The trace Cayley-Hamilton theorem

Transcription

1 The trace Cayley-Hamilto theorem Darij Griberg Jauary 10, 2019 Cotets 1. Itroductio 2 2. Notatios ad theorems Notatios The mai claims The proofs Propositio 2.2 ad Corollary Remiders o the adjugate Polyomials with matrix etries: a trivial lemma Proof of the Cayley-Hamilto theorem Derivatios ad determiats The derivative of the characteristic polyomial Proof of the trace Cayley-Hamilto theorem A corollary Applicatio: Nilpotecy ad traces A ilpotecy criterio A coverse directio More o the adjugate Fuctoriality The evaluatio homomorphism The adjugate of a product Determiat ad adjugate of a adjugate The adjugate of A as a polyomial i A Miors of the adjugate: Jacobi s theorem Aother applicatio of the ti + A strategy

2 The trace Cayley-Hamilto theorem page 2 1. Itroductio Let K be a commutative rig. The famous Cayley-Hamilto theorem says that if χ A det ti A K [t] is the characteristic polyomial of a -matrix A K, the χ A A 0. Speakig more explicitly, it meas that if we write this polyomial χ A i the form χ A i0 c i t i with c i K, the i0 c i A i 0. Various proofs of this theorem are well-kow we will preset oe i this ote, but it could ot be ay farther from beig ew. A less stadard fact, which I call the trace Cayley-Hamilto theorem, states that where i0 kc k + k Tr A i c k i 0 for every k N 1 c i t i is χ A as before, ad where we set c i 0 for every i < 0. I the case of k, this ca easily be obtaied from the Cayley-Hamilto theorem i0 c i A i 0 by multiplyig by A k ad takig traces 1 ; o such simple proof exists i the geeral case, however. The result itself is ot ew the k case, for example, is [LomQui16, Chapter III, Exercise 14], ad is well-kow e.g. to algebraic combiatorialists; however, it is hard to fid a expository treatmet. Whe the groud rig K is a field, it is possible to prove the trace Cayley- Hamilto theorem by expressig both Tr A i ad the c j through the eigevalues of A ideed, Tr A i is the sum of the i-th powers of these eigevalues, whereas c j is 1 j times their j-th elemetary symmetric fuctio; the idetity 1 the boils dow to the Newto idetities for said eigevalues. However, of course, the use of eigevalues i this proof requires K to be a field. There are ways to adapt this proof to the case whe K is a commutative rig. Oe is to apply the method of uiversal idetities see, e.g., [LomQui16, Chapter III, Exercise 14]; the method is also explaied i [Corad09] to reduce the geeral case to the case whe K is a field 2. Aother is to build up the theory of eigevalues for square matrices over a arbitrary commutative rig K; this is ot as simple as for fields, but doable see [Laksov13]. I this ote, I shall give a proof of both the Cayley-Hamilto ad the trace Cayley- Hamilto theorems usig a trick whose use i provig the former is well-kow see, e.g., [Heffer14, Chapter Five, Sectio IV, Lemma 1.9]. The trick is to observe that the adjugate matrix adj ti A ca be writte as D 0 t 0 + D 1 t D 1 t 1 for some matrices D 0, D 1,..., D 1 K ; the, a telescopig sum establishes the Cayley-Hamilto theorem. The same trick ca be used for the trace 1 The details are left to the iterested reader. The kc k term o the left had side appears off, but it actually is harmless: I the k case, it ca be rewritte as Tr A 0 c ad icorporated ito the sum, whereas i the k > case, it simply vaishes. 2 This relies o the observatio that 1 for a give k is a polyomial idetity i the etries of A.

3 The trace Cayley-Hamilto theorem page 3 Cayley-Hamilto theorem, although it requires more work; i particular, a itermediate step is ecessary, establishig that the derivative of the characteristic polyomial χ A det ti A is Tr adj ti A. I hope that this writeup will have two uses: makig the trace Cayley-Hamilto theorem more accessible, ad demostratig that the trick just metioed ca serve more tha oe purpose. Next, I shall show a applicatio of the trace Cayley-Hamilto theorem, aswerig a questio from [m.se ]. Fially, I shall discuss several other properties of the adjugate matrix. 2. Notatios ad theorems 2.1. Notatios Before we state the theorems that we will be occupyig ourselves with, let us agree o the otatios. Defiitio 2.1. Throughout this ote, the word rig will mea associative rig with uity. We will always let K deote a commutative rig with uity. The word matrix shall always mea matrix over K, uless explicitly stated otherwise. As usual, we let K [t] deote the polyomial rig i the idetermiate t over K. If f K [t] is a polyomial ad is a iteger, the [t ] f will deote the coefficiet of t i f. If is egative or greater tha the degree of f, the this coefficiet is uderstood to be 0. Let N deote the set {0, 1, 2,...}. If N ad m N, ad if we are give a elemet a i,j K for every i, j {1, 2,..., } {1, 2,..., m}, the we use the otatio a i,j for the 1 i, 1 j m m-matrix whose i, j-th etry is a i,j for all i, j {1, 2,..., } {1, 2,..., m}. For every N, we deote the idetity matrix by I. For every N ad m N, we deote the m zero matrix by 0 m. If A is ay -matrix, the we let det A deote the determiat of A, ad we let Tr A deote the trace of A. Recall that the trace of A is defied to be the sum of the diagoal etries of A. We cosider K as a subrig of K [t]. Thus, for every N, every -matrix i K ca be cosidered as a matrix i K [t] The mai claims We shall ow state the results that we will prove further below. We begi with a basic fact:

4 The trace Cayley-Hamilto theorem page 4 Propositio 2.2. Let N. Let A K ad B K be two -matrices. Cosider the matrix ta + B K [t]. a The, det ta + B K [t] is a polyomial of degree i t. b We have [ t 0] det ta + B det B. c We have [t ] det ta + B det A. Defiitio 2.3. Let N. Let A K be a -matrix. The, we cosider A as a matrix i K [t] as well as explaied above; thus, a matrix ti A K [t] is defied. We let χ A deote the polyomial det ti A K [t]; we call χ A the characteristic polyomial of A. We otice that the otio of the characteristic polyomial is ot stadardized across the literature. Our defiitio of χ A is idetical with the defiitio i [Kapp2016, V.3] except that we use t istead of X as the idetermiate, but the defiitio i [Heffer14, Chapter Five, Sectio II, Defiitio 3.9] is differet it defies χ A to be det A ti istead. The two defiitios differ merely i a sig amely, oe versio of the characteristic polyomial is 1 times the other, whece ay statemet about oe of them ca easily be traslated ito a statemet about the other; evertheless this discrepacy creates some occasios for cofusio. I shall, of course, use Defiitio 2.3 throughout this ote. Corollary 2.4. Let N. Let A K. a The, χ A K [t] is a polyomial of degree i t. b We have [ t 0] χ A 1 det A. c We have [t ] χ A 1. Of course, combiig parts a ad c of Corollary 2.4 shows that, for every N ad A K, the characteristic polyomial χ A is a moic polyomial of degree. Let me ow state the mai two theorems i this ote: Theorem 2.5 Cayley-Hamilto theorem. Let N. Let A K. The, χ A A 0. Here, χ A A deotes the result of substitutig A for t i the polyomial χ A. It does ot deote the result of substitutig A for t i the expressio det ti A; i particular, χ A A is a -matrix, ot a determiat! Theorem 2.6 trace Cayley-Hamilto theorem. Let N. Let A K. For every j Z, defie a elemet c j K by c j [ t j] χ A. The, kc k + k Tr A i c k i 0 for every k N.

