$\mathbf {The \ Problem \ is}:$ Suppose, $A$ be a $2×2$ real matrix with $tr(A) =0$ and $det(A) =-1. $ Then show that :
$(a)$ $\mathbb R^2$ has a basis consisting of eigenvectors of $A .$
$(b)$ Suppose, $T$ be the real matrix wrt the above basis $\beta$ such that $TA = AT.$ Prove that, $T$ is a diagonal matrix wrt that basis.
$\mathbf {My \ approach} :$ Actually, the eigenvalues of $A$ are $1$ and $-1 .$ Then, $A$ is diagonalisable and hence part $(a)$ is done . Name that ordered basis $\beta =\{v_1,v_2\} .$
Then, wrt the basis $\beta$, the matrix of $T$ be $T = \begin{pmatrix} x & z \\ y & w \\ \end{pmatrix}$ where $v_1 =\begin{pmatrix} x \\ y \\ \end{pmatrix}$ ; and $v_2 = \begin{pmatrix} z \\ w \\ \end{pmatrix}$ such that $Av_1 = v_1$ and $Av_2 = -v_2$, then we get, $TA = AT =\begin{pmatrix} x & -z \\ y & -w \\ \end{pmatrix}$; then we can get some system of equations, but how to show that $T$ is diagonal ???
A little help is greatly appreciated.
Let $v_{-1}, v_1$ be the eigenvectors corresponding to the eigenvalues $\lambda_{-1} = -1, \lambda_1 = 1$. Note that $\ker (A -\lambda_k I)$ is one dimensional.
Note that $A T v_k = TA v_k = \lambda_k T v_k$ and so $T v_k \in \ker (A -\lambda_k I)$. In particular, $Tv_k = \beta_k v_k$.
Hence in the basis $v_{-1},v_1$ we see that $T$ has the form $\operatorname{diag}(\beta_{-1}, \beta_1)$.