ℝ} . Is 𝐻 a subspace of ℝ ? Solution: We have infinite set of points, and 𝐻 lives inside ℝ . We try write each coordinate of 𝐻 as a linear combination of the free variables 𝑎, 𝑏 and 𝑐 . 𝑎 = 𝐹𝑖𝑥𝑒𝑑 𝑁𝑢𝑚𝑏𝑒𝑟 ∙ 𝑎 + 𝐹𝑖𝑥𝑒𝑑 𝑁𝑢𝑚𝑏𝑒𝑟 ∙ 𝑏 + 𝐹𝑖𝑥𝑒𝑑 𝑁𝑢𝑚𝑏𝑒𝑟 ∙ 𝑐 𝑎 is not a linear combination of 𝑎, 𝑏 and 𝑐 . We assume that 𝑤 = (1,1,0,1) ∈ 𝐻 , and 𝑎 = 1, 𝑏 = 𝑐 = 0 . If 𝛼 = −2 , then −2 ∙ 𝑤 = −2 ∙ (1,1,0,1) = (−2, −2,0, −2) ∉ 𝐻 . Since it is impossible to write each coordinate of 𝐻 as a linear combination of the free variables 𝑎, 𝑏 and 𝑐 , then we conclude that 𝐻 is not a subspace of ℝ . Example 2.4.7 Form a basis for ℝ . Solution: We just need to select any random four independent points, and then we form a matrix with four independent rows as follows: [2 30 5 0 41 10 00 0 2 30 𝜋 𝑒 ] Note: 𝜋 𝑒 is a number. Let’s assume the following: 𝑣 = (2,3,0,4) 𝑣 = (0,5,1,1) 𝑣 = (0,0,2,3) 𝑣 = (0,0,0, 𝜋 𝑒 ) Thus, the basis for ℝ = {𝑣 , 𝑣 , 𝑣 , 𝑣 } , and 𝑆𝑝𝑎𝑛{𝑣 , 𝑣 , 𝑣 , 𝑣 } = ℝ . Example 2.4.8 Form a basis for ℝ that contains the following two independent points: (0,2,1,4) and (0, −2,3, −10) . Solution: We need to add two more points to the given one so that all four points are independent. Let’s assume the following: 𝑣 = (0,2,1,4) 𝑣 = (0, −2,3, −10) 𝑣 = (0,0,4, −6) This is a random point. 𝑣 = (0,0,0,1000) This is a random point. Then, we need to write these vectors as a matrix. [0 20 −2 1 43 −100 00 0 4 −6 0 1000]
Each point is a row-operation. We need to reduce this matrix to Semi-Reduced Matrix. Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: 79
80 M. Kaabar [0 20 −2 1 43 −100 00 0 4 −6 0 1000] 𝑅 + 𝑅 → 𝑅 [0 23 0 1 45 300 00 0 4 −6 0 1000] This is a Semi-Reduced Matrix. Thus, the basis for ℝ is {(0,2,1,4) , (0, −2,3, −10), (3,0,5,30), (0,0,0,1000)} . Example 2.4.9 Given the following: 𝐷 = 𝑆𝑝𝑎𝑛{(1,1,1,1), (−1, −1,0,0), (0,0,1,1)} Is (1,1,2,2) ∈ 𝐷 ? Solution: We have infinite set of points, and 𝐷 lives inside ℝ . There are two different to solve this example: The First Way: Let’s assume the following: 𝑣 = (1,1,1,1) 𝑣 = (−1, −1,0,0) 𝑣 = (0,0,1,1) We start asking ourselves the following question: Question: Can we find 𝛼 , 𝛼 and 𝛼 such that (1,1,2,2) = 𝛼 ∙ 𝑣 + 𝛼 ∙ 𝑣 + 𝛼 ∙ 𝑣 ? Answer: Yes but we need to solve the following system of linear equations: − 𝛼 + 0 ∙ 𝛼 − 𝛼 + 0 ∙ 𝛼 + 𝛼 + 𝛼 Using what we have learned from chapter 1 to solve the above system of linear equations, we obtain: 𝛼 = 𝛼 = 𝛼 = 1 Hence, Yes: (1,1,2,2) ∈ 𝐷.
The Second Way (Recommended): We first need to find 𝑑𝑖𝑚 (𝐷) , and then a basis for 𝐷 . We have to write 𝑣 , 𝑣 and 𝑣 as a matrix. [ 1 1−1 −10 0 1 10 01 1] Each point is a row-operation. We need to reduce this matrix to Semi-Reduced Matrix. Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: [ 1 1−1 −10 0 1 10 01 1] 𝑅 + 𝑅 → 𝑅 [1 10 00 0 1 11 11 1] −𝑅 + 𝑅 → 𝑅 [1 10 00 0 1 11 10 0] This is a Semi-Reduced Matrix. Since there is a zero-row in the Semi-Reduced Matrix, then these elements are dependent. Thus, dim(𝐷) = 2 . Thus, Basis for 𝐷 is {(1,1,1,1), (0,0,1,1)} , and 81
82 M. Kaabar
𝐷 = 𝑆𝑝𝑎𝑛{(1,1,1,1), (0,0,1,1)} . Now, we ask ourselves the following question: Question: Can we find 𝛼 , 𝛼 and 𝛼 such that (1,1,2,2) = 𝛼 ∙ (1,1,1,1) + 𝛼 ∙ (0,0,1,1) ? Answer: Yes: + 𝛼 + 𝛼 Thus, 𝛼 = 𝛼 = 𝛼 = 1 . Hence, Yes: (1,1,2,2) ∈ 𝐷.
1. Let
𝑀 = 𝑆𝑝𝑎𝑛{(1, −1,1), (−1,0,1), (0, −1,2)} a. Find a basis for 𝑀 . b. Is (1, −3,5) ∈ 𝑀 ? Why? 2. Let 𝐷 = {(𝑥, 𝑦, 𝑧, 𝑡) ∈ ℝ |𝑥, 𝑦, 𝑧, 𝑡 ∈ ℝ, 𝑥 + 2𝑧 + 3𝑡 = 0, 𝑎𝑛𝑑 𝑦 − 𝑧 + 𝑡 = 0} a. Show that 𝐷 is a subspace of ℝ . b. Find a basis for 𝐷 . c. Write 𝐷 as a 𝑆𝑝𝑎𝑛 . 3. Let
𝐸 = {[ 𝑚𝑚 + 𝑤𝑤 ] |𝑚, 𝑤 ∈ ℝ} a. Show that 𝐸 is a subspace of ℝ . b. Find a basis for 𝐸 . c. Write 𝐸 as a 𝑆𝑝𝑎𝑛 . 4. Find a basis for the subspace 𝐹 where 𝐹 = {[ 𝑤 − 𝑠 + 3𝑢 4𝑤 + 3𝑠 − 9𝑢 2𝑤8𝑤 + 2𝑠 − 6𝑢 5𝑤 0 ] |𝑤, 𝑠, 𝑢 ∈ ℝ}
5. Determine the value(s) of 𝑥 such that the points (1,0,5), (1,2,4), (1,4, 𝑥) are dependent. 6. Let 𝐷 = 𝑆𝑝𝑎𝑛{[2 −10 1 ] , [1 −10 0.5] , [5 −30 2.5]} a. Find 𝑑𝑖𝑚 (𝐷) . b. Find a basis 𝐴 for 𝐷 . c. Is 𝐶 = [−2 00 −1] ∈ 𝐷 ? Why? d. If the answer to part c is yes, then write 𝐶 as a linear combination of the elements in 𝐴 . Otherwise, write the basis 𝐴 as a 𝑆𝑝𝑎𝑛 . 7. Let
𝐾 = 𝑆𝑝𝑎𝑛{(1, −1,0), (2, −1,0), (1,0,0)} . Find 𝑑𝑖𝑚 (𝐾) . 8. Find a basis for the subspace of ℝ spanned by {(2,9, −2,53), (−3,2,3, −2), (8, −3, −8,17), (0, −3,0,15)} . 9. Does the 𝑆𝑝𝑎𝑛{(−2,1,2), (2,1, −1), (2,3,0)} equal to ℝ ? 83
84 M. Kaabar
Chapter 3 Homogeneous Systems
In this chapter, we introduce the homogeneous systems, and we discuss how they are related to what we have learned in chapter 2. We start with an introduction to null space and rank. Then, we study one of the most important topics in linear algebra which is linear transformation. At the end of this chapter we discuss how to find range and kernel, and their relation to sections 3.1 and 3.2.
