OOctonionic Quadratic Equations
T.Kalpa Madhawa
Department of MathematicsSouthern Illinois University
January 1, 2021
Abstract
There are four division algebras over R , namely real numbers, complexnumbers, quaternions, and octonions. Lack of commutativity and associa-tivity make it difficult to investigate algebraic and geometric properties ofoctonions. It does not make sense to ask, for example, whether the equation x + 1 = 0 is solvable, without specifying the field in which we want thesolutions to be lie. The equation x + 1 = 0 has no solutions in R , whichis to say, there are no real numbers satisfying this equation. On the otherhand, there are complex numbers which do satisfy this equation in the field C of all complex numbers. How about if we extend the same idea to othertwo normed division algebras quaternions and octonions. Liping Huang andWasin So [6] derive explicit formulas for computing the roots of quaternionicquadratic equations. We extend their work to octonionic case and solvemonic left octonionic quadratic equation of the form x + bx + c = 0 , where a, b are octonions in general.[ We called this form of quadratic equation asleft octonion quadratic equation because we can consider x + xb + c = 0 as adifferent case due to non-commutativity of octonions]. Finally, we representthe left spectrum of × octonionic matrix as a set of solutions to a corre-sponding octonionic quadratic equation, which is an application of derivingexplicit formulas for computing the roots of octonionic quadratic equations. Keywords— octonionic quadratic equation, left spectrum.
Quaternions arose historically from Sir William Rowan Hamilton’s attempts in the mid-nineteenth century to generalize complex numbers in some way that would applicable tothree-dimensional space. [9] In complex numbers we have square root of -1 called i , whathappens if we include another, independent, square root of -1? Call it j . Then the bigquestion is, what is ij ? William Hamilton eventually proposed that k = ij should be yetanother square root of -1, and that the multiplication table should be cyclic, that is ij = k = − ji,jk = i = − kj,ki = j = − ik, a r X i v : . [ m a t h . G M ] D ec igure 1.1: The quaternionic multiplication table We refer to i , j , and k as imaginary quaternionic units. Notice that these units anticom-mute. This multiplication table is shown schematically in Figure . multiplying two ofthese quaternionic units together in the direction of the arrow yields the third; goingagainst the arrow contributes an additional minus sign. The quaternions are denoted by H ; the H is for "Hamilton", they are spanned by the identity element 1 and three imagi-nary units, that is , a quaternion q can be represented as four real numbers ( q , q , q , q ) ,usually written q = q + q i + q j + q k (1.1)which can be thought of as a point or vector in R . since (1 . can be written in the form q = ( q + q i ) + ( q + q i ) j (1.2)we see that a quaternion can be viewed as a pair of complex numbers, we can write H = C ⊕ C j in direct analogy to the construction of C from R . The quaternionic conjugate ¯ q of a quaternion q is obtained via the(real) linear map which reverses the sign of eachimaginary unit, so that ¯ q = q − q i − q j − q k (1.3)if q is given by (1 . . Conjugation leads directly to the norm of quaternion | q | , defined by | q | = q ¯ q = q + q + q + q . (1.4)again, the only quaternion with norm zero is zero, and every nonzero quaternion has aunique inverse, namely q − = ¯ q/ | q | (1.5)uaternionic conjugation satisfies the identity pq = ¯ q ¯ p from which it follows that the normsatisfies | pq | = | p || q | . Squaring both sides and expanding the result in terms of componentsyields the − sqaures rule, ( p q − p q − p q − p q ) + ( p q + p q − p q + p q ) + ( p q + p q + p q − p q ) + ( p q − p q + p q + p q ) = ( p + p + p + p )( q + q + q + q ) (1.6)which is not quite as obvious as the − squars rule. This identity implies that thequaternions form a normed division algebra, that is, not only are there inverses, but thereare no zero divisors-if a product is zero, one of the factors must be zero. It is importantto realize that ± i, ± j , and ± k are not the only quaternionic square roots of − , ny imaginary quaternion squares to a negative number, so it is only necessary to chooseits norm to be one in order to get a square root of − . The imaginary quaternions ofnorm one form a − dimensional sphere( q = 0 ); in the above notation, this is the set ofpoints q + q + q = 1 (1.7)any such unit imaginary quaternion u can be used to construct a complex subalgebra of H , which we will also denote by C , namely C = a + bu with a, b ∈ R . Furthermore, we can use the identity to write e uθ = cos θ + u sin θ This means that any quaternion can be written in the form q = re uθ where r = | q | and u denotes the direction of the imaginary part of q. .2 Algebra of octonions In analogy to the previous construction of C and H , an octonion x can be thought of asa pair of quaternions, so that O = H ⊕ H l we will denote i times l simply as il , and similarly with j and k . It is easy to see that l , il, jl, and kl all square to − ; there are now seven independent imaginary units, and wecould write x = x + x i + x j + x k + x l + x il + x jl + x kl where x m ∈ R f or i = 1 , , ..., . (1.8)which can be thought of as a point or vector in R . The real part of x is just x ; theimaginary part of x is everything else. Algebraically, we could define Re ( x ) = ( x + ¯ x ) / (1.9)Im ( x ) = ( x − ¯ x ) / (1.10)The imaginary part is differs slightly from the standard usage of these terms for complexnumbers, where Im ( z ) normally refers to a real numbers, the coefficient of i . This con-vention is not possible here, since the imaginary part has seven degrees of freedom, andcan be thought of as a vector in R . The full multiplication table is summarized in Figure . by means of the − point projective plane. Each point corresponds to an imaginaryunit. Figure 1.2: The octonionic multiplication table4 ach line corresponds to a quaternionic triple, much like { i, j, k } , with the arrow givingthe orientation. For Example, kl = kllkl = kklk = l (1.11)and each of these products anticommutes, that is, reversing the order contributes a minussign. We define the octonionic conjugate ¯ x of an octonion x as the (real) linear map whichreverses the sign of each imaginary unit. Thus, ¯ x = x − x i − x j − x k − x l − x il − x jl − x kl (1.12)if x is given by (1 . . Direct computation shows that xy = ¯ y ¯ x (1.13)The norm of an octonion | x | is defined by | x | = x ¯ x = x + x + x + x + x + x + x + x (1.14)Again, the only octonion with norm zero is zero, and every nonzero octonion has a uniqueinverse, namely x − = ¯ x/ | x | (1.15)as with the other division algebras, the norm satisfies the identity | xy | = | x || y | (1.16)writing out this expression in terms of components yields the − squares rule, which is nolonger at all obvious. The octonions therefore also form a normed division algebra.A remarkable property of the octonions is that they are not associative, For Example ( ij )( l ) = +( k )( l ) = + kl ( i )( jl ) = ( i )( jl ) = − kl (1.17) Octonionic quadratic formulas.
We will start with two lemmas.
Lemma 2.1.
Let
