Nilpotents Leave No Trace -- A Matrix Mystery for Pandemic Times
NNILPOTENTS LEAVE NO TRACEA MATRIX MYSTERY FOR PANDEMIC TIMES
ERIC L. GRINBERG
Abstract.
Reopening a cold case, inspector Echelon, high-ranking in the Row OperationsCenter, is searching for a lost linear map, known to be nilpotent. When a partially decom-posed matrix is unearthed, he reconstructs its reduced form, finding it singular. But wereits roots nilpotent? Early In the Investigation
In teaching Linear Algebra, the first topic often is rowreduction [1, 7], including
Row Reduced Echelon Form (RREF); its applicability is broad and growing. Anothertopic, surprisingly popular with beginning students, is nilpotent matrices . One naturally wonders about theirintersection. For instance, one would expect to find abook exercise asking:
What can be said about the RREFof a nilpotent matrix?
In the early days of the Covid-19 pandemic, as test de-livery went remote, demand grew for new, Internet-resistant problems. A limited literaturesearch for the Nilpotent-RREF connection came up short, suggesting potential for take-homefinal exam questions, hence the note at hand. We’ll first explore examples sufficient to settlethe 3 × Stumbling On Evidence
We refer to [1, 2] for general background on RREF and rank. Recall that a square matrix M is nilpotent if some power of M , say M k , is the zero matrix; the smallest such k is calledthe niloptent index or just index of M . For instance, the rightmost matrix in (2) below isnilpotent, of index 3. Indeed, every strictly upper-triangular matrix (square, with zeros on Key words and phrases: Nilpotent matrix, singular matrix, row reduced echelon form, RREF, null space,kernel, Jorge Luis Borges, mystery. a r X i v : . [ m a t h . HO ] J a n ERIC L. GRINBERG and below the diagonal) is nilpotent. Examining each general type of 3 × × a b , c , . (1)In (1) the entries a, b, c are fixed but unspecified and unrestricted constants.When working with matrices and their components we’ll follow a left to right and up todown convention. Thus first means leftmost , etc. We enumerate the rows of a matrix usingRoman numberals. Hence II is the second row from the top.The second and third matrices in (1) are strictly upper triangular, hence nilpotent. Callthe first matrix F . In F , if b = 0, interchange rows I and III to obtain a strictly lowertriangular matrix, hence a nilpotent matrix. If b (cid:54) = 0, perform III → III − b I (subtract b times row I from row III and make that the new row
III ) to obtain a b − b − ab − . This matrix squares to zero, hence is nilpotent.Next, we present the RREFs of 3 × a b , a
00 0 10 0 0 , . (2)The third matrix is strictly upper triangular, hence nilpotent. For the second matrix, ex-change rows I and III , then perform I → I − ( a ) II to obtain − a a . This matrix cubes to zero, hence is nilpotent. As with all 3 × T , takes a bitmore doing. It can be row reduced to the following matrix, which is nilpotent of index 3: ILPOTENTS LEAVE NO TRACE 3 − − a − ba − b − b − a . (3)We can get from T , the leftmost matrix in (2) to (3) by the following row operations: II ↔ III ; II → II − ba I ; I → ( − I ; III → III − b − b II. (4)This tacitly assumes that a, b are both nonzero. If both a and b are zero, row swaps turn T into a strictly lower triangular, and hence nilpotent matrix.If b = 0 and a (cid:54) = 0 then (3) is still nilpotent and row equivalent to (3), even though the stepswe took to get there involve a zero denominator. In case a = 0, perform row reduction steps II → III ; I → II ; II → II − bIII, obtaining the following nilpotent matrix: − b b b . But (3) and (4) and all the row manipulations beg the question: how did we come up withthese constructs? 3.