In this section, we first give an introduction to homogeneous systems, and we discuss how to find the null space and rank of homogeneous systems. In addition, we explain how to find row space and column space. Definition 3.1.1 Homogeneous System is a 𝑚 × 𝑛 system of linear equations that has all zero constants. (i.e. the following is an example of homogeneous system): { 2𝑥 + 𝑥 − 𝑥 + 𝑥 = 03𝑥 + 5𝑥 + 3𝑥 + 4𝑥 = 0−𝑥 + 𝑥 − 𝑥 = 0 Imagine we have the following solution to the homogeneous system: 𝑥 = 𝑥 = 𝑥 = 𝑥 = 0 . Then, this solution can be viewed as a point of ℝ 𝑛 (here is ℝ ) : (0,0,0,0) Result 3.1.1 The solution of a homogeneous system 𝑚 ×𝑛 can be written as {(𝑎 , 𝑎 , 𝑎 , 𝑎 , … , 𝑎 𝑛 |𝑎 , 𝑎 , 𝑎 , 𝑎 , … , 𝑎 𝑛 ∈ ℝ} . Result 3.1.2 All solutions of a homogeneous system 𝑚 × 𝑛 form a subset of ℝ 𝑛 , and it is equal to the number of variables. Result 3.1.3 Given a homogeneous system 𝑚 × 𝑛 . We write it in the matrix-form: 𝐶 [ 𝑥 𝑥 𝑥 ⋮𝑥 𝑛 ] = [ 000⋮0] where 𝐶 is a coefficient. Then, the set of all solutions in this system is a subspace of ℝ 𝑛 . Proof of Result 3.1.3 We assume that 𝑀 =(𝑚 , 𝑚 , … , 𝑚 𝑛 ) and 𝑊 = (𝑤, 𝑤 , … , 𝑤 𝑛 ) are two solutions to the above system. We will show that 𝑀 +𝑊 is a solution. We write them in the matrix-form:
𝐶 [ 𝑚 𝑚 𝑚 ⋮𝑚 𝑛 ] = [ 000⋮0] and [ 𝑤 𝑤 𝑤 ⋮𝑤 𝑛 ] = [ 000⋮0]
86 M. Kaabar Now, using algebra:
𝑀 + 𝑊 = 𝐶 [ 𝑚 𝑚 𝑚 ⋮𝑚 𝑛 ] + 𝐶 [ 𝑤 𝑤 𝑤 ⋮𝑤 𝑛 ] = [ 000⋮0] By taking 𝐶 as a common factor, we obtain: 𝐶 ( [ 𝑚 𝑚 𝑚 ⋮𝑚 𝑛 ] + [ 𝑤 𝑤 𝑤 ⋮𝑤 𝑛 ] ) = [ 000⋮0] 𝐶 [ 𝑚 + 𝑤 𝑚 + 𝑤 𝑚 + 𝑤 ⋮𝑚 𝑛 + 𝑤 𝑛 ] = [ 000⋮0] Thus,
𝑀 + 𝑊 is a solution. Fact 3.1.1 If 𝑀 = (𝑚 , 𝑚 , … , 𝑚 𝑛 ) is a solution, and 𝛼 ∈ℝ , then 𝛼𝑀 = (𝛼𝑚 , 𝛼𝑚 , … , 𝛼𝑚 𝑛 ) is a solution. Fact 3.1.2 The only system where the solutions form a vector space is the homogeneous system. Definition 3.1.2 Null Space of a matrix, say 𝐴 is a set of all solutions to the homogeneous system, and it is denoted by 𝑁𝑢𝑙𝑙(𝐴) or 𝑁(𝐴) . Definition 3.1.3 Rank of a matrix, say 𝐴 is the number of independent rows or columns of 𝐴 , and it is denoted by 𝑅𝑎𝑛𝑘(𝐴) . Definition 3.1.4 Row Space of a matrix, say 𝐴 is the 𝑆𝑝𝑎𝑛 of independent rows of 𝐴 , and it is denoted by 𝑅𝑜𝑤(𝐴) . Definition 3.1.5 Column Space of a matrix, say 𝐴 is the 𝑆𝑝𝑎𝑛 of independent columns of 𝐴 , and it is denoted by 𝐶𝑜𝑙𝑢𝑚𝑛(𝐴) . Example 3.1.1 Given the following matrix:
𝐴 = [1 −1 2 0 −10 1 2 0 20 0 0 1 0 ] . a.
Find
𝑁𝑢𝑙𝑙(𝐴) . b.
Find 𝑑𝑖𝑚 (𝑁𝑢𝑙𝑙(𝐴)) . c.
Rewrite
𝑁𝑢𝑙𝑙(𝐴) as 𝑆𝑝𝑎𝑛 . d.
Find
𝑅𝑎𝑛𝑘(𝐴) . e.
Find
𝑅𝑜𝑤(𝐴) . Solution: Part a: To find the null space of 𝐴 , we need to find the solution of 𝐴 as follows: Step 1: Write the above matrix as an Augmented-Matrix, and make all constants’ terms zeros. (1 −1 2 0 −10 1 2 0 20 0 0 1 0 |000) Step 2: Apply what we have learned from chapter 1 to solve systems of linear equations use Row-Operation Method. 87
88 M. Kaabar (1 −1 2 0 −10 1 2 0 20 0 0 1 0 |000) 𝑅 + 𝑅 → 𝑅 (1 0 4 0 10 1 2 0 20 0 0 1 0|000) This is a Completely-Reduced Matrix. Step 3: Read the solution for the above system of linear equations after using Row-Operation. 𝑥 + 4𝑥 + 𝑥 = 0𝑥 + 2𝑥 + 2𝑥 = 0𝑥 = 0 Free variables are 𝑥 and 𝑥 . Assuming that 𝑥 , 𝑥 ∈ ℝ . Then, the solution of the above homogeneous system is as follows: 𝑥 = −4𝑥 − 𝑥 𝑥 = −2𝑥 − 2𝑥 𝑥 = 0 Thus, according to definition 3.1.2,
𝑁𝑢𝑙𝑙(𝐴) = {(−4𝑥 − 𝑥 , −2𝑥 − 2𝑥 , 𝑥 , 0, 𝑥 )|𝑥 , 𝑥 ∈ ℝ} . Part b: It is always true that 𝑑𝑖𝑚(𝑁𝑢𝑙𝑙(𝐴)) = 𝑑𝑖𝑚(𝑁(𝐴)) = 𝑇ℎ𝑒 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐹𝑟𝑒𝑒 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 Here, 𝑑𝑖𝑚(𝑁𝑢𝑙𝑙(𝐴)) = 2 . Definition 3.1.6 The nullity of a matrix, say 𝐴 is the dimension of the null space of 𝐴 , and it is denoted by 𝑑𝑖𝑚(𝑁𝑢𝑙𝑙(𝐴)) or 𝑑𝑖𝑚 (𝑁(𝐴)) . Part c: We first need to find a basis for 𝑁𝑢𝑙𝑙(𝐴) as follows: To find a basis for
𝑁𝑢𝑙𝑙(𝐴) , we play a game called (ON-OFF GAME) with the free variables 𝑥 and 𝑥 . 𝑥 𝑥 𝑃𝑜𝑖𝑛𝑡 (−4, −2,1,0,0) (−1, −2,0,0,1) The basis for
𝑁𝑢𝑙𝑙(𝐴) = {(−4, −2,1,0,0), (−1, −2,0,0,1)} . Thus,
𝑁𝑢𝑙𝑙(𝐴) = 𝑆𝑝𝑎𝑛{(−4, −2,1,0,0), (−1, −2,0,0,1)} . Part d: To find the rank of matrix 𝐴 , we just need to change matrix 𝐴 to the Semi-Reduced Matrix. We already did that in part a. Thus, 𝑅𝑎𝑛𝑘(𝐴) = 3.
Part e: To find the row space of matrix 𝐴 , we just need to write the 𝑆𝑝𝑎𝑛 of independent rows. Thus,
𝑅𝑜𝑤(𝐴) = 𝑆𝑝𝑎𝑛{(1, −1,2,0, −1), (0,1,2,0,2), (0,0,0,1,0)}.
It is also a subspace of ℝ . Result 3.1.4 Let 𝐴 be 𝑚 × 𝑛 matrix. Then, 𝑅𝑎𝑛𝑘(𝐴) + 𝑑𝑖𝑚(𝑁(𝐴)) = 𝑛 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶𝑜𝑙𝑢𝑚𝑛𝑠 𝑜𝑓 𝐴 . Result 3.1.5 Let 𝐴 be 𝑚 × 𝑛 matrix. The geometric meaning of 𝑅𝑜𝑤(𝐴) = 𝑆𝑝𝑎𝑛{𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑅𝑜𝑤𝑠} “lives” inside ℝ 𝑛 . 89
90 M. Kaabar Result 3.1.6 Let 𝐴 be 𝑚 × 𝑛 matrix. The geometric meaning of 𝐶𝑜𝑙𝑢𝑚𝑛(𝐴) = 𝑆𝑝𝑎𝑛{𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝐶𝑜𝑙𝑢𝑚𝑛𝑠} “lives” inside ℝ 𝑚 . Result 3.1.7 Let 𝐴 be 𝑚 × 𝑛 matrix. Then, 𝑅𝑎𝑛𝑘(𝐴) = 𝑑𝑖𝑚(𝑅𝑜𝑤(𝐴)) = 𝑑𝑖𝑚 (𝐶𝑜𝑙𝑢𝑚𝑛(𝐴)) . Example 3.1.2 Given the following matrix:
𝐵 = [ 1 1 1−1 −1 −10 0 0 1 10 20 0] . a.
Find
𝑅𝑜𝑤(𝐵) . b.
Find
𝐶𝑜𝑙𝑢𝑚𝑛(𝐵) . c.