B, E, and D be real numbers such that1. D (cid:54) = 0 , and2. B < implies B < E Then the cubic equation y + 2 By + ( B − E ) y − D = 0 has exactly one positive solution.Proof. Let f ( y ) = y + 2 By + ( B − E ) y − D = 0 Note that f (0) = − D < and lim y → + ∞ f ( y ) = + ∞ . According to the intermediatevalue Theorem: if f is a continuous function over an interval [a,b], then f takes all valuesbetween f ( a ) and f ( b ) . Since above cubic polynomial is a continuous function, its graphmust intersect the x-axis at some finite points greater than zero. So the equation has atleast one positive root. Now let us prove that f has only one positive root.Suppose that f has three real roots, r , r and r . Take r > be the positive root wefound above. We must show that r , r < . Then we have the result.we know, r .r .r = D > this implies the product of r and r is positive. Therefore, r and r should be in samesign( both positive or negative). Let’s assuming r , r > ,if B < implies r + r + r = − B > But r .r + r .r + r .r = B − E and should be positive. This is a contradiction due to part of Lemma 2.1 . thus, wehave only one positive solution to f ( y ) = 0 . emma 2.2. Let
B, E, and D be real numbers such that1. E ≥ , and2. B < implies B < E Then the real system N − ( B + T ) N + E = 0 , (2.1) T + ( B − N ) T + D = 0 , (2.2) has at most two solutions ( T, N ) satisfying T ∈ R and N ≥ as follows.1. T = 0 , N = ( B ± (cid:112) B − E ) / provided that D = 0 , B ≥ E. T = ± (cid:112) √ E − B, N = √ E provided that D = 0 , B < E. T = ±√ z, N = ( T + BT + D ) / T provided that D (cid:54) = 0 and z is the unique positiveroot of the real polynomial z + 2 Bz + ( B − E ) z − D .Proof. (a) if D = 0 by (2.2) then we have T = 0 and (2.1) gives N = ( B ± (cid:112) B − E ) / (b) If D = 0 and B < E , by (2.2)we have T ( T − (2 N − B )) = 0 Note that N − B > Therefore, T = ±√ N − B By (2.1) we have, N − ( B + 2 N − B ) N + E = 0 N = E implies N = √ E Hence, T = ± (cid:113) √ E − B and N = √ E (c) If D (cid:54) = 0 from (2.2) N = ( T + BT + D ) / T plug this N value to (2.1) we have (cid:16) T + BT + D T (cid:17) − ( B + T ) (cid:16) T + BT + D T (cid:17) + E = 0 , (cid:16) T + BT + D (cid:17) − T ( B + T ) (cid:16) T + BT + D (cid:17) + 4 T E = 0 , (cid:16) T + BT + D (cid:17) − T ( B + T ) (cid:16) T + BT + D (cid:17) + 4 T E = 0 ,T + 2 BT + ( B − E ) T − D = 0 , Let T = ±√ z Then, by
Lemma 2.1 . we have z as unique positive root satisfies cubicequation z + 2 Bz + ( B − E ) z − D . heorem 2.3. The solutions of left octonionic quadratic equation x + bx + c = 0 can beobtained by formulas according to the following cases. case 1. if b, c ∈ R and b ≥ c , then x = − b ± √ b − c case 2. if b, c ∈ R and b < c, then x = 12 ( − b + α i + β j + γ k + δ l + p il + q jl + r kl ) where α + β + γ + δ + p + q + r = 4 c − b and α, β, γ, δ, p, q, r ∈ R case 3. if b ∈ R and c / ∈ R , then x = − b ± ρ ∓ c ρ i ∓ c ρ j ∓ c ρ k ∓ c ρ l ∓ c ρ il ∓ c ρ jl ∓ c ρ kl where c = c + c i + c j + c k + c l + c il + c jl + c kl and ρ = (cid:115)(cid:18) b − c + (cid:113) ( b − c ) + 16( c + c + c + c + c + c + c ) (cid:19) / case 4. if b / ∈ R , then x = − Re b − ( b (cid:48) + T ) − ( c (cid:48) − N ) , where b (cid:48) = b − Re b = Im b, c (cid:48) = c − (cid:0) (Re b ) / b − (Re b ) / (cid:1) , and ( T, N ) is chosen as follows . T = 0 , N = ( B ± √ B − E ) / provided that D = 0 , B ≥ E. T = ± (cid:112) √ E − B, N = √ E provided that D = 0 , B < E. T = ±√ z, N = ( T + BT + D ) / T provided that D (cid:54) = 0 and z is the unique positiveroot of the real polynomial z + 2 Bz + ( B − E ) z − D ,where B = b (cid:48) ¯ b (cid:48) + c (cid:48) + ¯ c (cid:48) = | b (cid:48) | +2Re c (cid:48) , E = c (cid:48) ¯ c (cid:48) = | c (cid:48) | , D = ¯ b (cid:48) c (cid:48) + ¯ c (cid:48) b (cid:48) = 2Re ( ¯ b (cid:48) c (cid:48) ) ,are real numbers. roof. Case 1. b, c ∈ R and b ≥ c. Note that x is a solution if and only if q − xq is alsoa solution for q (cid:54) = 0 , and hence, there are