Transpose map in $M(2,\mathbb{R})$

Question

Let $T$ the transpose map $T(A)=A^t$ for $A\in M(2,\mathbb{R})$. I want to find a basis such that $T$ is diagonal. I considered $T$ as a map from $R^4\rightarrow R^4$ where $T$ can be represented by $$ \begin{pmatrix} 1 & 0 & 0& 0\\0&0&0&1\\0&0&1&0\\0&1&0&0\end{pmatrix} $$ The characteristic polynomial is $(t-1)^3(t+1)$ and by finding basis for each of the two eigenspaces I got $\{(1,0,0,0),(0,0,1,0),(0,1,0,-1),(0,1,0,1)\}$.

Is this correct?

Answer 1

Presumably, you took as your basis of $M(2,\Bbb R)$ to be $$ \mathcal B = \left\{ \pmatrix{1&0\\0&0}, \pmatrix{0&1\\0&0}, \pmatrix{0&0\\0&1}, \pmatrix{0&0\\1&0} \right\} $$ or something to that effect. If so, then your representation of $T$ is correct, as is everything you've said about this transformation.

Answer 2

If we let $E_{ij} = e_i^T e_j$, you could take the basis of eigenvectors $E_{11}, E_{22}, E_{12}+E_{21}, E_{12}-E_{21}$.

The corresponding eigenvalues are $1,1,1,-1$ from which the result follows.