5 The trace Cayley-Hamilto theorem page 5 Theorem 2.5 is as has already bee said well-kow ad a corerstoe of liear algebra. It appears with proofs i [Berha11], [Garrett09, 28.10], [Heffer14, Chapter Five, Sectio IV, Lemma 1.9], [Kapp2016, Theorem 5.9], [Mate16, 4, Theorem 1], [Sage08, Secode méthode 3], [Shurma15], [Straub83], [BroWil89, Theorem 7.10], [Zeilbe85, 3] ad i may other sources 3. The proof we will give below will essetially repeat the proof i [Heffer14, Chapter Five, Sectio IV, Lemma 1.9]. Theorem 2.6 is a less kow result. It appears i [LomQui16, Chapter III, Exercise 14] with a sketch of a proof ad i [Zeilbe85, Exercise 5] without proof; its particular case whe K is a field also teds to appear i represetatio-theoretical literature mostly left as a exercise to the reader. We will prove it similarly to Theorem 2.5; this proof, to my kowledge, is ew. 3. The proofs 3.1. Propositio 2.2 ad Corollary 2.4 Let us ow begi provig the results stated above. As a warmup, we will prove the rather trivial Propositio 2.2. We first recall how the determiat of a matrix is defied: For ay N, let S deote the -th symmetric group i.e., the group of all permutatios of {1, 2,..., }. If N ad σ S, the 1 σ deotes the sig of the permutatio σ. If N, ad if A a i,j is a -matrix, the 1 i, 1 j det A σ S 1 σ a i,σi. 2 We prepare for the proof of Propositio 2.2 by statig a simple lemma: Lemma 3.1. Let N. Let x 1, x 2,..., x be elemets of K. Let y 1, y 2,..., y be elemets of K. Defie a polyomial f K [t] by f tx i + y i. a The, f is a polyomial of degree. b We have [t ] f x i. 3 All the sources we are citig with the possible exceptio of [Garrett09, 28.10] prove Theorem 2.5 i full geerality, although some of them do ot state Theorem 2.5 i full geerality ideed, they ofte state it uder the additioal requiremet that K be a field. There are other sources which oly prove Theorem 2.5 i the case whe K is a field. The ote [Sage08] gives four proofs of Theorem 2.5 for the case whe K C; the first of these proofs works for every field K, whereas the secod works for ay commutative rig K, ad the third ad the fourth actually require K C. Note that some authors declie to call Theorem 2.5 the Cayley-Hamilto theorem; they istead use this ame for some related result. For istace, Heffero, i [Heffer14], uses the ame Cayley-Hamilto theorem for a corollary.

6 The trace Cayley-Hamilto theorem page 6 c We have [ t 0] f y i. Proof of Lemma 3.1. Obvious by multiplyig out the product desires a formal proof, by a straightforward iductio over. tx i + y i or, if oe Proof of Propositio 2.2. Write the -matrix A i the form A a i,j 1 i, 1 j. Thus, a i,j K for every i, j {1, 2,..., } 2 sice A K. Write the -matrix B i the form B b i,j 1 i, 1 j. Thus, b i,j K for every i, j {1, 2,..., } 2 sice B K. For every σ S, defie a polyomial f σ K [t] by The followig holds: f σ ta i,σi + b i,σi. 3 Fact 1: For every σ S, the polyomial f σ is a polyomial of degree. [Proof of Fact 1: Let σ S. The, Lemma 3.1 a applied to a i,σi, b i,σi ad f σ istead of x i, y i ad f shows that f σ is a polyomial of degree. This proves Fact 1.] From A a i,j 1 i, 1 j ad B b i,j, we obtai ta + B 1 i, 1 j tai,j + b i,j 1 i, 1 j. Hece, det ta + B 1 σ ta i,σi + b i,σi σ S f σ by 3 by 2, applied to K [t], ta + B ad tai,j + b i,j σ S 1 σ f σ. istead of K, A ad a i,j Hece, det ta + B is a K-liear combiatio of the polyomials f σ for σ S. Sice all of these polyomials are polyomials of degree by Fact 1, we thus coclude that det ta + B is a K-liear combiatio of polyomials of degree. Thus, det ta + B is itself a polyomial of degree. This proves Propositio 2.2 a.

7 The trace Cayley-Hamilto theorem page 7 b We have [ t 0] det ta + B σ S 1 σ f σ [ t 0] 1 σ f σ 1 σ σ S σ S σ S 1 σ b i,σi. [ t 0] f σ b i,σi by Lemma 3.1 c applied to a i,σi, b i,σi ad f σ istead of x i, y i ad f Comparig this with det B σ S 1 σ b i,σi by 2, applied to B ad bi,j istead of A ad a i,j, we obtai [ t 0] det ta + B det B. This proves Propositio 2.2 b. c We have [t ] det ta + B [t 1 σ f σ σ S ] σ S 1 σ σ S 1 σ f σ a i,σi. σ S 1 σ [t ] f σ a i,σi by Lemma 3.1 b applied to a i,σi, b i,σi ad f σ istead of x i, y i ad f Comparig this with 2, we obtai [t ] det ta + B det A. This proves Propositio 2.2 c. Proof of Corollary 2.4. The defiitio of χ A yields χ A det ti A ti + A det ti + A. Hece, Corollary 2.4 follows from Propositio 2.2 applied to I ad A istead of A ad B. For part b, we eed the additioal observatio that det A 1 det A. Let me state oe more trivial observatio as a corollary: Corollary 3.2. Let N. Let A K. For every j Z, defie a elemet c j K by c j [ t j] χ A. The, χ A c k t k.

8 The trace Cayley-Hamilto theorem page 8 Proof of Corollary 3.2. For every k Z, the defiitio of c k yields [ c k t k] [ χ A t k] χ A. 4 We kow that χ A K [t] is a polyomial of degree i t by Corollary 2.4 a. Hece, χ A [t k] χ A t k c k t k. c k by 4 This proves Corollary Remiders o the adjugate Let us ow briefly itroduce the adjugate of a matrix ad state some of its properties. We first recall the defiitios mostly quotig them from [Gribe15, Chapter 6]: Defiitio 3.3. Let N ad m N. Let A a i,j be a m- 1 i, 1 j m matrix. Let i 1, i 2,..., i u be some elemets of {1, 2,..., }; let j 1, j 2,..., j v be some elemets of {1, 2,..., m}. The, we defie sub j 1,j 2,...,j v i 1,i 2,...,i u A to be the u v-matrix a ix,j y 1 x u, 1 y v. Defiitio 3.4. Let N. Let a 1, a 2,..., a be objects. Let i {1, 2,..., }. The, a 1, a 2,..., â i,..., a shall mea the list a 1, a 2,..., a i 1, a i+1, a i+2,..., a that is, the list a 1, a 2,..., a with its i-th etry removed. Thus, the hat over the a i meas that this a i is beig omitted from the list. For example, 1 2, 2 2,..., 5 2,..., , 2 2, 3 2, 4 2, 6 2, 7 2, 8 2. Defiitio 3.5. Let N ad m N. Let A be a m-matrix. For every i {1, 2,..., } ad j {1, 2,..., m}, we let A i, j be the 1 m 1- matrix sub 1,2,...,ĵ,...,m 1,2,...,î,..., A. Thus, A i, j is the matrix obtaied from A by crossig out the i-th row ad the j-th colum. Defiitio 3.6. Let N. Let A be a -matrix. We defie a ew - matrix adj A by adj A 1 i+j det A j, i. 1 i, 1 j This matrix adj A is called the adjugate of the matrix A.

