Say $V = M_{2 \times 2}(\mathbb{R})$, and $$T\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = \begin{bmatrix} c & d \\ a & b \end{bmatrix}$$
How to find the eigenvalues of T and a ordered basis $B$ for $V$ that $[V]_B$ is a diagonal matrix?
My approach is to find the matrix representation of this transfromation, and find the characteristic polynomial. However, I am stuck in finding such a matrix representation.
I guess the matrix would be $4 \times 4$, but it does not make sense for $M_{4 \times 4}(\mathbb{R}) \times M_{2 \times 2}(\mathbb{R}) = M_{2 \times 2}(\mathbb{R})$, if we follow the "definition" of a transformation matrix?
The operator $\;T\;$ simply changes rows, so;
$$\begin{pmatrix}c&d\\a&b\end{pmatrix}=\color{red}T\begin{pmatrix}a&b\\c&d\end{pmatrix}=\color{red}{\begin{pmatrix}0&1\\1&0\end{pmatrix}}\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
Thus, this operator's action on some basis (say the usual one) of $\;M_{2\times2}(\Bbb R)\;$ isw pretty simple:
$$\begin{align*}&\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix}\\{}\\ &\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&1\end{pmatrix}\\{}\\ &\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&0\end{pmatrix}\\{}\\ &\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix}=\begin{pmatrix}0&1\\0&0\end{pmatrix}\end{align*}$$
and thus
$$[T]=\begin{pmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{pmatrix}$$
This is a symmetric matrix and thus diagonalizable. Check it characteristic polynomial is
$$(x^2-1)^2=(x-1)^2(x+1)^2\;,\;\;\text{with eigenvalues}\;\;\pm1$$
each of geometric multiplicity two, and thus its JCF is
$$[T]_J=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&\!-1&0\\0&0&0&\!-1\end{pmatrix}$$