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.