Is there an invertible matrix that transposes? Quick question:
I was asked if there exists an invertible matrix $P$ over the complex numbers such that for any matrix $A$:
$PAP^{-1} = A^{T}$
I don't know how to prove it, but I don't think this is true. I know every matrix is similair to its transpose, but it can't be the same matrix $P$ for all matrices...So my gut feeling tells me no, but how do I show it?
 A: (promoted from a comment)
This solution was essentially given by the asker himself in a comment to the question.
First note that for $1\times 1$ matrices any invertible $P$ trivially works.
Then suppose such a magical $P$ also existed in general. Then for any matrices $A$ and $B$,
$$AB=P^{−1}A^TPP^{−1}B^TP=P^{−1}A^TB^TP=P^{−1}(BA)^TP=BA$$
where at first we used that $P$ works for both $A$ and $B$, and in the last equality we used that $P$ works for $BA$ too.
For 1×1 matrices this is no contradiction, but for matrices of greater size multiplication is not commutative.
A: As said in the comments, this solution seems to work.
Assume there is such a magical $P$ so that for any matrix $A$: $PAP^{-1}=A^{T}$
we can also write it as: $PA=A^{T}P$ (simply multiply by $P$ in the right side).
now lets look at $PABP^{-1}=(AB)^{T}=B^{T}A^{T}$.
$PABP^{-1}=A^{T}PBP^{-1} = A^{T}B^{T}$
Since matrix multiplication is not commutative, $A^{T}B^{T}=B^{T}A^{T}$ does not hold for all matrices. contradiction.
A: First, if you let $A=P,$ then you see that $P$ is symmetric. That means that it has an orthogonal basis of eigenvectors $e_1, \dots, e_n.$ Let $A$ be a matrix which sends $e_1$ to $e_2,$ $e_2$ to $e_3,$ etc ($e_n$ can go to $0.$) What happens to your matrices $P^{-1} A P$ and $A^t$ when applied to that basis?
EDIT As @julien points out, this argument works well if the ground field is the reals OR instead of transpose we use the hermitian adjoint (conjugate transpose). The argument does not work over $\mathbb{C}$ as stated.
A: You may also prove that, if $PSP^{-1}=S$ for all symmetric matrices $S$, then $P$ must be a scalar multiple of the identity matrix $I$ and hence $PAP^{-1}\equiv A$ for any (symmetric or not) matrix $A$. Since $A\ne A^T$ in general, the answer to your question is negative when $n\ge2$.