9 The trace Cayley-Hamilto theorem page 9 The mai property of the adjugate is the followig fact: Theorem 3.7. Let N. Let A be a -matrix. The, A adj A adj A A det A I. Recall that I deotes the idetity matrix. Expressios such as adj A A ad det A I have to be uderstood as adj A A ad det A I, respectively. Theorem 3.7 appears i almost ay text o liear algebra that cosiders the adjugate; for example, it appears i [Heffer14, Chapter Four, Sectio III, Theorem 1.9], i [Kapp2016, Propositio 2.38], i [BroWil89, Theorem 4.11] ad i [Gribe15, Theorem 6.100]. Agai, most of these sources oly state it i the case whe K is a field, but the proofs give apply i all geerality. Differet texts use differet otatios. The source that is closest to my otatios here is [Gribe15], sice Theorem 3.7 above is a verbatim copy of [Gribe15, Theorem 6.100]. Let us state a simple fact: Lemma 3.8. Let N. Let u ad v be two elemets of {1, 2,..., }. Let λ ad µ be two elemets of K. Let A ad B be two -matrices. The, λa + µb u, v λa u, v + µb u, v. Proof of Lemma 3.8. Obvious. Next, we prove a crucial, if simple, result: Propositio 3.9. Let N. Let A K be a -matrix. The, there exist matrices D 0, D 1,..., D 1 i K such that adj ti A 1 t k D k i K [t]. Here, of course, the matrix D k o the right had side is uderstood as a elemet of K [t]. Proof of Propositio 3.9. Fix u, v {1, 2,..., } 2. The, Propositio 2.2 a applied to 1, I u, v ad A u, v istead of, A ad B shows that det t I u, v + A u, v K [t] is a polyomial of degree 1 i t. I other words, there exists a -tuple d u,v,0, d u,v,1,..., d u,v, 1 K such that 1 det t I u, v + A u, v d u,v,k t k.

10 The trace Cayley-Hamilto theorem page 10 Cosider this d u,v,0, d u,v,1,..., d u,v, 1. But Lemma 3.8 applied to K [t], t, 1, I ad A istead of K, λ, µ, A ad B yields ti A u, v t I u, v + A u, v after some simplificatios. Thus, 1 det ti A u, v det t I u, v + A u, v d u,v,k t k. 5 Now, forget that we fixed u, v. Thus, for every u, v {1, 2,..., } 2, we have costructed a -tuple d u,v,0, d u,v,1,..., d u,v, 1 K satisfyig 5. Now, the defiitio of adj ti A yields adj ti A 1 i+j Comparig this with det ti A j, i 1 d j,i,k t k by 5, applied to u,vj,i 1 i+j 1 d j,i,k t k 1 t k 1 i+j d j,i,k 1 t k 1 i+j d j,i,k 1 i, 1 j 1 i, 1 j 1 i, 1 j 1 t k 1 i+j d j,i,k. 1 i, 1 j 1 t k 1 i+j d j,i,k, 1 i, 1 j we obtai adj ti A 1 t k 1 i+j d j,i,k. Hece, there exist 1 i, 1 j matrices D 0, D 1,..., D 1 i K such that amely, D k proves Propositio 3.9. adj ti A 1 i+j d j,i,k 1 t k D k 1 i, 1 j i K [t] for every k {0, 1,..., 1}. This 3.3. Polyomials with matrix etries: a trivial lemma

11 The trace Cayley-Hamilto theorem page 11 Lemma Let N ad m N. Let B 0, B 1,..., B m K m+1 ad C 0, C 1,..., C m K m+1 be two m + 1-tuples of matrices i K. Assume that m t k B k The, B k C k for every k {0, 1,..., m}. m t k C k i K [t]. Proof of Lemma For every k {0, 1,..., m}, write the matrix B k K i the form B k b k,i,j 1 i, 1 j, ad write the matrix C k K i the form C k c k,i,j 1 i, 1 j. m m Now, t k B k t k b k,i,j sice B k b k,i,j 1 i, 1 j 1 i, 1 j m m k {0, 1,..., m}. Similarly, t k C k t k c k,i,j. Thus, 1 i, 1 j m t k b k,i,j 1 i, 1 j I other words, m t k B k m t k m b k,i,j t k c k,i,j for every m m t k C k t k c k,i,j. 1 i, 1 j for every i, j {1, 2,..., } 2. Comparig coefficiets o both sides of this equality, we obtai b k,i,j c k,i,j for every k {0, 1,..., m} for every i, j {1, 2,..., } 2. Now, every k {0, 1,..., m} satisfies B k b k,i,j c k,i,j This proves Lemma i, 1 j c k,i,j 1 i, 1 j C k Proof of the Cayley-Hamilto theorem We are ow fully prepared for the proof of the Cayley-Hamilto theorem. However, we are goig to orgaize the crucial part of this proof as a lemma, so that we ca use it later i our proof of the trace Cayley-Hamilto theorem.

12 The trace Cayley-Hamilto theorem page 12 Lemma Let N. Let A K. For every j Z, defie a elemet c j K by c j [ t j] χ A. Let D 0, D 1,..., D 1 be matrices i K such that adj ti A 1 t k D k i K [t]. 6 Thus, a -tuple D 0, D 1,..., D 1 of matrices i K is defied. Exted this -tuple to a family D k k Z of matrices i K by settig The: a We have χ A D k 0 for every k Z \ {0, 1,..., 1}. 7 c k t k. b For every iteger k, we have c k I D k 1 AD k. c Every k N satisfies k c k i A i D 1 k. i0 Proof of Lemma a Lemma 3.11 a is just Corollary 3.2. b We have ad t k D k 1 t 0 D }{{ 0 1 } D 1 0 by 7 + k1 t k D k 1 t k 1 D k 1 k1 t k+1 D k+1 1 tt k } {{ } D k here, we have substituted k + 1 for k i the sum 1 tt k 1 D k t t k D k t adj ti A 8 adjti A by 6 t k D k t D + 0 by 7 1 t k D k 1 t k D k adj ti A by 6. 9 But Theorem 3.7 applied to K [t] ad ti A istead of K ad A shows that ti A adj ti A adj ti A ti A det ti A I.

13 The trace Cayley-Hamilto theorem page 13 Thus, i particular, so that ti A adj ti A det ti A I χ A I, χ A by the defiitio of χ A χ A I ti A adj ti A t adj ti A t k D k 1 by 8 t k D k 1 A adj ti A t k D k by 9 t k AD k t k D k 1 AD k. t k D k 1 A t k D k t k AD k Thus, t k D k 1 AD k χ A c k t k by Lemma 3.11 a t k c k I. I c k t k I Lemma 3.10 applied to m, B k D k 1 AD k ad C k c k I thus shows that D k 1 AD k c k I for every k {0, 1,..., }. 10 Now, let k be a iteger. We must prove that c k I D k 1 AD k. If k {0, 1,..., }, the this follows from 10. Thus, we WLOG assume that k / {0, 1,..., }. Hece, k 1 Z \ {0, 1,..., 1}, so that 7 applied to k 1 istead of k yields D k 1 0. Also, k / {0, 1,..., } leads to k Z \ {0, 1,..., 1}; therefore, 7 yields D k 0. Now, D }{{ k 1 A D } k O the other had, c k 0 4. Hece, c k I 0. Compared with D k 1 AD k 0, this yields c k I D k 1 AD k. Hece, c k I D k 1 AD k is prove. I other words, Lemma 3.11 b is prove. 4 Proof. Recall that χ A is a polyomial of degree by Corollary 2.4 a. Hece, [ sice k / {0, 1,..., }. Now, 4 yields c k t k] χ A 0. 0 [ t k] χ A 0