Find
𝑅𝑎𝑛𝑘(𝐵) . Solution: Part a: To find the row space of 𝐵 , we need to change matrix 𝐵 to the Semi-Reduced Matrix as follows: [ 1 1 1−1 −1 −10 0 0 1 10 20 0] 𝑅 + 𝑅 → 𝑅 𝑅 + 𝑅 → 𝑅 [1 1 10 0 00 0 0 1 11 30 0] This is a Semi-Reduced Matrix. To find the row space of matrix 𝐵 , we just need to write the 𝑆𝑝𝑎𝑛 of independent rows. Thus,
𝑅𝑜𝑤(𝐵) =𝑆𝑝𝑎𝑛{(1,1,1,1,1), (0,0,0,1,3)}.
Part b: To find the column space of 𝐵 , we need to change matrix 𝐵 to the Semi-Reduced Matrix. We already did that in part a. Now, we need to locate the columns in the Semi-Reduced Matrix of 𝐵 that contain the leaders, and then we should locate them to the original matrix 𝐵 . [1 1 10 0 00 0 0 1 11 30 0] Semi-Reduced Matrix [ 1 1 1−1 −1 −10 0 0 1 10 20 0]
Matrix 𝐵 Each remaining columns is a linear combination of the first and fourth columns. Thus,
𝐶𝑜𝑙𝑢𝑚𝑛(𝐵) = 𝑆𝑝𝑎𝑛{(1, −1,0), (1,0,0)} . Part c: To find the rank of matrix 𝐵 , we just need to change matrix 𝐴 to the Semi-Reduced Matrix. We already did that in part a. Thus, 𝑅𝑎𝑛𝑘(𝐴) = 𝑑𝑖𝑚(𝑅𝑜𝑤(𝐵)) = 𝑑𝑖𝑚 (𝐶𝑜𝑙𝑢𝑚𝑛(𝐵)) = 2.
We start this section with an introduction to polynomials, and we explain how they are similar to ℝ 𝑛 as vector spaces. At the end of this section we discuss a new concept called linear transformation. 91
92 M. Kaabar Before discussing polynomials, we need to know the following mathematical facts: Fact 3.2.1 ℝ 𝑛×𝑚 = ℝ 𝑛×𝑚 = 𝑀 𝑛×𝑚 (ℝ) is a vector space. Fact 3.2.2 ℝ is equivalent to ℝ as a vector space. (i.e. [1 2 30 1 1] is equivalent to (1,2,3,0,1,1) ). Fact 3.2.3 ℝ is equivalent to ℝ as a vector space. (i.e. [1 23 01 1] is equivalent to (1,2,3,0,1,1) ). After knowing the above facts, we introduce polynomials as follows: 𝑃 𝑛 = 𝑆𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙𝑠 𝑜𝑓 𝑑𝑒𝑔𝑟𝑒𝑒 < 𝑛. The algebraic expression of polynomials is in the following from: 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎 𝑥 + 𝑎 𝑎 𝑛 , 𝑎 𝑛−1 and 𝑎 are coefficients. 𝑛 and 𝑛 − 1 are exponents that must be positive integers whole numbers. 𝑎 is a constant term. The degree of polynomial is determined by the highest power (exponent). We list the following examples of polynomials: 𝑃 = 𝑆𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙𝑠 𝑜𝑓 𝑑𝑒𝑔𝑟𝑒𝑒 < 2 (i.e.
3𝑥 + 2 ∈ 𝑃 , ,
10 ∈ 𝑃 , √3 ∈ 𝑃 but √3√𝑥 ∉ 𝑃 ). 𝑃 = 𝑆𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙𝑠 𝑜𝑓 𝑑𝑒𝑔𝑟𝑒𝑒 < 4 (i.e. + 4 ∈ 𝑃 ). If 𝑃(𝑥) = 3 , then 𝑑𝑒𝑔(𝑃(𝑥)) = 0 . √𝑥 + 3 is not a polynomial. Result 3.2.1 𝑃 𝑛 is a vector space. Fact 3.2.4 ℝ = 𝑀 (ℝ) as a vector space same as ℝ . Result 3.2.2 𝑃 𝑛 is a vector space, and it is the same as ℝ 𝑛 . (i.e. 𝑎 + 𝑎 𝑥 + ⋯ + 𝑎 𝑛−1 𝑥 𝑛−1 ↔ (𝑎 , 𝑎 , … , 𝑎 𝑛−1 ) . Note: The above form is in an ascending order. Result 3.2.3 𝑑𝑖𝑚 (𝑃 𝑛 ) = 𝑛. Fact 3.2.5 𝑃 = 𝑆𝑝𝑎𝑛{3 𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑃𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙𝑠, 𝑎𝑛𝑑 𝐸𝑎𝑐ℎ 𝑜𝑓 𝐷𝑒𝑔𝑟𝑒𝑒 < 3} . (i.e. 𝑃 = 𝑆𝑝𝑎𝑛{1, 𝑥, 𝑥 } ). Example 3.2.1 Given the following polynomials: − 2, −5𝑥, 6𝑥 − 10𝑥 − 4 . a. Are these polynomials independent? b.
Let
𝐷 = 𝑆𝑝𝑎𝑛{3𝑥 − 2, −5𝑥, 6𝑥 − 10𝑥 − 4} . Find a basis for 𝐷 . Solution: Part a: We know that these polynomial live in 𝑃 , and as a vector space 𝑃 is the same as ℝ . According to result 3.2.2, we need to make each polynomial equivalent to ℝ 𝑛 as follows: 93
94 M. Kaabar − 2 = −2 + 0𝑥 + 3𝑥 ↔ (−2,0,3) −5𝑥 = 0 − 5𝑥 + 0𝑥 ↔ (0, −5,0) − 10𝑥 − 4 = −4 − 10𝑥 + 6𝑥 ↔ (−4, −10,6) Now, we need to write these vectors as a matrix. [−2 0 30 −5 0−4 −10 6]
Each point is a row-operation. We need to reduce this matrix to Semi-Reduced Matrix. Then, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: [−2 0 30 −5 0−4 −10 6] −2𝑅 + 𝑅 → 𝑅 [−2 0 30 −5 00 −10 0] −2𝑅 + 𝑅 → 𝑅 [−2 0 30 −5 00 0 0] This is a Semi-Reduced Matrix. Since there is a zero-row in the Semi-Reduced Matrix, then these elements are dependent. Thus, the answer to this question is NO. Part b: Since there are only 2 vectors survived after checking for dependency in part a, then the basis for 𝐷 (0, −5,0) ↔ −5𝑥 . Result 3.2.4 Given 𝑣 , 𝑣 , … , 𝑣 𝑘 points in ℝ 𝑛 where 𝑘 < 𝑛 . Choose one particular point, say 𝑄 , such that 𝑄 = 𝑐 𝑣 + 𝑐 𝑣 + ⋯ + 𝑐 𝑘 𝑣 𝑘 where 𝑐 , 𝑐 , … , 𝑐 𝑘 are constants. If 𝑐 , 𝑐 , … , 𝑐 𝑘 are unique, then 𝑣 , 𝑣 , … , 𝑣 𝑘 are independent. Note: The word “unique” in result 3.2.4 means that there is only one value for each of 𝑐 , 𝑐 , … , 𝑐 𝑘 . Proof of Result 3.2.4 By using proof by contradiction, we assume that 𝑣 = 𝛼 𝑣 + 𝛼 𝑣 + ⋯ + 𝛼 𝑘 𝑣 𝑘 where 𝛼 , 𝛼 , … , 𝛼 𝑘 are constants. Our assumption means that it is dependent. Using algebra, we obtain: 𝑄 = 𝑐 𝛼 𝑣 + 𝑐 𝛼 𝑣 + ⋯ + 𝑐 𝛼 𝑘 𝑣 𝑘 + 𝑐 𝑣 + ⋯ + 𝑐 𝑘 𝑣 𝑘 . 𝑄 = (𝑐 𝛼 +𝑐 )𝑣 + (𝑐 𝛼 + 𝑐 )𝑣 + ⋯ + (𝑐 𝛼 𝑘 + 𝑐 𝑘 )𝑣 𝑘 +0𝑣 . Thus, none of them is a linear combination of the others which means that they are linearly independent. This is a contradiction. Therefore, our assumption that 𝑣 , 𝑣 , … , and 𝑣 𝑘 were linearly dependent is false. Hence, 𝑣 , 𝑣 , … , and 𝑣 𝑘 are linearly independent. Result 3.2.5 Assume 𝑣 , 𝑣 , … , 𝑣 𝑘 are independent and 𝑄 ∈ 𝑆𝑝𝑎𝑛{𝑣 , 𝑣 , … , 𝑣 𝑘 } . Then, there exists unique number 𝑐 , 𝑐 , … , 𝑐 𝑘 such that 𝑄 = 𝑐 𝑣 + 𝑐 𝑣 + ⋯ +𝑐 𝑘 𝑣 𝑘 . Linear Transformation: Definition 3.2.1 𝑇: 𝑉 → 𝑊 where 𝑉 is a domain and 𝑊 is a co-domain. 𝑇 is a linear transformation if for every 𝑣 , 𝑣 ∈ 𝑉 and 𝛼 ∈ ℝ , we have the following: 𝑇(𝛼𝑣 + 𝑣 ) = 𝛼𝑇(𝑣 ) + 𝑇(𝑣 ) . 95
96 M. Kaabar Example 3.2.2 Given
𝑇: ℝ → ℝ where ℝ is a domain and ℝ is a co-domain. 𝑇((𝑎 , 𝑎 )) = (3𝑎 + 𝑎 , 𝑎 , −𝑎 ) . a. Find
𝑇((1,1)) . b.
