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I have to diagonalize the endomorphism $fin mathrm{End(M_2}(mathbb{R}))$ 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!

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 $4times 4$). 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 $$f(a,b,c,d) = (a,b,3c-a,3d-b).$$

Hint Let $lambda$ an eigenvalue and $Ane 0$ an associated eigenvector so

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

Since $Ane0$ so $begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}$ should not be invertible hence $lambdain{1,3}$. Now for every value of $lambda$ solve the equation $f(A)=lambda A$ for $A=begin{pmatrix}a &b\ c & dend{pmatrix}$ to find the associated eigenspace.