14 The trace Cayley-Hamilto theorem page 14 c Let k N. The, k c k i A i i0 i k i k i k i k i k k +i c k k +i A c i here, we have substituted k + i for i i the sum c i A k +i A k +i c i I A k +i D i 1 AD i A k +i D i 1 A k +i AD i This proves Lemma 3.11 c. A k +i D i 1 A k +i c i I }{{ } i k D i 1 AD i by Lemma 3.11 b, applied to i istead of k A k +i A D i A k +i+1 A k +i+1 D i+1 1 A k +i D i 1 A k +i+1 D i+1 1 A k + k A 0 I D k 1 A k ++1 D +1 1 D 0 by 7 by the telescope priciple D k 1 D 1 k. Proof of Theorem 2.5. For every j Z, defie a elemet c j K by c j [ t j] χ A. Propositio 3.9 shows that there exist matrices D 0, D 1,..., D 1 i K such that 1 adj ti A t k D k i K [t]. Cosider these D 0, D 1,..., D 1. Thus, a -tuple D 0, D 1,..., D 1 of matrices i K is defied. Exted this -tuple to a family D k k Z of matrices i K by settig D k 0 for every k Z \ {0, 1,..., 1}. Thus, i particular, D 1 0. Lemma 3.11 a shows that χ A c k t k c i t i. Substitutig A for t i i0

15 The trace Cayley-Hamilto theorem page 15 this equality, we obtai χ A A c i A i D 1 by Lemma 3.11 c, applied to k i0 D 1 0. This proves Theorem Derivatios ad determiats Now, let us make what seems to be a detour, ad defie K-derivatios of a K- algebra 5 : Defiitio Let L be a K-algebra. A K-liear map f : L L is said to be a K-derivatio if it satisfies f ab a f b + f a b for every a L ad b L. 11 The otio of a K-derivatio is a particular case of the otio of a k-derivatio defied i [Gribe16a, Defiitio 1.5]; specifically, it is obtaied from the latter whe settig k K, A L ad M L. This particular case will suffice for us. Examples of K-derivatios aboud there are several i [Gribe16a], but the oly oe we will eed is the followig: Propositio Let : K [t] K [t] be the differetiatio operator i.e., the map that seds every polyomial f K [t] to the derivative of f. The, : K [t] K [t] is a K-derivatio. Proof of Propositio This follows from the fact that ab a b + a b for ay two polyomials a ad b the well-kow Leibiz law. A fudametal fact about K-derivatios is the followig: Propositio Let L be a K-algebra. Let f : L L be a K-derivatio. Let N, ad let a 1, a 2,..., a L. The, f a 1 a 2 a a 1 a 2 a i 1 f a i a i+1 a i+2 a. 5 See [Gribe16a, Covetio 1.1] for what we mea by a K-algebra. I a utshell, we require K-algebras to be associative ad uital, ad we require the multiplicatio map o a K-algebra to be K-biliear.

16 The trace Cayley-Hamilto theorem page 16 This propositio is a particular case of [Gribe16a, Theorem 1.14] obtaied by settig k K, A L ad M L; it is also easy to prove 6. What we are goig to eed is a formula for how a derivatio acts o the determiat of a matrix. We first itroduce a otatio: Defiitio Let N ad m N. Let L ad M be rigs. Let f : L M be ay map. The, f m will deote the map from L m to M m which seds every matrix a i,j 1 i, 1 j m L m to the matrix f a i,j 1 i, 1 j m M m. I other words, f m is the map which takes a m-matrix i L m, ad applies f to each etry of this matrix. Theorem Let L be a commutative K-algebra. Let f : L L be a K- derivatio. Let N. Let A L. The, f det A Tr f A adj A. Provig Theorem 3.16 will take us a while. Let us begi by statig three lemmas: Lemma Let N ad m N. Let A a i,j 1 i, 1 j m K m ad B b i,j 1 i m, 1 j Km. The, Tr AB m j1 a i,j b j,i. m Proof of Lemma The defiitio of AB yields AB a i,k b k,j k1 1 i, 1 j sice A a i,j 1 i, 1 j m ad B b i,j. Hece, 1 i m, 1 j Tr AB m k1 a i,k b k,i m j1 a i,j b j,i here, we have reamed the summatio idex k as j i the secod sum. This proves Lemma Lemma Let L be a commutative K-algebra. Let f : L L be a K- derivatio. Let N, ad let a 1, a 2,..., a L. The, f a 1 a 2 a k1 f a k i {1,2,...,}; i k a i. 6 First oe should show that f 1 0 by applyig 11 to a 1 ad b 1. The, oe ca prove Propositio 3.14 by straightforward iductio o.

17 The trace Cayley-Hamilto theorem page 17 Proof of Lemma Propositio 3.14 yields f a 1 a 2 a k1 This proves Lemma a 1 a 2 a i 1 f a i a i+1 a i+2 a a 1 a 2 a k 1 f a k a k+1 a k+2 a }{{ } f a k a 1 a 2 a k 1 a k+1 a k+2 a sice L is commutative here, we have reamed the summatio idex i as k f a k a 1 a 2 a k 1 a k+1 a k+2 a k1 i {1,2,...,}; i k a i k1 f a k i {1,2,...,}; i k a i. Lemma Let N. Let A a i,j 1 i, 1 j p {1, 2,..., } ad q {1, 2,..., }. The, be a -matrix. Let 1 σ a i,σi 1 p+q det A p, q. σ S ; i {1,2,...,}; σpq i p Lemma 3.19 is [Gribe15, Lemma 6.84]; it is also easy to prove it is the mai step i the proof of the Laplace expasio formula for the determiat. Proof of Theorem Write the matrix A L i the form A a i,j 1 i, 1 j. Hece, f A f a i,j 1 i, 1 j by the defiitio of f. The defiitio of adj A shows that adj A 1 i+j det A j, i. Hece, Lemma 1 i, 1 j 3.17 applied to L,, f A, f a i,j, adj A ad 1 i+j det A j, i istead of K, m, A, a i,j, B ad b i,j yields Tr f A adj A j1 k1 j1 k1 j1 f a i,j 1 j+i det A i, j f a k,j 1 j+k 1 k+j det A k, j here, we have reamed the summatio idex i as k i the outer sum f a k,j 1 k+j det A k, j. 12

18 The trace Cayley-Hamilto theorem page 18 But the map f is a K-derivatio, ad thus is K-liear. Now, 2 applied to L istead of K yields det A 1 σ a i,σi. Applyig f to both sides of this σ S equality, we fid f det A f σ S 1 σ σ S 1 σ f a i,σi a i,σi a 1,σ1 a 2,σ2 a,σ 1 σ f a 1,σ1 a 2,σ2 a,σ σ S 1 σ f σ S k1 k1 k1 fa k,σk i {1,2,...,}; i k a i,σi by Lemma 3.18, applied to a i,σi istead of a i a k,σk 1 σ f a k,σk σ S a i,σi i {1,2,...,}; i k i {1,2,...,}; i k sice the map f is K-liear a i,σi. 13

19 The trace Cayley-Hamilto theorem page 19 But every k {1, 2,..., } satisfies σ S j {1,2,...,} σ S ; σkj sice σk {1,2,...,} for each σ S j {1,2,...,} σ S ; σkj j {1,2,...,} σ S ; σkj j {1,2,...,} j1 Hece, 13 becomes f det A f a k,j 1 σ f 1 σ f a k,j σ S ; σkj 1 σ f a k,σk a k,σk a k,j sice σkj 1 σ i {1,2,...,}; i k i {1,2,...,}; i k a i,σi i {1,2,...,}; i k a i,σi i {1,2,...,}; i k } {{ } 1 k+j deta k, j by Lemma 3.19, applied to L, k ad j istead of K, p ad q f a k,j 1 k+j det A k, j. j1 k1 k1 j1 1 σ f a k,σk σ S by 12. This proves Theorem i {1,2,...,}; i k a i,σi fa k,j 1 k+j deta k, j j1 a i,σi a i,σi f a k,j 1 k+j det A k, j Tr f A adj A 3.6. The derivative of the characteristic polyomial The characteristic polyomial χ A of a square matrix A is, first of all, a polyomial; ad a polyomial has a derivative. We shall have eed for a formula for this derivative:

