Let $T$ be the linear operator on $M_{n\times n}(\mathbb{R})$ definded by $T(A)=A^t$. I am struggling to find an order basis such that [T]_B is a diagonal matrix

It is easy to see that the eigenvalue of $T$ is $\pm 1$

This basically just changes order of elements of the matrix.

It might be useful to check $2\times2$ case first.

I am not sure about that case either

You should get that $T(A)=A$ for $A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$.

Similarly $T(A)=-A$ for $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.

It is clear that $T(D)=D$ for any diagonal matrix, including $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $\begin{pmatrix}0&0\\0&1\end{pmatrix}$.

I guess it should be clear that similar pattern works for $n\times n$.

BTW there are some posts about this on the main, I was able to find this: The eigenvectors of the transpose operator.

I don't get $A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$

What is the transpose of this matrix?

A

Well, then you have $T(A)=A$, so it is an eigenvector for the eigenvalue $A$.

I'd guess the hint suggested in the linked questions explains it well.

$A^T=A$ is the condition that describes symmetric matrices - so you want basis of the subspace of symmetric matrices.

Similarly, $A^T=-A$ describes antisymmetric matrices.

I see now, as $n$ getting larger, the order basis will get more elements

@MartinSleziak, thanks the help, I had stuck at this question for an hour