Find
𝑇((1,0)) . c.
Show that 𝑇 is a linear transformation. Solution: Part a: Since 𝑇((𝑎 , 𝑎 )) = (3𝑎 + 𝑎 , 𝑎 , −𝑎 ) , then 𝑎 = 𝑎 = 1 . Thus, 𝑇((1,1)) = (3(1) + 1,1, −1) =(4,1, −1) . Part b: Since
𝑇((𝑎 , 𝑎 )) = (3𝑎 + 𝑎 , 𝑎 , −𝑎 ) , then 𝑎 =1 and 𝑎 = 0 . Thus, 𝑇((1,0)) = (3(1) + 0,0, −1) =(3,0, −1) . Part c: Proof: We assume that 𝑣 = (𝑎 , 𝑎 ) , 𝑣 = (𝑏 , 𝑏 ) , and 𝛼 ∈ ℝ . We will show that 𝑇 is a linear transformation. Using algebra, we start from the Left-Hand-Side (LHS): 𝛼𝑣 + 𝑣 = (𝛼𝑎 + 𝑏 , 𝛼𝑎 + 𝑏 ) 𝑇(𝛼𝑣 + 𝑣 ) = 𝑇((𝛼𝑎 + 𝑏 , 𝛼𝑎 + 𝑏 )) 𝑇(𝛼𝑣 + 𝑣 ) = (3𝛼𝑎 + 3𝑏 + 𝛼𝑎 + 𝑏 , 𝛼𝑎 + 𝑏 , −𝛼𝑎 − 𝑏 ) Now, we start from the Right-Hand-Side (RHS): 𝛼𝑇(𝑣 ) + 𝑇(𝑣 ) = 𝛼𝑇(𝑎 , 𝑎 ) + 𝑇(𝑏 , 𝑏 ) 𝛼𝑇(𝑣 ) + 𝑇(𝑣 ) = 𝛼(3𝑎 + 𝑎 , 𝑎 , −𝑎 ) + (3𝑏 + 𝑏 , 𝑏 , −𝑏 ) = (3𝛼𝑎 + 𝛼𝑎 , 𝛼𝑎 , −𝛼𝑎 ) + (3𝑏 + 𝑏 , 𝑏 , −𝑏 ) = (3𝛼𝑎 + 𝛼𝑎 + 3𝑏 + 𝑏 , 𝛼𝑎 + 𝑏 , −𝛼𝑎 − 𝑏 ) Thus, 𝑇 is a linear transformation. Result 3.2.6 Given 𝑇: ℝ 𝑛 → ℝ 𝑚 . Then, 𝑇((𝑎 , 𝑎 , 𝑎 , … , 𝑎 𝑛 )) = Each coordinate is a linear combination of the 𝑎 𝑖 ′𝑠 . Example 3.2.3 Given 𝑇: ℝ → ℝ where ℝ is a domain and ℝ is a co-domain. a. If 𝑇((𝑥 , 𝑥 , 𝑥 )) = (−3𝑥 + 6𝑥 , −10𝑥 , 13, −𝑥 ) , is 𝑇 a linear transformation? b. If 𝑇((𝑥 , 𝑥 , 𝑥 )) = (−3𝑥 + 6𝑥 , −10𝑥 , 0, −𝑥 ) , is 𝑇 a linear transformation? Solution: Part a: Since 13 is not a linear combination of 𝑥 , 𝑥 and 𝑥 . Thus, 𝑇 is not a linear transformation. Part b: Since 0 is a linear combination of 𝑥 , 𝑥 and 𝑥 . Thus, 𝑇 is a linear transformation. Example 3.2.4 Given 𝑇: ℝ → ℝ where ℝ is a domain and ℝ is a co-domain. If 𝑇((𝑎 , 𝑎 )) = (𝑎 + 𝑎 , −𝑎 ) , is 𝑇 a linear transformation? Solution: Since 𝑎 + 𝑎 is not a linear combination of 𝑎 and 𝑎 . Hence, 𝑇 is not a linear transformation. Example 3.2.5 Given 𝑇: ℝ → ℝ . If
𝑇(𝑥) = 10𝑥 , is 𝑇 a linear transformation? Solution: Since it is a linear combination of 𝑎 such that 𝛼𝑎 = 10𝑥 . Hence, 𝑇 is a linear transformation. Example 3.2.6 Find the standard basis for ℝ . Solution: The standard basis for ℝ is the rows of 𝐼 . 97
98 M. Kaabar Since 𝐼 = [1 00 1] , then the standard basis for ℝ is {(1,0), (0,1)} . Example 3.2.7 Find the standard basis for ℝ . Solution: The standard basis for ℝ is the rows of 𝐼 . Since 𝐼 = [1 0 00 1 00 0 1] , then the standard basis for ℝ is {(1,0,0), (0,1,0), (0,0,1)} . Example 3.2.8 Find the standard basis for 𝑃 . Solution: The standard basis for 𝑃 is {1, 𝑥, 𝑥 } . Example 3.2.9 Find the standard basis for 𝑃 . Solution: The standard basis for 𝑃 is {1, 𝑥, 𝑥 , 𝑥 } . Example 3.2.10 Find the standard basis for ℝ =𝑀 (ℝ) . Solution: The standard basis for ℝ = 𝑀 (ℝ) is {[1 00 0] , [0 10 0] , [0 01 0] , [0 00 1]} because ℝ =𝑀 (ℝ) = ℝ as a vector space where standard basis for ℝ = 𝑀 (ℝ) is the rows of 𝐼 = [1 00 1 0 00 00 00 0 1 00 1] that are represented by matrices. Example 3.2.11 Let 𝑇: ℝ → ℝ be a linear transformation such that 𝑇(2,0) = (0,1,4)
𝑇(−1,1) = (2,1,5)
Find
𝑇(3,5) . Solution: The given points are (2,0) and (−1,1) . These two points are independent because of the following: [ 2 0−1 1] 12 𝑅 + 𝑅 → 𝑅 [2 00 1] Every point in ℝ is a linear combination of (2,0) and (−1,1) . There exists unique numbers 𝑐 and 𝑐 such that (3,5) = 𝑐 (2,0) + 𝑐 (−1,1) . − 𝑐 Now, we substitute 𝑐 = 5 in − 𝑐 , we obtain: − 5 𝑐 = 4 Hence, (3,5) = 4(2,0) + 5(−1,1) . 𝑇(3,5) = 𝑇(4(2,0) + 5(−1,1))
𝑇(3,5) = 4𝑇(2,0) + 5𝑇(−1,1)
𝑇(3,5) = 4(0,1,4) + 5(2,1,5) = (10,9,41)
Thus,
𝑇(3,5) = (10,9,41) . Example 3.2.12 Let
𝑇: ℝ → ℝ be a linear transformation such that
𝑇(1) = 3 . Find
𝑇(5) . Solution: Since it is a linear transformation, then
𝑇(5) = 𝑇(5 ∙ 1) = 5𝑇(1) = 5(3) = 15 . If it is not a linear transformation, then it is impossible to find
𝑇(5) .
100 M. Kaabar
In this section, we discuss how to find the standard matrix representation, and we give examples of how to find kernel and range. Definition 3.3.1 Given
𝑇: ℝ 𝑛 → ℝ 𝑚 where ℝ 𝑛 is a domain and ℝ 𝑚 is a co-domain. Then, Standard Matrix Representation is a 𝑚 × 𝑛 matrix. This means that it is 𝑑𝑖𝑚(𝐶𝑜 − 𝐷𝑜𝑚𝑎𝑖𝑛) × 𝑑𝑖𝑚 (𝐷𝑜𝑚𝑎𝑖𝑛) matrix. Definition 3.3.2 Given 𝑇: ℝ 𝑛 → ℝ 𝑚 where ℝ 𝑛 is a domain and ℝ 𝑚 is a co-domain. Kernel is a set of all points in the domain that have image which equals to the origin point, and it is denoted by 𝐾𝑒𝑟(𝑇) . This means that
𝐾𝑒𝑟(𝑇) = 𝑁𝑢𝑙𝑙 𝑆𝑝𝑎𝑐𝑒 𝑜𝑓 𝑇 . Definition 3.3.3 Range is the column space of standard matrix representation, and it is denoted by
𝑅𝑎𝑛𝑔𝑒(𝑇) . Example 3.3.1 Given
𝑇: ℝ → ℝ where ℝ is a domain and ℝ is a co-domain. 𝑇((𝑥 , 𝑥 , 𝑥 )) = (−5𝑥 , 2𝑥 + 𝑥 , −𝑥 , 0) a. Find the Standard Matrix Representation. b.
Find
𝑇((3,2,1)) . c.
Find
𝐾𝑒𝑟(𝑇) . d.