20 The trace Cayley-Hamilto theorem page 20 Theorem Let N. Let A K. Let : K [t] K [t] be the differetiatio operator i.e., the map that seds every polyomial f K [t] to the derivative of f. The, χ A Tr adj ti A. Proof of Theorem Propositio 3.13 shows that : K [t] K [t] is a K-derivatio. Now, cosider the map : K [t] K [t] defied accordig to Defiitio It is easy to see that tb A B 14 for ay -matrix B K 7. Applyig this to B I, we obtai ti A I. The defiitio of χ A yields χ A det ti A. Applyig the map to both sides of this equality, we obtai χ A det ti A Tr ti A adj ti A I by Theorem 3.16 applied to K [t], ad ti A istead of L, f ad A This proves Theorem Tr I adj ti A Tr adj ti A. adjti A 7 Proof. Let B K be a -matrix. Write the matrix B i the form B b i,j 1 i, 1 j. Write the matrix A i the form A a i,j 1 i, 1 j. Both matrices A ad B belog to K ; thus, every i, j {1, 2,..., } 2 satisfies a i,j K ad b i,j K ad therefore tb i,j a i,j bi,j sice is the differetiatio operator. Now, t B b i,j 1 i, 1 j t b i,j A a i,j 1 i, 1 j 1 i, 1 j a i,j Hece, the defiitio of the map yields qed. tb A tb i,j a i,j b i,j 1 i, 1 j tb i,j a i,j 1 i, 1 j 1 i, 1 j. b i,j 1 i, 1 j B,

21 The trace Cayley-Hamilto theorem page 21 We ca use Theorem 3.20 to obtai the followig result: Propositio Let N. Let A K. For every j Z, defie a elemet c j K by c j [ t j] χ A. Let D 0, D 1,..., D 1 be matrices i K satisfyig 6. Thus, a -tuple D 0, D 1,..., D 1 of matrices i K is defied. Exted this -tuple to a family D k k Z of matrices i K by settig 7. The, every k Z satisfies Tr D k k + 1 c k Proof of Propositio Let : K [t] K [t] be the differetiatio operator i.e., the map that seds every polyomial f K [t] to the derivative of f. Lemma 3.11 a yields χ A equality, we obtai χ A c k t k kc k t k 1 1 k1 c k kt k 1 k1 c k t k. Applyig the map to both sides of this k + 1 c k+1 t k sice is the differetiatio operator here, we have substituted k + 1 for k i the sum. Comparig this with we obtai χ A Tr adj ti A Tr 1 t k D k by 6 1 t k 1 D k 1 Tr D k t k by Theorem 3.20 t k Tr D k 1 k + 1 c k+1 t k. 1 Tr D k t k, This is a idetity betwee two polyomials i K [t]. Comparig coefficiets o both sides of this idetity, we coclude that Tr D k k + 1 c k+1 for every k {0, 1,..., 1}. 16 Now, let k Z. We must prove 15.

22 The trace Cayley-Hamilto theorem page 22 If k {0, 1,..., 1}, the 15 follows immediately from 16. Hece, for the rest of this proof, we WLOG assume that we do t have k {0, 1,..., 1}. We do t have k {0, 1,..., 1}. Thus, k Z \ {0, 1,..., 1}. Hece, 7 yields D k 0, so that Tr D k Tr 0 0. Recall agai that k Z \ {0, 1,..., 1}. I other words, we have either k < 0 or k. Thus, we are i oe of the followig two cases: Case 1: We have k < 0. Case 2: We have k. Let us first cosider Case 1. I this case, we have k < 0. If k 1, the 15 holds 8. Hece, for the rest of this proof, we WLOG assume that k 1. Combiig k < 0 with k 1, we obtai k < 1. Hece, k + 1 < 0. The defiitio of c k+1 yields c k+1 χ A [ t k+1] χ A 0 t k+1 t k+1 sice k + 1 < 0, but χ A is a polyomial. Hece, k + 1 c k Comparig this with Tr D k 0, we obtai Tr D k k + 1 c k+1. Hece, 15 is prove i Case 1. Let us ow cosider Case 2. I this case, we have k. Thus, k >. But χ A is a polyomial of degree. Hece, [t m ] χ A 0 for every iteger m >. Applyig this to m k + 1, we obtai [ t k+1] χ A 0 sice k + 1 >. The defiitio of c k+1 yields c k+1 Hece, k + 1 c k+1 0 χ A [ t k+1] χ A 0. t k+1 t k+1 0. Comparig this with Tr D k 0, we obtai Tr D k k + 1 c k+1. Hece, 15 is prove i Case 2. We have ow prove 15 i each of the two Cases 1 ad 2. Thus, 15 always holds. Thus, Propositio 3.21 is prove Proof of the trace Cayley-Hamilto theorem Now, we ca fially prove the trace Cayley-Hamilto theorem itself: Proof of Theorem 2.6. Propositio 3.9 shows that there exist matrices D 0, D 1,..., D 1 i K such that adj ti A 1 t k D k i K [t]. 8 Proof. Assume that k 1. The, k + 1 0, so that k + 1 c k+1 0. Comparig this with Tr D k 0, we obtai Tr D k k + 1 c k+1 ; hece, 15 holds, qed. 0

23 The trace Cayley-Hamilto theorem page 23 Cosider these D 0, D 1,..., D 1. Thus, a -tuple D 0, D 1,..., D 1 of matrices i K is defied. Exted this -tuple to a family D k k Z of matrices i K by settig D k 0 for every k Z \ {0, 1,..., 1}. Now, let k N. The, Propositio 3.21 applied to 1 k istead of k yields Thus, Tr D 1 k 1 k + 1 c 1 k+1 k c k. k c k sice 1 k+1k k c k Tr D 1 k k c k i A i i0 by Lemma 3.11 c k c k i Tr A i c k 0 Tr i0 c k k Tr c k i A i i0 A 0 I + k c k i Tr A i here, we have split off the added for i 0 from the sum k c k Tr I + c k i Tr A i k c k + Tr A i c k i. TrA i c k i Solvig this equatio for k Tr A i c k i, we obtai k Tr A i c k i k c k c k kc k c k c k c k kc k c k kc k. Addig kc k to both sides of this equatio, we obtai kc k + k proves Theorem A corollary Tr A i c k i 0. This The followig fact which ca also be easily prove by other meas follows readily from Theorem 2.6:

24 The trace Cayley-Hamilto theorem page 24 Corollary Let N. Let A K. The, [ t 1] χ A Tr A. Proof of Corollary For every j Z, defie a elemet c j K by c j [ t j] χ A. The defiitio of c 1 yields c 1 [ t 1] χ A. The defiitio of c 0 yields c 0 [t 0] χ A [t ] [t ] χ A 1 by Corollary 2.4 c. Theorem 2.6 applied to k 1 yields 1c Tr A i c 1 i 0. Thus, 1 1c 1 Tr A i c 1 i Tr A 1 c 1 1 Tr A. A c 0 1 TrA 1 c 1 1 Comparig this with 1c 1 c 1 [ t 1] χ A, we obtai [ t 1] χ A Tr A. This proves Corollary Applicatio: Nilpotecy ad traces 4.1. A ilpotecy criterio As a applicatio of Theorem 2.6, let us ow prove the followig fact geeralizig [m.se ]: Corollary 4.1. Let N. Let A K. Assume that Tr A i 0 for every i {1, 2,..., }. 17 a The,!A 0. b If K is a commutative Q-algebra, the A 0. c We have!χ A!t. d If K is a commutative Q-algebra, the χ A t. Proof of Corollary 4.1. For every j Z, defie a elemet c j K by c j [ t j] χ A. The defiitio of c 0 yields c 0 [t 0] χ A [t ] χ A 1 by Corollary 2.4 c. [t ] We ow claim that kc k 0 for every k {1, 2,..., }. 18 [Proof of 18: Let k {1, 2,..., }. The, every i {1, 2,..., k} satisfies i {1, 2,..., } ad therefore also Tr A i 0 19

