If $A$ and $B$ are similar, can the same be said of $A^t$ and $B^t$?

Question

If $A$ and $B$ are similar, can the same be said of $A^t$ and $B^t$?

$A$ similar to $B$ $\iff \exists P$ such that $A = PBP^{-1}$

Apply the transpose on both sides

$A^t = (PBP^{-1})^t = (P^{-1})^tB^tP^t = (P^t)^{-1}B^tP^t$

Can I conclude that they are similar?

Answer 1

A different and more high-brow approach: Every matrix is similar to its transpose. Hence $A^t$ is similar to $A$ is similar to $B$ is similar to $B^t$. By transitivity of similarity, the claim follows.

Answer 2

Your reasoning looks good. In case you're not certain about $(P^t)^{-1} = (P^{-1})^t$:

$$I = P P^{-1} \iff I^t =(PP^{-1})^t = (P^{-1})^t P ^t.$$

Since $I = I^t,$ we have that $(P^{-1})^t$ is a left inverse for $P^t$. Use $I = P^{-1}P$ for the right inverse and you have shown that it's ok to exchange transposing and taking the inverse (for an invertible matrix).