Find
𝑅𝑎𝑛𝑔𝑒(𝑇) . Solution: Part a: According to definition 3.3.1, the Standard Matrix Representation, let’s call it 𝑀 , here is . We know from section 3.2 that the standard basis for domain (here is ℝ ) is {(1,0,0), (0,1,0), (0,0,1)} . We assume the following: 𝑣 = (1,0,0) 𝑣 = (0,1,0) 𝑣 = (0,0,1) Now, we substitute each point of the standard basis for domain in
𝑇((𝑥 , 𝑥 , 𝑥 )) = (−5𝑥 , 2𝑥 + 𝑥 , −𝑥 , 0) as follows: 𝑇((1,0,0)) = (−5,0, −1,0)
𝑇((0,1,0)) = (0,2,0,0)
𝑇((0,0,1)) = (0,1,0,0)
Our goal is to find 𝑀 so that 𝑇((𝑥 , 𝑥 , 𝑥 )) = 𝑀 [𝑥 𝑥 𝑥 ] . 𝑀 = [−50 02 01−1 0 00 0 0]
This is the Standard Matrix Representation. The first, second and third columns represent
𝑇(𝑣 ), 𝑇(𝑣 ) and 𝑇(𝑣 ) . Part b: Since ((𝑥 , 𝑥 , 𝑥 )) = 𝑀 [𝑥 𝑥 𝑥 ] , then 𝑇((3,2,1)) = [−50 02 01−1 0 00 0 0] [321]
102 M. Kaabar
𝑇((3,2,1)) = 3 ∙ [−50−10 ] + 2 ∙ [0200] + 1 ∙ [0100] = [−155−30 ] [−155−30 ] is equivalent to (−15,5, −3,0) . This lives in the co-domain. Thus,
𝑇((3,2,1)) = (−15,5, −3,0) . Part c: According to definition 3.3.2,
𝐾𝑒𝑟(𝑇) is a set of all points in the domain that have image = (0,0,0,0) . Hence,
𝑇((𝑥 , 𝑥 , 𝑥 )) = (0,0,0,0) . This means the following: 𝑀 [𝑥 𝑥 𝑥 ] = [0000] [−50 02 01−1 0 00 0 0] [𝑥 𝑥 𝑥 ] = [0000] Since
𝐾𝑒𝑟(𝑇) = 𝑁𝑢𝑙𝑙(𝑀) , then we need to find
𝑁(𝑀) as follows: (−50 02 01−1 0 00 0 0|0000) − 15 𝑅 ( 10 02 01−1 0 00 0 0|0000) 𝑅 + 𝑅 → 𝑅 (10 02 010 0 00 0 0|0000) 𝑅 (10 01 00.50 0 00 0 0 |0000) This is a Completely-Reduced Matrix. Now, we need to read the above matrix as follows: 𝑥 = 0 𝑥 + 12 𝑥 = 0 To write the solution, we need to assume that 𝑥 ∈ ℝ (𝐹𝑟𝑒𝑒 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒) . Hence, 𝑥 = 0 and 𝑥 = − 𝑥 . 𝑁(𝑀) = {(0, − 𝑥 , 𝑥 )|𝑥 ∈ ℝ} . By letting 𝑥 = 1 , we obtain: 𝑁𝑢𝑙𝑙𝑖𝑡𝑦(𝑀) = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐹𝑟𝑒𝑒 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 = 1 , and
𝐵𝑎𝑠𝑖𝑠 = {(0, − , 1)} Thus,
𝐾𝑒𝑟(𝑇) = 𝑁(𝑀) = 𝑆𝑝𝑎𝑛{(0, − , 1)} . Part d: According to definition 3.3.3, 𝑅𝑎𝑛𝑔𝑒(𝑇) is the column space of 𝑀 . Now, we need to locate the columns in the Completely-Reduced Matrix in part c that contain the leaders, and then we should locate them to the original matrix as follows: (10 01 00.50 0 00 0 0 |0000) Completely-Reduced Matrix (−50 02 01−1 0 00 0 0|0000)
Orignial Matrix 103
104 M. Kaabar Thus,
𝑅𝑎𝑛𝑔𝑒(𝑇) = 𝑆𝑝𝑎𝑛{(−5,0, −1,0), (0,2,0,0)} . Result 3.3.1 Given
𝑇: ℝ 𝑛 → ℝ 𝑚 where ℝ 𝑛 is a domain and ℝ 𝑚 is a co-domain. Let 𝑀 be a standard matrix representation. Then, 𝑅𝑎𝑛𝑔𝑒(𝑇) = 𝑆𝑝𝑎𝑛{𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝐶𝑜𝑙𝑢𝑚𝑛𝑠𝑜𝑓 𝑀} . Result 3.3.2 Given
𝑇: ℝ 𝑛 → ℝ 𝑚 where ℝ 𝑛 is a domain and ℝ 𝑚 is a co-domain. Let 𝑀 be a standard matrix representation. Then, 𝑑𝑖𝑚(𝑅𝑎𝑛𝑔𝑒(𝑇)) = 𝑅𝑎𝑛𝑘(𝑀) = Number of Independent Columns. Result 3.3.3 Given 𝑇: ℝ 𝑛 → ℝ 𝑚 where ℝ 𝑛 is a domain and ℝ 𝑚 is a co-domain. Let 𝑀 be a standard matrix representation. Then, 𝑑𝑖𝑚(𝐾𝑒𝑟(𝑇)) = 𝑁𝑢𝑙𝑙𝑖𝑡𝑦(𝑀) . Result 3.3.4 Given 𝑇: ℝ 𝑛 → ℝ 𝑚 where ℝ 𝑛 is a domain and ℝ 𝑚 is a co-domain. Let 𝑀 be a standard matrix representation. Then, 𝑑𝑖𝑚(𝑅𝑎𝑛𝑔𝑒(𝑇)) + 𝑑𝑖𝑚 (𝐾𝑒𝑟(𝑇)) =𝑑𝑖𝑚 (𝐷𝑜𝑚𝑎𝑖𝑛) . Example 3.3.2 Given 𝑇: 𝑃 → ℝ . 𝑇(𝑓(𝑥)) = ∫ 𝑓(𝑥)𝑑𝑥 is a linear transformation. a. Find
𝑇(2𝑥 − 1) . b.
Find
𝐾𝑒𝑟(𝑇) . c.
Find
𝑅𝑎𝑛𝑔𝑒(𝑇) . Solution: Part a:
𝑇(2𝑥 − 1) = ∫ (2𝑥 − 1)𝑑𝑥 = 𝑥 − 𝑥 |𝑥 = 1𝑥 = 0 = 0. Part b: To find
𝐾𝑒𝑟(𝑇) , we set equation of
𝑇 = 0 , and 𝑓(𝑥) = 𝑎 + 𝑎 𝑥 ∈ 𝑃 . Thus, 𝑇(𝑓(𝑥)) = ∫ (𝑎 + 𝑎 𝑥)𝑑𝑥 = 𝑎 𝑥 + 𝑎 𝑥 |𝑥 = 1𝑥 = 0 = 0 𝑎 + 𝑎 𝑎 = − 𝑎 Hence,
𝐾𝑒𝑟(𝑇) = {− 𝑎 + 𝑎 𝑥|𝑎 ∈ ℝ} . We also know that 𝑑𝑖𝑚 (𝐾𝑒𝑟(𝑇) = 1 because there is one free variable. In addition, we can also find basis by letting 𝑎 be any real number not equal to zero, say 𝑎 = 1 , as follows: 𝐵𝑎𝑠𝑖𝑠 = {− 12 + 𝑥}
Thus,
𝐾𝑒𝑟(𝑇) = 𝑆𝑝𝑎𝑛{− + 𝑥} . Part c: It is very easy to find range here. 𝑅𝑎𝑛𝑔𝑒(𝑇) = ℝ because we linearly transform from a second degree polynomial to a real number. For example, if we linearly transform from a third degree polynomial to a second degree polynomial, then the range will be 𝑃 .
1. Given
𝑇: ℝ → ℝ such that 𝑇((𝑥 , 𝑥 , 𝑥 )) = [𝑥 𝑥 𝑥 𝑥 ] is a linear transformation. a. Find the standard matrix representation of 𝑇 . b. Find
𝐾𝑒𝑟(𝑇) . c.
Find a basis for
𝑅𝑎𝑛𝑔𝑒(𝑇) and write
𝑅𝑎𝑛𝑔𝑒(𝑇) as a
𝑆𝑝𝑎𝑛 . 105
106 M. Kaabar 2. Let
𝑇: ℝ → ℝ be a linear transformation such that 𝑇(1,0) = 4 , 𝑇(2, −2) = 2 . Find the standard matrix representation of 𝑇 . 3. Given 𝑇: 𝑃 → ℝ such that 𝑇(𝑎 + 𝑏𝑥 + 𝑐𝑥 ) =∫ (𝑎 + 𝑏𝑥 + 𝑐𝑥 )𝑑𝑥 . Find 𝐾𝑒𝑟(𝑇) . 4. Given
𝑇: ℝ → ℝ is a linear transformation such that 𝑇(𝑥, 𝑦, 𝑧, 𝑤) = (𝑥 + 𝑦 + 𝑧 − 2𝑤, −2𝑤, 𝑤) . a.
Find the standard matrix representation of 𝑇 . b. Find 𝑑𝑖𝑚 (𝐾𝑒𝑟(𝑇)) . c.
