Can matrix transposition be considered as a change of basis?

Question

For a given field $\mathbb{K}$ and positive integer $n$, can matrix transposition in $\mathcal{M}_n(\mathbb{K})$ be considered as a change of basis. In other terms, are there $P,Q\in\mathcal{M}_n(\mathbb{K})$ such that for every $A\in\mathcal{M}_n(\mathbb{K})$, $A^\top=PAQ$ ?

Answer 1

Consider the case of $A=Q^{-1}$: we get $$ (Q^{-1})^T=PQ^{-1}Q=P $$ which is the same as $Q=(P^{T})^{-1}$, so $A^T=PA(P^T)^{-1}$ or $$ (PA)^T=PA $$ for all $A$.

This is clearly impossible: take $B$ a non symmetric matrix and $A=P^{-1}B$.

Answer 2

No, because that would make $\ker(A^T) = \ker(A Q) = Q^{-1}(\ker(A))$ depend only on $\ker(A)$. But $\ker(A^T)$ can be any subspace of $\mathbb K^n$ with the same dimension as $\ker(A)$.

Answer 3

Well there are really two questions here. First $V=M_n(K)$ is a vector space, it happens to be matrices but still a $K$-vector space. And transpose is $K$-linear and invertible (it is its own inverse). So yes, as a vector space transposition is a linear invertible function $V\to V$ and therefore a change of basis. If you think of the unit vectors $E_{ij}$ then it is in fact just a permutation of the ordering of these, i.e. $E_{ij}^t=E_{ji}$. That is one reason it is tempting to think of this as transform as concerning a change of basis.

Now the way the second part of your question goes is not true. The change of basis cannot be performed by simply multiply $PAQ$ (as demonstrated by @egreg). The defect at the heart of this is that $(AB)^t=B^t A^t$. Composing $A$ with other elements of the ring $M_(K)$ will remain in the ring, not suddenly take you to its opposite ring. (Opposite rings are where you purposefully swap the order of multiplication.)

But there is a sense in which you are close to correct. It is through the Skolem-Neother Theorem. It says

If $\varphi:M_n(K)\to M_n(K)$ is a $K$-linear ring isomorphism then there is an invertible matrix $X$ such that $\varphi(A)=XAX^{-1}$.

As a corollary, if $\varphi:M_n(K)\to M_n(K)$ is any $K$-linear isomorphism where $\varphi(AB)=\varphi(B)\varphi(A)$ then there is an invertible matrix $X$ such that $\varphi(A)=XA^t X^{-1}$.

So in sense no transpose is conjugation, but any two transposes will differ only by conjugation.