I know it's been answered before (at least to the case with $n$ different eigenvalues) but I didn't find a proof for the general case, and I would like some help with this question.
We are given linear transforms $S,T: V\to V$ where $V$ is some vector space.
We are given that $S$ and $T$ commute, $ST=TS$, and that they are diagonalizable:
$T=PD_1P^{-1}$ and $S=KD_2K^{-1}$, where $D_1, D_2$ are diagonal and $K,P$ are invertible.
We are asked to show that $S$ and $T$ have a common eigenspace.
My solution
Maybe I understood the question wrong, but what I tried to do is show that if $v$ is an eigenvector of $S$ then it is also an eigenvector of $T$.
let $Sv=\lambda v$.
$STv=TSv=T\lambda v=\lambda Tv$ which implies that $Tv$ is an eigenvector of $S$ with eigenvalue $\lambda$.
Why does that mean that $v$ is an eigenvalue of $T$?
Another possible way to solve this question is write:
$PD_1P^{-1}KD_2K^{-1} = KD_2K^{-1}PD_1P^{-1}$ and get that $P=K$ but I don't know how to do that either.
Best Answer
The result which's known for two (or more) diagonalizable matrices that commute is that are simultaneous diagonalizable, that's they are diagonalizable in the same basis. Let's prove it by induction on the dimension $\dim E$ (the result is trivial if $\dim E=1$):
If $S$ or $T$ is an homothetie then the result is obvious. Now assume that neither $S$ nor $T$ is an homothetie and since $S$ is diagonalizable then $$E=\bigoplus_{\lambda\in\mathrm{sp}(S)}E_\lambda(S)$$ where $E_\lambda(S)$ is the eigenspace of $S$ associated to the eigenvalue $\lambda$. Since $S$ isn't an homothetie then $$\forall\lambda\in \mathrm{sp}(S)\;\;\dim E_\lambda(S)\le\dim E-1$$ and since $ST=TS$ then $ E_\lambda(S)$ is invariant by $T$. Let $T'=T_{| E_\lambda(S)}$ the restriction of $T$ to $ E_\lambda(S)$ so by hypothesis $S$ and $T'$ are simultaneous diagonalizable on $ E_\lambda(S)$ that's there's a basis $B_\lambda$ of $ E_\lambda(S)$ in which $T'$ is diagonal. In the basis $B=\cup B_\lambda$ the two matrices $S$ and $T$ are diagonal.