Find
𝑅𝑎𝑛𝑔𝑒(𝑇) . 5. Given
𝑇: 𝑃 → ℝ such that 𝑇(𝑔(𝑥)) = [𝑔(−1) 𝑔(0)𝑔(−1) 𝑔(0)] is a linear transformation. a.
Find the standard matrix representation of 𝑇 . b. Find
𝐾𝑒𝑟(𝑇) and write
𝐾𝑒𝑟(𝑇) as a
𝑆𝑝𝑎𝑛 . c.
Find a basis for
𝑅𝑎𝑛𝑔𝑒(𝑇) and write
𝑅𝑎𝑛𝑔𝑒(𝑇) as a
𝑆𝑝𝑎𝑛 . 6. Given
𝑇: 𝑃 → ℝ is a linear transformation such that 𝑇(1) = 6, 𝑇(𝑥 + 𝑥) = −5 , and 𝑇(𝑥 + 2𝑥 + 1) = 4 . a. Find
𝑇(𝑥), 𝑇(𝑥 )and 𝑇(5𝑥 + 3𝑥 + 8) . b. Find the standard matrix representation of 𝑇 . c. Find
𝐾𝑒𝑟(𝑇) and write
𝐾𝑒𝑟(𝑇) as a
𝑆𝑝𝑎𝑛 . hapter 4 Characteristic Equation of Matrix In this chapter, we discuss how to find eigenvalues and eigenvectors of a matrix. Then, we introduce the diagonalizable matrix, and we explain how to determine whether a matrix is diagonalizable or not. At the end of this chapter we discuss how to find diagonal matrix and invertible matrix.
In this section, we give an example explaining the steps for finding eigenvalues and eigenvectors. Example 4.1.1 Given 𝑛 × 𝑛 matrix, say 𝐴 , Let 𝛼 ∈ ℝ . Can we find a point 𝑄 in ℝ 𝑛 , 𝑄 = (𝑎 , 𝑎 , … , 𝑎 𝑛 ) , such that 𝑄 ≠ (0,0,0,0, … ,0) , and 𝐴 𝑛×𝑛 𝑄 = 𝐴 [ 𝑎 𝑎 𝑎 ⋮𝑎 𝑛 ] = 𝛼 [ 𝑎 𝑎 𝑎 ⋮𝑎 𝑛 ] ? 107
108 M. Kaabar Solution: If such 𝛼 and such 𝑄 exist, we say 𝛼 is an eigenvalue of 𝐴 , and we say 𝑄 is an eigenvector of 𝐴 corresponds to the eigenvalue 𝛼 . 𝐴 [ 𝑎 𝑎 𝑎 ⋮𝑎 𝑛 ] = 𝛼 [ 𝑎 𝑎 𝑎 ⋮𝑎 𝑛 ] 𝐴 [ 𝑎 𝑎 𝑎 ⋮𝑎 𝑛 ] − 𝛼 [ 𝑎 𝑎 𝑎 ⋮𝑎 𝑛 ] = [ 000⋮0] [𝐴 − 𝛼𝐼 𝑛 ] [ 𝑎 𝑎 𝑎 ⋮𝑎 𝑛 ] = [ 000⋮0] We conclude that such 𝛼 and (𝑄 ≠ origin ) exist if and only if 𝑑𝑒𝑡(𝐴 − 𝛼𝐼 𝑛 ) = 0 . Note: 𝑑𝑒𝑡(𝐴 − 𝛼𝐼 𝑛 ) is called characteristic polynomial of 𝐴 . Example 4.1.2 Given the following matrix: 𝐴 = [2 0 10 1 −20 0 −1] a. Find all eigenvalues of 𝐴 . b. For each eigenvalue, find the corresponding eigenspace. Solution: Part a: To find all eigenvalues, we set characteristic polynomial of
𝐴 = 0 . This means that 𝑑𝑒𝑡(𝐴 − 𝛼𝐼 𝑛 ) = 0 . Then, we do the following: 𝑑𝑒𝑡(𝐴 − 𝛼𝐼 𝑛 ) = 0 𝑑𝑒𝑡 ([2 0 10 1 −20 0 −1] − 𝛼 [1 0 00 1 00 0 1]) = 0 𝑑𝑒𝑡 ([2 0 10 1 −20 0 −1] − [𝛼 0 00 𝛼 00 0 𝛼]) = 0 𝑑𝑒𝑡 ([2 − 𝛼 0 10 1 − 𝛼 −20 0 −1 − 𝛼]) = 0 𝑑𝑒𝑡 ([2 − 𝛼 0 10 1 − 𝛼 −20 0 −1 − 𝛼]) = (2 − 𝛼) ∙ (1 − 𝛼) ∙ (−1 − 𝛼) = 0 Thus, the eigenvalues of 𝐴 are: 𝛼 = 2𝛼 = 1𝛼 = −1 Part b: Since the first eigenvalue is 2, then the eigenspace of 𝐴 that corresponds to the eigenvalue 2 equals to the set of all points in ℝ such that (𝑝𝑜𝑖𝑛𝑡 𝐴) = 𝛼 ∙ 𝑝𝑜𝑖𝑛𝑡 = set of all eigenvectors of 𝐴 that correspond to 𝛼 = (2 + 𝑜𝑟𝑖𝑔𝑖𝑛 𝑝𝑜𝑖𝑛𝑡) . To find eigenspace that corresponds to 𝛼 = 2 , say 𝐸 , we need to find 𝐸 =𝑁(𝐴 − 2𝐼 ) . Note: 𝑁 here represents 𝑁𝑢𝑙𝑙 . Now, we do the following: 𝐸 = 𝑁(𝐴 − 2𝐼 ) = 𝑁 ([2 0 10 1 −20 0 −1] − 2 [1 0 00 1 00 0 1]) 𝐸 = 𝑁(𝐴 − 2𝐼 ) = 𝑁 ([2 − 2 0 10 1 − 2 −20 0 −1 − 2]) 𝐸 = 𝑁(𝐴 − 2𝐼 ) = 𝑁 [0 0 10 −1 −20 0 −3]
110 M. Kaabar Thus, we form the augmented-matrix for the homogeneous system as follows: (0 0 10 −1 −20 0 −3|000)
It is very easy to solve it: 𝑎 = 0𝑎 = −2𝑎 = 0𝑎 = 0 𝑎 ∈ ℝ (Free-Variable), and 𝑎 = 𝑎 = 0 . 𝐸 = {(𝑎 , 0,0)|𝑎 ∈ ℝ} . Now, we can select any value for 𝑎 ≠ 0 . Let’s select 𝑎 = 1 . We can also find 𝑑𝑖𝑚(𝐸 ) = 1 . Hence, 𝐸 = 𝑆𝑝𝑎𝑛{(1,0,0)} . Similarly, we can find the eigenspaces that correspond to the other eigenvalues: 𝛼 = 1 and 𝛼 = −1 . To find eigenspace that corresponds to 𝛼 = 1 , say 𝐸 , we need to find 𝐸 = 𝑁(𝐴 − 1𝐼 ) . Note: 𝑁 here represents 𝑁𝑢𝑙𝑙 . Now, we do the following: 𝐸 = 𝑁(𝐴 − 1𝐼 ) = 𝑁 ([2 0 10 1 −20 0 −1] − 1 [1 0 00 1 00 0 1]) 𝐸 = 𝑁(𝐴 − 1𝐼 ) = 𝑁 ([2 − 1 0 10 1 − 1 −20 0 −1 − 1]) 𝐸 = 𝑁(𝐴 − 1𝐼 ) = 𝑁 [1 0 10 0 −20 0 −2] Thus, as we did previously, we form the augmented-matrix for the homogeneous system as follows: (1 0 10 0 −20 0 −2|000)
It is very easy to solve it: 𝑎 = −𝑎 𝑎 = 0 𝑎 ∈ ℝ (Free-Variable), and 𝑎 = 𝑎 = 0 . 𝐸 = {(0, 𝑎 , 0)|𝑎 ∈ ℝ} . Now, we can select any value for 𝑎 ≠ 0 . Let’s select 𝑎 = 1 . We can also find 𝑑𝑖𝑚(𝐸 ) = 1 . Hence, 𝐸 = 𝑆𝑝𝑎𝑛{(0,1,0)} . Finally, to find eigenspace that corresponds to 𝛼 = −1 , say 𝐸 −1 , we need to find 𝐸 −1 = 𝑁(𝐴 + 1𝐼 ) . Note: 𝑁 here represents 𝑁𝑢𝑙𝑙 . Now, we do the following: 𝐸 −1 = 𝑁(𝐴 + 1𝐼 ) = 𝑁 ([2 0 10 1 −20 0 −1] + 1 [1 0 00 1 00 0 1]) 𝐸 −1 = 𝑁(𝐴 + 1𝐼 ) = 𝑁 ([2 + 1 0 10 1 + 1 −20 0 −1 + 1]) 𝐸 −1 = 𝑁(𝐴 + 1𝐼 ) = 𝑁 [3 0 10 2 −20 0 0 ] Thus, as we did previously, we form the augmented-matrix for the homogeneous system as follows: (3 0 10 2 −20 0 0 |000)
