Every matrix over a field is similar to its transpose

Let A be a matrix over a field. Prove that A is similar to A^\mathsf{T}.

We begin with a simple lemma.

Lemma: Let A be an invertible matrix. Then A^\mathsf{T} is invertible, with (A^\mathsf{T})^{-1} = (A^{-1})^\mathsf{T}. Proof: Recall that (AB)^\mathsf{T} = B^\mathsf{T} A^\mathsf{T}. Now A^\mathsf{T}(A^{-1})^\mathsf{T} = (A^{-1}A)^\mathsf{T} = I^\mathsf{T} = I. \square

Let A be a square matrix over a field. By Theorem 21 on page 479 in D&F, there exist invertible matrices P and Q such that D = P(xI-A)Q is in Smith Normal Form. Recall that D is diagonal, and that the nonunit diagonal entries are precisely the invariant factors of A. Now D^\mathsf{T} = D, and in particular D = Q^\mathsf{T}(xI-A)^\mathsf{T}P^\mathsf{T} = Q^\mathsf{T}(xI-A^\mathsf{T})P^\mathsf{T}. Now P^\mathsf{T} and Q^\mathsf{T} are invertible, and so D is the Smith Normal Form of xI-A^\mathsf{T}. In particular, A and A^\mathsf{T} have the same invariant factors, and thus are similar.

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