at most two solutions, both are real − b ± √ b − c Case 2. b, c ∈ R and b < c. Note that x is a solution if and only if q − xq is also asolution for q (cid:54) = 0 , and there are at least two complex solutions − b ± √ c − b i Hence, the solution set is (cid:40) q − (cid:16) − b + √ c − b i (cid:17) q : q (cid:54) = 0 (cid:41) Let R = 4 c − b > and q ∈ O (cid:40) q − (cid:16) − b + R i (cid:17) q : q (cid:54) = 0 (cid:41)(cid:40) − b + q − R i q q (cid:54) = 0 (cid:41)(cid:40) − b + Rq − i q q (cid:54) = 0 (cid:41) Rq − iq = α i + β j + γ k + δ l + p il + q jl + r kl ∈ Im O and Rq − iq = − Rq − iq Therefore, | Rq − iq | = ( Rq − iq )( Rq − iq ) = − ( Rq − iq )( Rq − iq ) = R = 4 c − b implies, α + β + γ + δ + p + q + r = 4 c − b Hence solution set is, = (cid:40)
12 ( − b + α i + β j + γ k + δ l + p il + q jl + r kl ) : α + β + γ + δ + p + q + r = 4 c − b (cid:41) How we will describe having infinitely many solutions?For Geometric view,Consider | x | = ( − b ) α β γ δ p q r b + 4 c − b c Conclusion : Solutions are set of all points on S ( - sphere) in 8-dimension with normequal to √ c (Note that c ∈ R ) . ase 3. b ∈ R and c / ∈ R . Let x = x + x i + x j + x k + x l + x il + x jl + x kl and c = c + c i + c j + c k + c l + c il + c jl + c kl Then x + bx + c = 0 becomes the real system ( x − x − x − x − x − x − x − x ) + bx + c = 0 (cid:16) x + b (cid:17) − x − x − x − x − x − x − x (cid:1) = b − c (2 x + b ) x i = − c i where i = 1 , , , , , , Since c is non-real (2 x + b ) is non-zero (2 x + b ) − x − x − x − x − x − x − x = ( b − c )(2 x + b ) − c − c − c − c − c − c − c = (2 x + b ) ( b − c )(2 x + b ) − ( b − c )(2 x + b ) + 4( c + c + c + c + c + c + c ) = 0(2 x + b ) = ( b − c ) ± (cid:112) ( b − c ) + 16( c + c + c + c + c + c + c )2(2 x + b ) = ± (cid:118)(cid:117)(cid:117)(cid:116) (cid:16) b − c + (cid:112) ( b − c ) + 16( c + c + c + c + c + c + c ) (cid:17) x = ( − b ± ρ )2 where ρ (cid:54) = 0 x = − b ± ρ ∓ c ρ i ∓ c ρ j ∓ c ρ k ∓ c ρ l ∓ c ρ il ∓ c ρ jl ∓ c ρ klCase 4. b / ∈ R . Rewrite the equation x + bx + c = 0 as y + b (cid:48) y + c (cid:48) = 0 , where y = x + (Re b ) / , b (cid:48) = b − Re b / ∈ R and c (cid:48) = c − (Re b/ (cid:0) b − (Re b/ (cid:1) . By [2], weobserve that the solution of the quadratic equation y + b (cid:48) y + c (cid:48) = 0 also satisfies y − T y + N = 0 , where N = ¯ yy ≥ and T = y + ¯ y ∈ R . Hence, ( b (cid:48) + T ) y + ( c (cid:48) − N ) = 0 , and so y = − ( b (cid:48) + T ) − ( c (cid:48) − N ) because T ∈ R and b (cid:48) / ∈ R implies that b (cid:48) + T (cid:54) = 0 To solve for T and N , we substitute y back into the definitions T = y + ¯ y and N = ¯ yy and simplify to obtain the real system N − ( B + T ) N + E = 0 ,T + ( B − N ) T + D = 0 , where B = b (cid:48) ¯ b (cid:48) + c (cid:48) + ¯ c (cid:48) = | b (cid:48) | + 2Re c (cid:48) , E = c (cid:48) ¯ c (cid:48) = | c (cid:48) | , D = ¯ b (cid:48) c (cid:48) + ¯ c (cid:48) b (cid:48) = 2Re ¯ b (cid:48) c (cid:48) arereal numbers. Note that E = | c (cid:48) | ≥ . if B < , then c (cid:48) + ¯ c (cid:48) < and B − E = | b (cid:48) | B + | b (cid:48) | ( c (cid:48) + ¯ c (cid:48) ) + ( c (cid:48) − ¯ c (cid:48) ) ≤ because of the face that ( c (cid:48) − ¯ c (cid:48) ) ≤ . It follows that B − E < , otherwise B − E = 0 and so | b (cid:48) | B = | b (cid:48) | ( c (cid:48) + ¯ c (cid:48) ) = ( c (cid:48) − ¯ c (cid:48) = 0 , i.e., b (cid:48) = 0 ∈ R , a contradiction. Hence, by lemma (2 . , such system can be solved explicitlyas claimed. Consequently, x = ( − Re b ) / − ( b (cid:48) + T ) − ( c (cid:48) − T ) . xample 1. Consider the equation x + 5 x + 6 = 0 , Find solutions x ∈ O Solution: b = 5 and c = 6 , therefore, b > c. From theorem 2.3, case 1. x = − ± √ − ×
62 = − ±
12 = − or − Example 2.