It is very easy to solve it: 𝑎 = − 𝑎 𝑎 = 𝑎 𝑎 ∈ ℝ (Free-Variable). 𝐸 −1 = {(− 𝑎 , 𝑎 , 𝑎 )|𝑎 ∈ ℝ} . Now, we can select any value for 𝑎 ≠ 0 . Let’s select 𝑎 = 1 . We can also find 𝑑𝑖𝑚(𝐸 −1 ) = 1 . Hence, 𝐸 −1 = 𝑆𝑝𝑎𝑛{(− , 1,1)} . 111
112 M. Kaabar
In this section, we explain the concept of diagonalization, and we give some examples explaining it, and how to find the diagonal and invertible matrices. Definition 4.2.1 𝐴 is diagonalizable if there exists an invertible matrix 𝐿 , and a diagonal matrix 𝐷 such that 𝐴 = 𝐿𝐷𝐿 −1 . Result 4.2.1 𝐴 is 𝑛 × 𝑛 diagonalizable matrix if and only if 𝑑𝑒𝑡(𝐴 − 𝛼𝐼 𝑛 ) is written as multiplication of linear equation, say 𝑑𝑒𝑡(𝐴 − 𝛼𝐼 𝑛 ) = some constants (𝑐 − 𝛼) ∙(𝑐 − 𝛼) ∙ … ∙ (𝑐 𝑘 − 𝛼) , and 𝑑𝑖𝑚(𝐸 𝑐 𝑖 ) = 𝑛 𝑖 (Multiplicity of the Eigenvalues) for 𝑖 ≤ 𝑖 ≤ 𝑘 . Example 4.2.1 Assume 𝐴 is matrix, and 𝑑𝑒𝑡(𝐴 − 𝛼𝐼 ) = (3 − 𝛼) ∙ (−2 − 𝛼) ∙ (4 − 𝛼) , and the eigenvalues of 𝐴 are 𝛼 = 3𝛼 = −2𝛼 = 4 Given 𝑑𝑖𝑚(𝐸 ) = 1, 𝑑𝑖𝑚(𝐸 −2 ) = 1 and 𝑑𝑖𝑚(𝐸 ) = 2 . Is 𝐴 diagonalizable matrix? Solution: It is not a diagonalizable matrix because 𝑑𝑖𝑚(𝐸 ) = 1 , and it must be equal to 2 instead of 1. Example 4.2.2 Assume 𝐴 is matrix, and 𝑑𝑒𝑡(𝐴 − 𝛼𝐼 ) = (2 − 𝛼) ∙ (3 − 𝛼) , and given 𝑑𝑖𝑚(𝐸 ) = 3 and 𝑑𝑖𝑚(𝐸 ) = 1 . Is 𝐴 diagonalizable matrix? Solution: According result 4.2.1, it is a diagonalizable matrix. Example 4.2.3 Given the following matrix: 𝐴 = [2 0 10 1 −20 0 −1] . Use example 4.1.2 from section 4.1 to answer the following questions: a. Is 𝐴 diagonalizable matrix? If yes, find a diagonal matrix 𝐷 , and invertible matrix 𝐿 such that 𝐴 = 𝐿𝐷𝐿 −1 . b. Find 𝐴 such that 𝐴 = 𝐿𝐷𝐿 −1 . c. Find 𝐴 such that 𝐴 = 𝐿𝐷𝐿 −1 . Solution: Part a: From example 4.1.2, we found the following: 𝐸 = 𝑆𝑝𝑎𝑛{(1,0,0)}𝐸 = 𝑆𝑝𝑎𝑛{(0,1,0)}𝐸 −1 = 𝑆𝑝𝑎𝑛{(− 13 , 1,1)} According to result 4.2.1, 𝐴 is a diagonalizable matrix. Now, we need to find a diagonal matrix 𝐷 , and invertible matrix 𝐿 such that 𝐴 = 𝐿𝐷𝐿 −1 . To find a diagonal matrix 𝐷 , we create a matrix, and we put eigenvalues on the main diagonal with 113
114 M. Kaabar repetition if there is a repetition, and all other elements are zeros. Hence, the diagonal matrix 𝐷 is as follows: 𝐷 = [2 0 00 −1 00 0 1]
To find an invertible matrix 𝐿 , we create a matrix like the one above but each eigenvalue above is represented by a column of the eigenspace that corresponds to that eigenvalue as follows: 𝐿 = [1 − 13 00 1 10 1 0]
Thus,
𝐴 = [2 0 10 1 −20 0 −1] = 𝐿𝐷𝐿 −1 = [1 −
00 1 10 1 0] [2 0 00 −1 00 0 1] [1 −
00 1 10 1 0] −1 Part b: To find 𝐴 such that 𝐴 = 𝐿𝐷𝐿 −1 , we do the following steps: 𝐴 = 𝐿𝐷𝐿 −1 𝐴 = (𝐿𝐷𝐿 −1 ) ∙ (𝐿𝐷𝐿 −1 ) ∙ … ∙ (𝐿𝐷𝐿 −1 ) 𝐴 = (𝐿𝐷 𝐿 −1 ) ∙ … ∙ (𝐿𝐷𝐿 −1 ) 𝐴 = 𝐿𝐷 𝐿 −1 Thus, 𝐴 = 𝐿𝐷 𝐿 −1 = 𝐿 [2 0 00 −1 00 0 1] 𝐿 −1 =𝐿 [2
00 0 1 ] 𝐿 −1 = 𝐿 [64 0 00 1 00 0 1] 𝐿 −1 Part c: To find 𝐴 such that 𝐴 = 𝐿𝐷𝐿 −1 , we do the following steps: 𝐴 = 𝐿𝐷𝐿 −1 𝐴 = (𝐿𝐷𝐿 −1 ) ∙ (𝐿𝐷𝐿 −1 ) ∙ … ∙ (𝐿𝐷𝐿 −1 ) 𝐴 = (𝐿𝐷 𝐿 −1 ) ∙ … ∙ (𝐿𝐷𝐿 −1 ) 𝐴 = 𝐿𝐷 𝐿 −1 Thus, 𝐴 = 𝐿𝐷 𝐿 −1 = 𝐿 [2 0 00 −1 00 0 1] 𝐿 −1 =𝐿 [2
00 0 1 ] 𝐿 −1 = 𝐿 [2 −1
1. Given the following matrix:
𝐶 = [1 0 0 00 1 1 100 00 −1 1 0 −1 ] . Is 𝐶 diagonalizable matrix? 2. Assume that 𝐴 is diagonalizable matrix, and given the following: 𝐸 = 𝑆𝑝𝑎𝑛{(2,1,0,0,1), (0,1,0,1,1), (0,0,2,2,0)}
115 116 M. Kaabar 𝑑𝑖𝑚(𝐸 ) = 3 𝐸 = 𝑆𝑝𝑎𝑛{(0,0,0,1,1), (0,0,0,0,10)} a. Find the characteristic polynomial of 𝐴 . b. Find a diagonal matrix 𝐷 such that 𝐴 = 𝐿𝐷𝐿 −1 . c. Find an invertible matrix 𝐿 such that 𝐴 = 𝐿𝐷𝐿 −1 . 3. Assume that 𝑊 is matrix, and 𝑊 − 𝛼𝐼 = (1 − 𝛼)(2 − 𝛼) . Given 𝑁(𝑊 − 𝐼) =𝑆𝑝𝑎𝑛{(1,2,0)} , and
𝑁(𝑊 − 2𝐼) = 𝑆𝑝𝑎𝑛{(2,0,3)} . Is 𝑊 diagonalizable matrix? Explain. Chapter 5 Matrix Dot Product
In this chapter, we discuss the dot product only in ℝ 𝑛 . In addition, we give some results about the dot product in ℝ 𝑛 . At the end of this chapter, we get introduced to a concept called “Gram-Schmidt Orthonormalization”. ℝ 𝑛 In this section, we first give an example of the dot product in ℝ 𝑛 , and then we give three important results related to this concept. Example 5.1.1 Assume that 𝐴 = (2,4,1,3) and
𝐵 = (0,1,2,5) where
𝐴, 𝐵 ∈ ℝ . Find 𝐴 ∙ 𝐵 . olution: To find
𝐴 ∙ 𝐵 , we need to do a simple vector dot product as follows:
𝐴 ∙ 𝐵 = (2,4,1,3) ∙ (0,1,2,5)
𝐴 ∙ 𝐵 = 2 ∙ 0 + 4 ∙ 1 + 1 ∙ 2 + 3 ∙ 5
𝐴 ∙ 𝐵 = 0 + 4 + 2 + 15
Thus,
𝐴 ∙ 𝐵 = 21