25 The trace Cayley-Hamilto theorem page 25 by 17. Now, Theorem 2.6 yields kc k + Solvig this equatio for kc k, we obtai kc k This proves 18.] Now, we claim that k k Tr A i c k i 0. Tr A i c k i 0 by 19 k 0c k i } {{ } !c k 0 for every k {1, 2,..., }. 20 [Proof of 20: Let k {1, 2,..., }. The product 1 2 cotais k as a factor, ad thus is a multiple of k; i other words,! is a multiple of k sice! 1 2. Hece,!c k is a multiple of kc k. Thus, 20 follows from 18.] Fially, we observe that!c k 0 for every k {0, 1,..., 1}. 21 [Proof of 21: Let k {0, 1,..., 1}. The, k {1, 2,..., }. Hece, 20 applied to k istead of k yields!c k 0. This proves 21.] Now, Corollary 3.2 yields χ A c k t k. Substitutig A for t i this equality, we obtai χ A A we obtai Hece,!χ A A! 1 c k A k. Multiplyig both sides of the latter equality by!, c k A k!c k A k 1!c k A k +! c A 0 c 0 1 by 21 here, we have split off the added for k from the sum 0A k +!A!A. 0 This proves Corollary 4.1 a.!a! χ A A 0 by Theorem

26 The trace Cayley-Hamilto theorem page 26 b Assume that K is a commutative Q-algebra. Corollary 4.1 a yields!a 1 0. Now,! Q, so that we ca multiply a -matrix i K by 1! sice K is a Q-algebra. We have 1!! This proves Corollary 4.1 b. 1 A A. Hece, A 1!!A 0 1! 0 0. c Multiplyig the equality χ A c k t k by!, we obtai!χ A! 1 0 c k t k!c k t k 1!c k 0 by 21 t k +! c t c 0 1 here, we have split off the added for k from the sum 0t k +!t!t. This proves Corollary 4.1 c. d Assume that K is a commutative Q-algebra. Corollary 4.1 c yields!χ A!t. 1 Now, sice K is! Q, so that we ca multiply ay polyomial i K [t] by 1! a Q-algebra. We have 1!! proves Corollary 4.1 d A coverse directio χ A χ A. Hece, χ A 1!!χ A!t 1!!t t. This The followig result i a sese, a coverse of Corollary 4.1 d also follows from Theorem 2.6: Corollary 4.2. Let N. Let A K. Assume that χ A t. The, Tr A i 0 for every positive iteger i. Proof of Corollary 4.2. For every j Z, defie a elemet c j K by c j [ t j] χ A. The, each positive iteger j satisfies c j [Proof of 22: Let j be a positive iteger. Thus, j 0, so that j ad thus [ t j ] t 0. I view of χ A t, this rewrites as [ t j] χ A 0. But the defiitio of c j yields c j [ t j] χ A 0. This proves 22.]

27 The trace Cayley-Hamilto theorem page 27 The defiitio of c 0 yields c 0 [t 0] χ A [t ] χ A 1 by Corollary 2.4 c. [t ] Now, we claim that Tr A p 0 for every positive iteger p. 23 [Proof of 23: We shall prove 23 by strog iductio o p: Iductio step: Fix a positive iteger k. Assume as the iductio hypothesis that 23 holds wheever p < k. We must ow prove that 23 holds for p k. From 22 applied to j k, we obtai c k 0. We have assumed that 23 holds wheever p < k. I other words, Now, Theorem 2.6 yields Hece, 0 k c k + 0 k 1 0c k i } {{ } 0 k Tr A p 0 for every positive iteger p < k. 24 kc k + Tr A i c k i k Tr A i c k i 0. k Tr A i k 1 c k i Tr A i 0 by 24, applied to pi here, we have split off the added for i k from the sum + Tr A k Tr A k. c k i + Tr A k c k k c 0 1 Thus, Tr A k 0. I other words, 23 holds for p k. This completes the iductio step. Thus, 23 is prove by strog iductio.] We have thus prove that Tr A p 0 for every positive iteger p. Reamig the variable p as i i this statemet, we coclude that Tr A i 0 for every positive iteger i. This proves Corollary More o the adjugate I shall ow discuss various other properties of the adjugate adj A of a square matrix A.

28 The trace Cayley-Hamilto theorem page Fuctoriality For ay N ad m N, a homomorphism f : L M betwee two rigs L ad M gives rise to a map f m : L m M m as defied i Defiitio We recall some classical properties of these maps f m : Propositio 5.1. Let L ad M be two commutative rigs. Let f : L M be a rig homomorphism. a For every N ad m N, the map f m : L m M m is a homomorphism of additive groups. b Every N satisfies f I I. c For every N, m N, p N, A L m ad B L m p, we have f p AB f m A f m p B. d For every N ad m N ad every A L m ad λ L, we have f m λa f λ f m A. Now, let me state the classical ad simple fact which is ofte somewhat icompletely subsumed uder the sloga rig homomorphisms preserve determiats ad adjugates : Propositio 5.2. Let L ad M be two commutative rigs. Let f : L M be a rig homomorphism. Let N. Let A L. a We have f det A det f A. b Ay two elemets u ad v of {1, 2,..., } satisfy f 1 1 A u, v f A u, v. c We have f adj A adj f A. Proof of Propositio 5.2. Provig Propositio 5.2 is completely straightforward, ad left to the reader The evaluatio homomorphism We shall apply the above to relate the determiat ad the adjugate of a matrix A with those of the matrix ti + A: Propositio 5.3. Let ε : K [t] K be the map which seds every polyomial p K [t] to its value p 0. It is well-kow that ε is a K-algebra homomorphism. Let N. Let A K. Cosider the matrix ti + A K [t]. The: a We have ε det ti + A det A. b We have ε adj ti + A adj A. c We have ε ti + A A. Proof of Propositio 5.3. We have ε tb + A A

29 The trace Cayley-Hamilto theorem page 29 for every B K 9. Applyig this to B I, we obtai ε ti + A A. This proves Propositio 5.3 c. a Propositio 5.2 a applied to K [t], K, ε ad ti + A istead of L, M, f ad A yields ε det ti + A det ε ti + A det A. A This proves Propositio 5.3 a. b Propositio 5.2 c applied to K [t], K, ε ad ti + A istead of L, M, f ad A yields ε adj ti + A adj ε ti + A adj A. A This proves Propositio 5.3 b. If A K is a square matrix, the the matrix ti + A K [t] has a property which the matrix A might ot have: amely, its determiat is regular. Let us first defie what this meas: Defiitio 5.4. Let A be a commutative rig. Let a A. The elemet a of A is said to be regular if ad oly if every x A satisfyig ax 0 satisfies x 0. Istead of sayig that a is regular, oe ca also say that a is cacellable, or that a is a o-zero-divisor. A basic property of regular elemets is the followig: Lemma 5.5. Let A be a commutative rig. Let a be a regular elemet of A. Let b ad c be two elemets of A such that ab ac. The, b c. Proof of Lemma 5.5. We have a b c ab ac ac ac 0. ac Now, recall that the elemet a of A is regular if ad oly if every x A satisfyig ax 0 satisfies x 0 by the defiitio of regular. Hece, every x A satisfyig ax 0 satisfies x 0 because the elemet a of A is regular. Applyig this to x b c, we obtai b c 0 sice a b c 0. Thus, b c. This proves Lemma 5.5. Regular elemets, of course, ca also be cacelled from matrix equatios: Lemma 5.6. Let N ad m N. Let a be a regular elemet of K. Let B K m ad C K m be such that ab ac. The, B C. 9 Proof. This equality is similar to 14, ad is prove aalogously.