Consider the equation x + 1 = 0 , Find solutions x ∈ O Solution: b = 0 and c = 1 , therefore, b < c. From theorem 2.3, case 2. x = α i + β j + γ k + δ l + p il + q jl + r kl where α + β + γ + δ + p + q + r = 1 and α, β, γ, δ, p, q, r ∈ R so there are infinitely many solutions to the equation x + 1 = 0 in O . Example 3.
Consider the equation x − x + l = 0 , Find solutions x ∈ O Solution: b = − and c = l , From theorem 2.3, case 3. x = 12 ± ρ ∓ l ρ where ρ = (cid:115) √ Example 4.
Consider the equation x + l x + j = 0 , Find solutions x ∈ O Solution: b = l / ∈ R and c = j , From theorem 2.3, case 4 x = − Re b − ( b (cid:48) + T ) − ( c (cid:48) − N ) since Re b = 0 we have x = − ( b (cid:48) + T ) − ( c (cid:48) − N ) where note that B < E Therefore, T = ± (cid:112) √ E − B = ± and N = √ E = 1 Thus, x = − ( l + 1) − ( j − or x = − ( l − − ( j − Finally, x = ( l − − j ) − l + jl − j − − j − l − jl or x = ( l + 1)(1 − j ) − l + jl + 1 − j − − j − l − jl Application × octonionic matrix We can find left spectrum of × octonionic matrices by solving corresponding octonionicquadratic equation. Let’s begin with following lemma Lemma 3.1.
For p and q ∈ O , σ l ( pI + qA ) = { p + qt : t ∈ σ l ( A ) } , where I is the identitymatrix.Proof. Let t be a left eigenvalue of A with eigenvector v , Av = tv consider ( pI + qA ) v = pIv + qAv = pv + qtv = ( p + qt ) v Therefore, p + qt where t ∈ σ l ( A ) represent the set of left spectrum of ( pI + qA ) . Theorem 3.2.
Let A = (cid:20) a bc d (cid:21) , where a, b, c, ∈ O .(i) if bc = 0 , then σ l ( A ) = { a, d } (ii) if bc (cid:54) = 0 , then σ l ( A ) = { a + bt : t + b − ( a − d ) t − b − c = 0 } Proof. (i) A is a triangular matrix, then results follows.(ii) Using Lemma 3.1 , we have σ l ( A ) = σ l (cid:20) aI + b (cid:20) b − c b − ( d − a ) (cid:21)(cid:21) Let t be any left eigenvalue of (cid:20) b − c b − ( d − a ) (cid:21) , then there exists non-zero vector v = (cid:20) xy (cid:21) such that (cid:20) b − c b − ( d − a ) (cid:21) (cid:20) xy (cid:21) = t (cid:20) xy (cid:21) y = tx, (3.1) b − cx + b − ( d − a ) y = ty (3.2)From equations (3.1) and (3.2), we have t + b − ( a − d ) t − b − c = 0 . (3.3)We can use Theorem (3 . to find roots of (3 . and explicitly find left spectrum of given × octonionic Matrix A . eferences [1] F. Zhang, Quaternions and matrices of quaternions , Linear Algebra Appl. 251(1997)21-57.[2] I. Niven,
Equations in quaternions , American Math. Monthly 48, 654-661, (1941).[3] Jerzy Kocik,
Through the Apollonian Window , Southern Illinois University.[4] John C. Baez,
The Octonions , Bulletin. American Mathematical Society, 39(2002),145-205.[5] John H. Conway and Derek A. Smith,
On Quaternions and Octonions , A.K. Peters,Canada, 2003.[6] L. Huang, W. So,
Quadratic formulas for quaternions , Applied Mathematics Letters,15(2002), 533-540.[7] L. Huang, W. So,
On left eigenvalues of a quaternionic matrix , Linear Algebra Appl.323, No. 1-3:105-116, 2001.[8] R.M.W. Wood,
Quaternionic eignenvalues , Bull. London Math. Soc. 17(1985) 137-138.[9] Tevian Dray. and Corinne, A. Manogue,
The Geometry of the Octonions , WorldScientific, 2015., WorldScientific, 2015.