Diagonalize $f(A)= \begin{pmatrix} 1 & 0 \\ -1 & 3 \end{pmatrix} A $

Question

I have to diagonalize the endomorphism $f\in \mathrm{End(M_2}(\mathbb{R}))$ defined by $f(A)= \begin{pmatrix} 1 & 0 \\ -1 & 3 \end{pmatrix} A $

I know I can rewrite it, if $A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, as $f(A)= \begin{pmatrix} a & b \\ 3c-a & 3d-b \end{pmatrix} $,

but I don't know how to continue. Could you help me? Thanks in advance!

Answer 1

Hint: Choose a basis for $\mathrm M_2(\mathbb R)$, say $\{E_{11},E_{12},E_{21},E_{22}\}$. Calculate $f(E_{ij})$ and write it in the basis. This will give you matrix for $f$ (it's $4\times 4$). Proceed to diagonalize it as usual.

It might help if you think of $\mathrm M_2(\mathbb R)\cong \mathbb R^4$. Then $f$ becomes $$f(a,b,c,d) = (a,b,3c-a,3d-b).$$

Answer 2

Hint Let $\lambda$ an eigenvalue and $A\ne 0$ an associated eigenvector so

$$f(A)=\lambda A\iff \begin{pmatrix}1-\lambda &0\\ -1 & 3-\lambda\end{pmatrix}A=0$$

Since $A\ne0$ so $\begin{pmatrix}1-\lambda &0\\ -1 & 3-\lambda\end{pmatrix}$ should not be invertible hence $\lambda\in\{1,3\}$. Now for every value of $\lambda$ solve the equation $f(A)=\lambda A$ for $A=\begin{pmatrix}a &b\\ c & d\end{pmatrix}$ to find the associated eigenspace.