MATH 210A, FALL 2017 Question 1. Let a and B Be Two N × N Matrices

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MATH 210A, FALL 2017 Question 1. Let a and B Be Two N × N Matrices MATH 210A, FALL 2017 HW 6 SOLUTIONS WRITTEN BY DAN DORE (If you find any errors, please email [email protected]) Question 1. Let A and B be two n × n matrices with entries in a field K. −1 Let L be a field extension of K, and suppose there exists C 2 GLn(L) such that B = CAC . −1 Prove there exists D 2 GLn(K) such that B = DAD . (that is, turn the argument sketched in class into an actual proof) Solution. We’ll start off by proving the existence and uniqueness of rational canonical forms in a precise way. d d−1 To do this, recall that the companion matrix for a monic polynomial p(t) = t + ad−1t + ··· + a0 2 K[t] is defined to be the (d − 1) × (d − 1) matrix 0 1 0 0 ··· 0 −a0 B C B1 0 ··· 0 −a1 C B C B0 1 ··· 0 −a2 C M(p) := B C B: : : C B: :: : C @ A 0 0 ··· 1 −ad−1 This is the matrix representing the action of t in the cyclic K[t]-module K[t]=(p(t)) with respect to the ordered basis 1; t; : : : ; td−1. Proposition 1 (Rational Canonical Form). For any matrix M over a field K, there is some matrix B 2 GLn(K) such that −1 BMB ' Mcan := M(p1) ⊕ M(p2) ⊕ · · · ⊕ M(pm) Here, p1; : : : ; pn are monic polynomials in K[t] with p1 j p2 j · · · j pm. The monic polynomials pi are 1 −1 uniquely determined by M, so if for some C 2 GLn(K) we have CMC = M(q1) ⊕ · · · ⊕ M(qk) for q1 j q1 j · · · j qm 2 K[t], then m = k and qi = pi for each i. Moreover, if ι: K,−! L is a field extension, then if we let ι(M) denote the matrix M considered as a matrix with coefficients in L, we have ι(M)can = ι(Mcan). In other words, the rational canonical form does not depend on which field we consider the coefficients of M to lie inside. Before we give the proof, let’s see why this implies the statement of the question. By the proposition, −1 −1 there exist matrices α; β 2 GLn(K) such that Acan = αAα and Bcan = βBβ . But then, letting ι: K,−! L be the field extension, we have: −1 −1 −1 ι(Bcan) = ι(β)ι(B)ι(β) = (ι(β)C)ι(A)(C ι(β) ) −1 Let γ = (ι(β)C) 2 GLn(L), so we have ι(Bcan) = γι(A)γ . By the uniqueness part of the proposition, this implies that ι(Bcan) = ι(A)can. By the part of the proposition regarding field extensions, we have ι(A)can = ι(Acan), so ι(Acan) = ι(Bcan). Since ι is injective, this implies that Acan = Bcan, i.e. that −1 −1 −1 −1 αAα = βBβ . Letting D = βα 2 GLn(K), this shows us that B = DAD . Now, let’s prove the proposition: 1 Q In particular, i pi is the characteristic polynomial of M and pm is the minimal polynomial of M. 1 Proof. This follows from the invariant factor form of the structure theorem for finitely generated modules over a principal ideal domain. Namely, if M is a n × n matrix, we may regard it as a linear transformation on V := Kn with respect to the standard ordered basis of Kn. Defining t · v = Mv, we may give V the structure of a K[t]-module. V is finitely generated as a K[t]-module (since it is finite-dimensional as a K-vector space), so by the invariant factor form of the structure theorem for finitely generated modules over a principal ideal domain, there is an isomorphism r ∼ ': K[t] ⊕ K[t]=(p1(t)) ⊕ K[t]=(p2(t)) ⊕ · · · ⊕ K[t]=(pm(t)) −! V (1) for p1; : : : ; pm 2 K[t] with p1 j p2 j · · · j pm, and this isomorphism is as a t-module. Note that r = 0, so we may omit the K[t]r factor: K[t] has infinite dimension as a K-vector space, but V has finite dimension n. Furthermore, this theorem tells us that the principal ideals (pi) are uniquely determined by the K[t]-module V (i.e. they are uniquely determined by M). Thus, the pi are unique up to multiplication by a unit of K[t]. We know that these units are exactly the non-zero elements of K, so if we add the requirement that the pi are monic, they are uniquely determined as elements of K[t]. If di = deg pi and e1; : : : ; em are the images of 1 2 K[t]=(pi(t)) under the canonical inclusions into d −1 d −1 dm−1 the direct sum, then b := fe1; t · e1; : : : ; t 1 · e1; e2; : : : ; t 2 · e2; : : : ; em; : : : ; t · emg is an ordered basis for the left-hand side. Since M(pi) is the matrix of the linear transformation t acting on K[t]=(pi(t)) with respect to the ordered basis 1; : : : ; tdi−1, we see that the matrix of the linear transformation t acting on the left-hand-side of (1) with respect to b is Mcan. Now, since ' is an isomorphism of K[t]-modules, we see that B := '(b) is an ordered basis of V and n that the matrix of t with respect to B is Mcan. Let f be the standard ordered basis of K ' V , and define a Pn matrix (Bij) by the equations fi = j=1 BjiBj , which are uniquely determined because B is a basis. Since Pn 0 f is also a basis, we may also uniquely write Bj = k=1 Bkjfk , so this tells us that n n X X 0 fj = BkiBijfk i=1 k=1 0 0 Since f is a basis, we may compare coefficients to see that B ·B = I, so B 2 GLn(K). Note that B ·B = I as well, i.e. B0 = B−1. This says that: 0 n 1 0 X 0 fj = BB fj = B @ BkjfkA = BBj k=1 −1 −1 We want to show that Mcan = BMB . This says that in the standard ordered basis f, BMB (fj) = Pn i=1(Mcan)ijfi. 2 We have, using the fact that Mcan is the matrix for M with respect to the basis B: −1 0 BMB (fj) = BMB (fj) 0 n 1 X 0 = BM @ BkjfkA k=1 = BM(Bj) 0 1 X = B @ (Mcan)ijBiA i X = (Mcan)ijB(Bi) i X = (Mcan)ijfi i −1 Now, assume that we have some C 2 GLn(K) with CMC = N := M(q1) ⊕ · · · ⊕ M(q`) with n q1 j q2 j · · · j qm and qi monic for all i. Let W be the K[t] module K with t acting by the matrix N with n respect to the standard basis f of K . Thus, W ' K[t]=(q1(t)) ⊕ · · · ⊕ K[t]=(q`(t)). We consider C as an ∼ P homomorphism V −! W , sending fj to i Cijfi. Then we have C(t · v) = CMv = NCv = t · (C(v)) for all v 2 V , so C is an homomorphism of K[t]-modules. Since C 2 GLn(K), it is bijective and thus an isomorphism of K[t]-modules. Thus, we have an isomorphism: K[t]=(p1(t)) ⊕ · · · ⊕ K[t]=(pm(t)) ' K[t]=(q1(t)) ⊕ · · · ⊕ K[t]=(q`(t)) Now, we may apply the uniqueness of the invariant factor form of the structure theorem for finitely generated modules over a principal ideal domain to conclude that ` = m and (pi) = (qi) as principal ideals. Thus, qi and pi differ by multiplication by a non-zero element of K, and since they are both assumed to be monic, we have pi = qi. Finally, we need to show that the rational canonical form is preserved by field extensions. This is a consequence of the above existence and uniqueness statements. Let ι: K,−! L be a field extension and consider the matrix ι(M) with coefficients in L. By the existence of rational canonical form, we know that −1 L there is a matrix B 2 GLn(K) with BMB = Mcan = i M(pi) with p1 j · · · j pm and pi 2 K[t] −1 L L monic. Applying ι, we have ι(B)ι(M)ι(B) = ι(Mcan) = i ι(M(pi)) = i M(ι(pi)), where ι(pi) is the polynomial pi 2 K[t] viewed as a polynomial with coefficients in L. Now, ι(pi) = qi is a monic polynomial in L[t], and the condition that pi j pi+1, i.e. pi+1 = pi · fi with fi 2 K[t], is also preserved by ι (since ι(pi+1) = ι(pi) · ι(fi) and ι(fi) 2 L[t]). Thus, ι(Mcan) is conjugate to ι(M) by ι(B) 2 GLn(L), L` and it is of the form i=1 M(qi) with qi monic and q1 j · · · j q`. Thus, by the above uniqueness statement, ι(Mcan) is in rational canonical form, so ι(Mcan) = ι(M)can. Note that this invariance under field extensions is not true for other normal forms of matrices, such as the normal form obtained by using the elementary divisor version of the structure theorem for principal ideal domains: if π(t) is irreducible as an element of K[t] but not of L[t], then the matrix corresponding to the K[t]-module K[t]=(πe) is not in its “elementary divisor form” as a matrix with coefficients in L.
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