Result 5.1.1 If 𝑊 and 𝑊 in ℝ 𝑛 and 𝑊 ≠ (0,0,0, … ,0) and 𝑊 ≠ (0,0,0, … ,0) , and 𝑊 ∙ 𝑊 = 0 , then 𝑊 and 𝑊 are independent. Result 5.1.2 If 𝑊 and 𝑊 in ℝ 𝑛 are independent, then may/maybe not 𝑊 ∙ 𝑊 = 0 . (i.e. Assume that 𝑊 = (1,1,1,1) and 𝑊 = (0,1,1,1) , then 𝑊 ∙ 𝑊 = 3 ) Result 5.1.3 If 𝑊 , 𝑊 , 𝑊 and 𝑊 in ℝ 𝑛 and none of them is (0,0,0, … ,0) , then we say that 𝑊 , 𝑊 , 𝑊 and 𝑊 are orthogonal if they satisfy the following conditions: 𝑊 ∙ 𝑊 = 0 𝑊 ∙ 𝑊 = 0 𝑊 ∙ 𝑊 = 0 𝑊 ∙ 𝑊 = 0 𝑊 ∙ 𝑊 = 0 𝑊 ∙ 𝑊 = 0
117 118 M. Kaabar
Result 5.1.4 Assume that
𝑊 = (𝑥 , 𝑥 , 𝑥 , … , 𝑥 𝑛 ) , then the squared-norm of 𝑊 is written as follows: ||𝑊|| = 𝑥 + 𝑥 + 𝑥 + ⋯ + 𝑥 𝑛2 = 𝑊 ∙ 𝑊 . In this section, we give one example that explains the concept of Gram-Schmidt Orthonormalization, and how it is related to what we have learned in chapters 2 and 3. Example 5.2.1 Given the following:
𝐴 =𝑆𝑝𝑎𝑛{(1,0,1,1), (0,1,0,1), (0,1,1,1)} . Find the orthogonal basis for 𝐴 . Hint: The orthogonal basis means Gram-Schmidt Orthonormalization. Solution: To find the orthogonal basis for 𝐴 , we need to do the following steps: Step 1: Find a basis for 𝐴 . [1 0 1 10 1 0 10 1 1 1] −𝑅 + 𝑅 → 𝑅 [1 0 1 10 1 0 10 0 1 0] This is the Semi-Reduced Matrix. Since we do now have a zero-row, then 𝑑𝑖𝑚(𝐴) = 3 . To write a basis for 𝐴 , it is recommended to choose rows in the above Semi-Reduced Matrix. Thus, a basis for 𝐴 is {(1,0,1,1), (0,1,0,1), (0,0,1,0)} . Step 2: Vector spaces assumptions. Since the basis for 𝐴 is {(1,0,1,1), (0,1,0,1), (0,0,1,0)} , we assume the following: 𝑣 = (1,0,1,1) 𝑣 = (0,1,0,1) 𝑣 = (0,0,1,0) Step 3: Use Gram-Schmidt Orthonormalization method. Let’s assume that the orthogonal basis for 𝐴 is 𝐵 Orthogonal = {𝑊 , 𝑊 , 𝑊 } . Now, we need to find 𝑊 , 𝑊 and 𝑊 . To find them, we need to do the following: 𝑊 = 𝑣 = (1,0,1,1) 𝑊 = 𝑣 − 𝛼 ∙ (𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑊 𝑖 ′ 𝑠 ) where 𝛼 = ( 𝑣 ∙𝑊 ||𝑊 || ) Thus, 𝑊 = 𝑣 − ( 𝑣 ∙𝑊 ||𝑊 || ) ∙ 𝑊 𝑊 = 𝑣 − (𝑣 ∙ 𝑊 ||𝑊 || ) ∙ 𝑊 = (0,1,0,1) − ((0,1,0,1) ∙ (1,0,1,1)||(1,0,1,1)|| ) ∙ (1,0,1,1)= (0,1,0,1) − 13 ∙ (1,0,1,1)= (0,1,0,1) − (13 , 0, 13 , 13) = (− 13 , 1, − 13 , 23)
120 M. Kaabar 𝑊 = 𝑣 − 𝛼 ∙ 𝑊 − 𝛼 ∙ 𝑊 where 𝛼 = ( 𝑣 ∙𝑊 ||𝑊 || ) and 𝛼 =( 𝑣 ∙𝑊 ||𝑊 || ) . Thus, 𝑊 = 𝑣 − ( 𝑣 ∙𝑊 ||𝑊 || ) ∙ 𝑊 − ( 𝑣 ∙𝑊 ||𝑊 || ) ∙ 𝑊 𝑊 = (0,0,1,0) − ( (0,0,1,0) ∙ (− 13 , 1, − 13 , 23)||(− 13 , 1, − 13 , 23)|| ) ∙ (− 13 , 1, − 13 , 23)− ((0,0,1,0) ∙ (1,0,1,1)||(1,0,1,1)|| ) ∙ (1,0,1,1) Thus, 𝑊 = (− , − , , − ) . Hence, the orthogonal basis for 𝐴 (Gram-Schmidt Orthonormalization) is {(1,0,1,1), (− , 1, − , ) , (− , − , , − )} .
1. Let
𝐷 = 𝑆𝑝𝑎𝑛{(0,0,1,1), (1,0,1,1), (1, −1,1,0)} . Find the orthogonal basis for 𝐷 . 2. Let 𝐸 = 𝑆𝑝𝑎𝑛{(1,1, −1,0), (0,1,1,1), (3,5, −1,2)} . Find the orthogonal basis for 𝐸 . nswers to Odd-Numbered Exercises 1.10 Exercises 𝑥 , 𝑥 , 𝑥 ∈ ℝ 𝑥 = 𝑥 + 𝑥 − 𝑥 + 15 𝑥 = − 𝑥 + 𝑥 + 𝑥 − 𝑥 = 𝑥 − 𝑥 + 5
3. a. [−3 3 −1 5] b. [740] c. −2
5. a. [1 02 1] [2 00 1] 𝐴 = 𝐴 b. [
00 1] [ 1 0−2 1] 𝐴 = 𝐴 𝐴 is invertible (non-singular), 𝐴 −1 = [1 1 𝑐 det(A) = (−1) det[ 2 2 1−2 2 −11 12 4 ]−1144 = − = − 𝑥 = 0𝑥 = −3𝑥 = 1
122 M. Kaabar
1. a. Basis for 𝑀 is {(1, −1,1), (−1,0,1)} . b. (1, −3,5) ∈ 𝑀 because there is a zero row that corresponds to (1, −3,5) which means that (1, −3,5) is dependent, and (1, −3,5) can be written as a linear combination. 3. a. Let 𝑣 = [ 𝑥𝑥 + 𝑦𝑦 ] ∈ 𝐸, and 𝑣 = [ 𝑎𝑎 + 𝑧𝑧 ] , then 𝑣 + 𝑣 = [ 𝑥 + 𝑎(𝑥 + 𝑎) + (𝑦 + 𝑧)𝑦 + 𝑧 ] ∈ 𝐸 . For 𝑚 ∈ ℝ and 𝑣 = [ 𝑥𝑥 + 𝑦𝑦 ] ∈ 𝐸 , then 𝛼𝑣 =[ 𝛼𝑥𝛼(𝑥 + 𝑦)𝛼𝑦 ] = [ 𝛼𝑥𝛼𝑥 + 𝛼𝑦𝛼𝑦 ] ∈ 𝐸 . Thus, 𝐸 is a subspace of ℝ . b. Basis for 𝐸 is {(1,1,0), (0,1,1)} . c. 𝐸 = 𝑆𝑝𝑎𝑛{(1,1,0), (0,1,1)} . 5. 𝑥 = 3 𝑑𝑖𝑚(𝐾) = 2
9. No, since the determinant of [−2 2 21 −1 32 −1 0] equals zero, then the elements {(−2,1,2), (2,1, −1), (2,3,0)} do not ℝ . .4 Exercises
1. a. Standard Matrix Representation of 𝑇 is [1 0 01 0 000 01 10] b. Basis for 𝐾𝑒𝑟(𝑇) is {(1,1,0,0), (0,0,0,1), (0,0,1,0)} . Thus, 𝐾𝑒𝑟(𝑇) is 𝑆𝑝𝑎𝑛{(1,1,0,0), (0,0,0,1), (0,0,1,0)} . c. Basis for
𝑅𝑎𝑛𝑔𝑒(𝑇) = {[1 10 0] , [0 00 1] , [0 01 0]} . Thus,
𝐾𝑒𝑟(𝑇) is 𝑆𝑝𝑎𝑛 {[1 10 0] , [0 00 1] , [0 01 0]} . 3.
𝐾𝑒𝑟(𝑇) = 𝑆𝑝𝑎𝑛{−0.5 + 𝑥, − + 𝑥 } . 5. a. Standard Matrix Representation of 𝑇 is [1 −1 11 0 011 −10 10 −10−10 ] b. Basis for 𝐾𝑒𝑟(𝑇) is {𝑥 + 𝑥 , −𝑥 + 𝑥 } . Thus, 𝐾𝑒𝑟(𝑇) is 𝑆𝑝𝑎𝑛{𝑥 + 𝑥 , −𝑥 + 𝑥 } . c. Basis for 𝑅𝑎𝑛𝑔𝑒(𝑇) = {[1 11 1] , [−1 0−1 0]} . Thus,
𝐾𝑒𝑟(𝑇) is 𝑆𝑝𝑎𝑛 {[1 11 1] , [−1 0−1 0]} .
1. Since 𝑑𝑖𝑚(𝐸 −1 ) ≠ 2 , 𝐶 is not diagonalizable. 3. 𝑊 is not diagonalizable because 𝑑𝑖𝑚(𝐸 ) must be 2 not 1. 123 124 M. Kaabar
1. The orthogonal basis for 𝐴 (Gram-Schmidt Orthonormalization) is {(0,0,1,1), (1,0,0,0), (0, −1, , − )} . ibliographyibliography