30 The trace Cayley-Hamilto theorem page 30 Proof of Lemma 5.6. Write the m-matrices B ad C i the forms B b i,j ad C c i,j 1 i, 1 j m. The, ab ab i,j 1 i, 1 j m ad ac ac i,j Hece, abi,j 1 i, 1 j m ab ac ac i,j I other words, 1 i, 1 j m. ab i,j ac i,j for every i, j {1, 2,..., } {1, 2,..., m}. 1 i, 1 j m 1 i, 1 j m. Thus, b i,j c i,j for every i, j {1, 2,..., } {1, 2,..., m} by Lemma 5.5, applied to b b i,j ad c c i,j. Hece, bi,j 1 i, 1 j m ci,j 1 i, 1 j m. Thus, B b i,j 1 i, 1 j m c i,j C. Lemma 1 i, 1 j m 5.6 is prove. Oe importat way to costruct regular elemets is the followig fact: Propositio 5.7. Let N. Let p K [t] be a moic polyomial of degree. The, the elemet p of K [t] is regular. Proof of Propositio 5.7. Propositio 5.7 is precisely [Gribe16b, Corollary 3.15]. Corollary 5.8. Let N. Let A K. Cosider the matrix ti + A K [t]. The, the elemet det ti + A of K [t] is regular. Proof of Corollary 5.8. Propositio 2.2 a applied to I ad A istead of A ad B yields that det ti + A K [t] is a polyomial of degree i t. Propositio 2.2 c applied to I ad A istead of A ad B yields that [t ] det ti + A det I 1. So we kow that the polyomial det ti + A K [t] is a polyomial of degree, ad that the coefficiet of t i this polyomial is [t ] det ti + A 1. I other words, the polyomial det ti + A K [t] is moic of degree. Thus, Propositio 5.7 applied to p det ti + A shows that the elemet det ti + A of K [t] is regular. This proves Corollary 5.8. A square matrix whose determiat is regular ca be cacelled from equatios, as the followig lemma shows: Lemma 5.9. Let N. Let A K. Assume that the elemet det A of K is regular. Let m N. a If B K m ad C K m are such that AB AC, the B C. b If B K m ad C K m are such that BA CA, the B C.

31 The trace Cayley-Hamilto theorem page 31 Proof of Lemma 5.9. Defie a elemet a of K by a det A. Recall that the elemet det A of K is regular. I other words, the elemet a of K is regular sice a det A. Theorem 3.7 yields A adj A adj A A det A I. a Let B K m ad C K m be such that AB AC. We must prove that B C. We have Thus, adj A A det A I ab adj A AB AC B det }{{ A} a I B ab. B adj A A C det }{{ A} a det A I I C ac. C Lemma 5.6 thus yields B C. This proves Lemma 5.9 a. b The proof of Lemma 5.9 b is similar to the proof of Lemma 5.9 a but ow we eed to work with BA adj A ad CA adj A istead of adj A AB ad adj A AC. The details are left to the reader The adjugate of a product Corollary 5.8 ca be put to use i several circumstaces. Here is a simple example: Theorem Let N. Let A ad B be two -matrices. The, adj AB adj B adj A. Theorem 5.10 is the statemet of [Gribe15, Exercise 6.33]; see [Gribe15, solutio of Exercise 6.33] for a proof of this theorem. We shall show a differet proof of it ow. We begi by showig a particular case of Theorem 5.10: Lemma Let N. Let A ad B be two -matrices. Assume that the elemets det A ad det B of K are regular. The, adj AB adj B adj A. Proof of Lemma Theorem 3.7 yields A adj A adj A A det A I. Theorem 3.7 applied to B istead of A yields B adj B adj B B det B I. Theorem 3.7 applied to AB istead of A yields AB adj AB adj AB AB det AB I.

32 The trace Cayley-Hamilto theorem page 32 Now, Comparig this with A B adj B adj A A det B I }{{ } det B I det B A adj A det B A adj A det B det A I det A I det A det B I. AB adj AB det AB I det A det B I, det A det B we obtai AB adj B adj A AB adj AB. Lemma 5.9 a applied to, B adj B adj A ad B adj AB istead of m, B ad C therefore yields B adj B adj A B adj AB sice the elemet det A of K is regular. Thus, Lemma 5.9 a applied to, B, adj B adj A ad adj AB istead of m, A, B ad C yields adj B adj A adj AB sice the elemet det B of K is regular. This proves Lemma We ow derive Theorem 5.10 from this lemma: Proof of Theorem Defie the K-algebra homomorphism ε : K [t] K as i Propositio 5.3. Defie two matrices à ad B i K [t] by à ti + A ad B ti + B. From à ti + A, we obtai ε adj à ε adj ti + A adj A by Propositio 5.3 b. Similarly, ε adj B adj B. From à ti + A, we obtai ε à ε ti + A A by Propositio 5.3 c. Similarly, ε B B. Corollary 5.8 shows that the elemet det ti + A of K [t] is regular. I other words, the elemet det à of K [t] is regular sice à ti + A. Similarly, the elemet det B of K [t] is regular. Lemma 5.11 applied to K [t], à ad B istead of K, A ad B thus yields adj à B adj B adj Ã. Applyig the map ε to both sides of this equality, we obtai ε adj à B ε adj B adj à ε adj B ε adj à adj B adj A by Propositio 5.1 c, applied to K [t], K, ε,,, adj B ad adj à istead of L, M, f, m, p, A ad B adj B adj A.

33 The trace Cayley-Hamilto theorem page 33 Hece, adj B adj A ε adj à B adj ε à B 25 by Propositio 5.2 c, applied to K [t], K, ε ad à B istead of L, M, f ad A. But Propositio 5.1 c applied to K [t], K, ε,,, à ad B istead of L, M, f, m, p, A ad B shows that ε à B ε à ε B AB. } {{ } } {{ } A B Hece, 25 becomes This proves Theorem adj B adj A adj ε à B adj AB. AB 5.4. Determiat ad adjugate of a adjugate Our ext target is the followig result: Theorem Let N. Let A be a -matrix. a If 1, the det adj A det A 1. b If 2, the adj adj A det A 2 A. Agai, we shall first prove it i a particular case: Lemma Let N. Let A be a -matrix. Assume that the elemet det A of K is regular. a If 1, the det adj A det A 1. b If 2, the adj adj A det A 2 A. Before we start provig Lemma 5.13, let us first recall the followig fact: If N, λ K ad C K, the det λc λ det C. 26 I fact, this is precisely [Gribe15, Propositio 6.12] applied to C istead of A. Proof of Lemma a Assume that 1. Theorem 3.7 yields A adj A adj A A det A I.