No Basis For An Investigation the facts which you have brought me are so indefinitethat we have no basis for an investigationSherlock Holmesin
The Adventure of the Dancing Men
We have a good working basis, however, on which to start.Sherlock Holmesin
A Study In Scarlet
Consider the matrix T = a b . Recall that the (right) null space of T is the solution set of the linear system T (cid:126)v = (cid:126)
0. Oneeasily checks that the span of the vector (cid:126)w ≡ (cid:0) − a − b (cid:1) t gives all solutions of this linearsystem. We will form a basis for R by extending the one element set { (cid:126)w } and use that tobuild a nilpotent matrix whose RREF is T . Using the familiar notation (cid:126)e ≡ (cid:0) (cid:1) t andanalogs for the canonical basis of R , we write (cid:126)u ≡ (cid:126)e and (cid:126)v = (cid:126)e . Then, if a is a nonzero ERIC L. GRINBERG scalar, { (cid:126)u, (cid:126)v, (cid:126)w } is a basis for R . There is a unique linear transformation H on R with theproperties H(cid:126)u = (cid:126)v ; H(cid:126)v = (cid:126)w ; H (cid:126)w = 0; we summarize these as follows: (cid:126)u −→ (cid:126)v −→ (cid:0) − a − b (cid:1) t −→ (cid:126) . (This is not an exact sequence, and not even trying to be one.) Let’s find M , the matrixrepresentation of H .The second column of M is the vector M(cid:126)e , which is already prescribed: it is (cid:126)e . The thirdcolumn of M is the vector M(cid:126)e , prescribed as (cid:126)w . What about the first column of M ? It is M(cid:126)e , but what’s that? We can express (cid:126)e in the basis { (cid:126)u, (cid:126)v, (cid:126)w } as follows: ( − a ) (cid:126)e = − a − b − + b , which we can rewrite as (cid:126)e = − a ( (cid:126)w − (cid:126)v + b(cid:126)u ) . Thus
M(cid:126)e = − a ( M (cid:126)w − M(cid:126)v + bM (cid:126)u ) = − a (cid:16) (cid:126) − (cid:126)w + b(cid:126)v (cid:17) = − − ba − ba , which matches the first column of (3), and thereby reproduces (3).This approach un-begs one question while begging another. The vector space basis procedurehere is guaranteed to produce a nilpotent matrix, but how did we know that this matrix willhave the requisite RREF, namely T ? We know that M and T have the same null space:span { (cid:126)w } . We now quote [7, chp. 2, p. 58] : Corollary (Hoffman-Kunze) . Let A and B be m × n matrices over the field F . Then A and B are row-equivalent if and only if they have the same row space. In our context, relating row equivalence to the null space is needed, and such a relation isimplicit in the literature, e.g., [7] again, or [2, VFSLS]. A recent posting [5] gives:
Corollary.
The null space of a matrix M determines the RREF and the row space of M .Hence if two matrices of the same size have the same null space, they are row equivalent. Remark.
Everyone knows many famous theorems and some famous lemmas, but there isa dearth of famous corollaries. In fact, the two best known to us are not mathematical,emanating from the Monroe Doctrine.Thus, since the nilpotent matrix M has the right (right) null space, it has the right RREFas well. 4. General Impressions
Never trust to general impressions, . . . , but concentrate yourself upon details.Sherlock Holmesin
A Case of Identity
ILPOTENTS LEAVE NO TRACE 5
I have had no proof yet of the existenceSherlock Holmesin
The Sign Of The Four
Theorem.
Every singular matrix is row equivalent to a nilpotent matrix.Proof.
Let M be a singular n × n matrix and take a basis of the (right) null space of M , { (cid:126)k , . . . , (cid:126)k (cid:96) } , where (cid:96) is the nullity of M . As M is singular, (cid:96) is greater or equal to 1. If (cid:96) = n then M is the zero vector, which is nilpotent, and we are done; assume (cid:96) is smaller than n . Extend { (cid:126)k , . . . , (cid:126)k (cid:96) } to { (cid:126)z , . . . (cid:126)z n − (cid:96) , (cid:126)k , . . . , (cid:126)k (cid:96) } , a basis of the vector space of all n × (cid:126)k j and “shifts” eachbasis vector (cid:126)z i to the next one, except for (cid:126)z n − (cid:96) , which is shifted to (cid:126)k . This corresponds toa matrix N with the following properties: N (cid:126)z i − = (cid:126)z i ; N (cid:126)z n − (cid:96) = (cid:126)k , for all suitable values of i . The matrix N is nilpotent of index n − (cid:96) + 1, with null spacespanned by { (cid:126)k , . . . , (cid:126)k (cid:96) } . Thus M and N share a null space. Hence, by the companioncorollary of the Hoffman-Kunze Corollary , they have the same RREF and are thereby rowequivalent. (cid:3) What’s It All About? The Aftermath
Nilpotency figures in the deepest moments of a first course in linear algebra [10]. It is par-ticularly accessible to beginning students. Experience indicates that they latch onto to thesubject, with curiosity and enthusiasm; ditto for RREF. Yet, in the literature, the two seldominteract. Why? The theorem may give a clue. In fact, our discussion shows that it’s not aboutnilpotency at all. It’s about the null space.6.
