Is A+B digonalizable if they share the same basis of eigenvectors I was given the following statement:

I know that the sum of two diagonalizable matrices is not allways diagonalizable,
but i'm not sure how the added element of the shared base contributes..
I would very much appreciate an explanation or some guidance..
 A: If they can be diagonalized using the same basis of eigenvectors, then $A = PD_AP^{-1}$ and $B = PD_BP^{-1}$ for some orthogonal matrix $P$ and diagonal matrices $D_A, D_B$. We then have
$$
A + B = PD_AP^{-1}+ PD_BP^{-1} = P(D_A+D_B)P^{-1}
$$
which is a diagonalisation.
Alternatively, let $v$ be an eigenvector of $A$ with eigenvalue $\lambda_A$ and at the same time an eigenvector of $B$ with eigenvalue $\lambda_B$. Then
$$
(A+B)v = Av + Bv = \lambda_Av + \lambda_Bv = (\lambda_A + \lambda_B)v
$$
so $v$ is also an eigenvector of $A + B$. Since the eigenvectors of $A$ and $B$ form a basis for the whole space, the same is then true for the eigenvectors of $A + B$, and thus $A+B$ is diagonalisable.
A: the same basis  is also a basis of $A+B$, because if $v$ is an element of this basis we have $(A+B)v=Av+Bv=\lambda v+\gamma v=(\lambda +\gamma)v$, so $v$ is an eigenvectors of $A+B$.
A: You can either see that, as matrices,
$A=PD_1P^{-1}, \quad B=PD_2P^{-1} \implies A+B=P(D_1+D_2)P^{-1},$
or simply verify that the eigenvectors of $A+B$ will be the same of $A$ and $B$, hence you'll have a basis of eigenvectors, and the linear map is thus diagonalizable.