34 The trace Cayley-Hamilto theorem page 34 Now, det A adj A det det A I det A det I det A I 1 by 26 applied to det A ad I istead of λ ad C Thus, det A det A det A 1. det A det A 1 det A adj A det A det adj A. Hece, Lemma 5.5 applied to A K, a det A, b det A 1 ad c det adj A yields det A 1 det adj A sice det A is a regular elemet of K. This proves Lemma 5.13 a. b Assume that 2. Thus, 1 1 ad 2 1. Now, Lemma 5.11 a yields det adj A det A 1 det A det A 1 1 det A 2 sice 1 1 det A det A 2. But Theorem 3.7 applied to adj A istead of A yields Now, Hece, adj A adj adj A adj adj A adj A det adj A I. A adj A adj adj A A det adj A I det adj A detadj A I det A det A 2 A. det A det A 2 A det A det A 2 A A adj A det A I adj adj A det A I adj adj A det A adj adj A. Hece, Lemma 5.6 applied to, det A, det A 2 A ad adj adj A istead of m, a, B ad C yields det A 2 A adj adj A sice det A is a regular elemet of K. This proves Lemma 5.13 b. Let us ow derive Theorem 5.12 from this lemma:

35 The trace Cayley-Hamilto theorem page 35 Proof of Theorem Defie the K-algebra homomorphism ε : K [t] K as i Propositio 5.3. Defie a matrix à K [t] by à ti + A. Corollary 5.8 shows that the elemet det ti + A of K [t] is regular. I other words, the elemet det à of K [t] is regular sice à ti + A. We have ε adj à ε adj ti + A adj A by Propositio 5.3 ti +A b. Also, ε det à ε det ti + A det A by Propositio 5.3 a. ti +A a Assume that 1. Lemma 5.13 a applied to K [t] ad à istead of K ad A yields det adj à det à 1. Now, Propositio 5.2 a applied to K [t], K, ε ad adj à istead of L, M, f ad A yields Hece, ε det adj à det ε adj à det adj A. adj A det adj A ε det 1 adj à ε det à det à 1 det ε à det A det A 1. 1 sice ε is a K-algebra homomorphism This proves Theorem 5.12 a. b Assume that 2. Lemma 5.13 b applied to K [t] ad à istead of K ad 2 A yields adj adj à det à Ã. We have ε à ε ti + A ti +A A by Propositio 5.3 c. Propositio 5.2 c applied to K [t], K, ε ad adj Ã

36 The trace Cayley-Hamilto theorem page 36 istead of L, M, f ad A yields Thus, ε adj adj à adj ε adj à adj adj A. adj A adj adj A ε adj adj à ε det à 2 à 2 ε det à ε à εdet à 2 A sice ε is a K-algebra homomorphism det ε à det A This proves Theorem 5.12 b. 2 det à à by Propositio 5.1 applied to K [t], K, 2 ε,, à ad det à istead of L, M, f, m, A ad λ 2 A det A 2 A The adjugate of A as a polyomial i A Next, let us show that the adjugate of a square matrix A is a polyomial i A with coefficiets that deped o A, but are scalars ot matrices: Theorem Let N. Let A K. For every j Z, defie a elemet c j K by c j [ t j] χ A. The, adj A i0 c 1 i A i. Oe cosequece of Theorem 5.14 is that every -matrix which commutes with a give -matrix A must also commute with adj A. We prepare for the proof of Theorem 5.14 with two really simple facts:

37 The trace Cayley-Hamilto theorem page 37 Lemma Let N. Let u ad v be two elemets of {1, 2,..., }. Let λ K. Let A be a -matrix. The, λa u, v λa u, v. Proof of Lemma This follows from Lemma 3.8 applied to µ 0 ad B A. Propositio Let be a positive iteger. Let A K ad λ K. The, adj λa λ 1 adj A. Proof of Propositio Recallig the defiitios of adj λa ad adj A ad usig Lemma 5.15, the reader ca easily reduce Propositio 5.16 to 26 applied to 1 ad A j, i istead of ad C. Now, let me show a slightly simpler variat of Theorem 5.14: Lemma Let be a positive iteger. Let A K. For every j Z, defie a elemet c j K by c j [ t j] χ A. The, adj A 1 c 1 i A i. i0 Proof of Lemma For every j Z, defie a elemet c j K by c j [ t j] χ A. Propositio 3.9 shows that there exist matrices D 0, D 1,..., D 1 i K such that 1 adj ti A t k D k i K [t]. 27 Cosider these D 0, D 1,..., D 1. Thus, a -tuple D 0, D 1,..., D 1 of matrices i K is defied. Exted this -tuple to a family D k k Z of matrices i K by settig 7. Lemma 3.11 c applied to k 1 yields 1 c 1 i A i D 1 1 D i0 O the other had, defie the K-algebra homomorphism ε : K [t] K as i Propositio 5.3. This homomorphism ε satisfies ε t 0. Also, it satisfies ε u u for every u K. Hece, the map ε : K [t] K defied as i Defiitio 3.15 satisfies ε F F for every F K. 29

38 The trace Cayley-Hamilto theorem page 38 But Propositio 5.1 a applied to L K [t], M K, f ε ad m yields that the map ε : K [t] K is a homomorphism of additive groups. Hece, 1 ε t k D k ε t k D k εt k ε D k by Propositio 5.1 d applied to K[t], K, ε,, D k ad t k istead of L, M, f, m, A ad λ t k ε εt k sice ε is a rig ε D k D k by 29 applied to FD k homomorphism k 1 1 ε t D k 0 k D k 0 0 D k D k 0 1 k1 0 sice k 1 here, we have split off the added for k 0 from the sum 1 sice 0 {0, 1,..., 1} D 0 + 0D k D k1 0 But applyig the map ε to both sides of the equality 27, we obtai by 30. Thus, 1 ε adj ti A ε t k D k D 0 D 0 ε adj ti A ε adj ti + A adj A ti + A by Propositio 5.3 b, applied to A istead of A. Hece, 28 becomes This proves Lemma c 1 i A i D 0 adj A. i0 Fially, we are ready to prove Theorem 5.14:

39 The trace Cayley-Hamilto theorem page 39 Proof of Theorem We must prove the equality adj A i0 c 1 i A i. This is a equality betwee two -matrices, ad thus obviously holds if 0. Hece, we WLOG assume that 0. Thus, is a positive iteger. Hece, Propositio 5.16 applied to λ 1 yields Therefore, adj A 1 1 This proves Theorem adj A 1 1 adj A. adj A 1 c 1 i A i i0 by Lemma i Miors of the adjugate: Jacobi s theorem c 1 i A i. A mior of a matrix A is defied to be a determiat of a square submatrix of A. A theorem due to Jacobi coects the miors of adj A for a square matrix A with the miors of A. Before we ca state this theorem, let us itroduce some otatios: Defiitio Let N ad m N. Let A a i,j be a 1 i, 1 j m m-matrix. Let i 1, i 2,..., i u be some elemets of {1, 2,..., }; let j 1, j 2,..., j v be some elemets of {1, 2,..., m}. The, we shall use sub j 1,j 2,...,j v A as a syoym i 1,i 2,...,i u for the u v-matrix sub j 1,j 2,...,j v Thus, for every i {1, 2,..., } u ad j i 1,i 2,...,i u A. {1, 2,..., m} v, a u v-matrix sub j i A is defied. Defiitio If I is a fiite set of itegers, the I shall deote the sum of all elemets of I. Thus, I i. i I Defiitio If I is a fiite set of itegers, the w I shall deote the list of all elemets of I i icreasig order with o repetitios. For example, w {3, 4, 8} 3, 4, 8. The followig fact is obvious: Remark Let N. Let I be a subset of {1, 2,..., }. The, w I {1, 2,..., } I. Now, we ca state Jacobi s theorem 10 : 10 This is [Gribe15, Corollary 7.256]. It also appears i [Prasolov, Theorem 2.5.2] i a differet form.