## 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.