Late Inspiration
In the course of the 2019-2020 academic-pandemic year this author developed the habit ofstaying up late, delving into the literature, mathematical, fictional and non-fictional. Tryingto compose Internet-lookup-resistant take-home final exam questions, he stumbled on thenilpotency-RREF pairing. At the same (late) time, he was reminded of the stories of JorgeLuis Borges, where mathematical points figure into detective stories. In
Death and TheCompass [4, 12] the mystery hinges on the twists of mathematical reckoning, its inverseand converse, and consequences; an equilateral triangle figures prominently. Recently wenoted a new paper about the
Morley triangle [6]. Morley’s is perhaps the most famous ofall equilateral triangles. Could a Borges-like story revolve around the Morley triangle, hewondered. In his half-dormant state he recalled Borges’
The Garden of Forking Paths , wherea secret message is transmitted by the first letter of the name of a person cited prominentlyin a news story. Could the nilpotency of a matrix serve a similar purpose? More generally,could a mathematical theorem give rise to a mystery story? (More generally still, is there
ERIC L. GRINBERG a functor from the category of mathematics to the category of mystery stories?) That ledto the present note. In the vein of the pandemic teachers were asked to exercise particularunderstanding and accommodation with students. In the same vein one hopes that the readerwill do similarly with the would-be lockdown literato responsible for this pandemic-producedessay.Note: The cartoon included in this essay is by Frank Cotham. It appeared in the New Yorkermagazine in 2007, and is included in the Cartoon Bank. Permission for use was obtained byarrangement with Cond´e Nast.
References [1] Beezer, R.A. (2014). Extended Echelon Form and Four Subspaces,
American Math Monthly , 121:7, 644-647.[2] Beezer, R.A. (2016).
A First Course In Linear Algebra , Open Source book, http://linear.ups.edu ,accessed May, 2020.[3] Borges, J.L. (1942) .
El jardin de senderos que se bifurcan.
Buenos Aires: Sur. Translated by D. A. Yates(1964), in
Labyrinths: Selected Stories & Other Writings , 19-29. New York: New Directions[4] Borges, J.L. (1942).
The Aleph and Other Stories
A Gauche perspective on row reduced echelon form and its uniqueness , arxiv.org/abs/2005.06275 [6] Grinberg E.L., and Orhon, M. (2020). Morley Trisectors and the Law of Sines with Reflections, arxiv.org/abs/2003.10438 , submitted to Amer. Math. Monthly .[7] Hoffman, K., Kunze, R. (1971).
Linear Algebra , 2nd ed. Upper Saddle River, NJ: Prentice-Hall.[8] Strang, G. (1993). The Fundamental Theorem of Linear Algebra.
Amer Math Monthly , 100(9), 848-855.doi:10.2307/2324660.[9] Strang, G. (2014) The Core Ideas in Our Teaching
Notices Am. Math.
Soc. 61, no. 10, 1243-1245,doi:10.1090/noti1174.[10] Trotter, H. F. (1961). A Canonical Basis for Nilpotent Transformations.
Amer. Math. Monthly yutsumura.com/tag/reduced-row-echelon-form/ [12] Zalcman, L. (1988). Death and the Calendar,
Hebrew University Studies in Literature and the Arts
Vol.16, 97-112.
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