When restriction of a diagonalizable linear operator to an eigenspace is also diagonalizable ?

Question

Let $T, S$ be diagonalizable linear operators on $\mathbb R^n$ such that $TS = ST$. Let $E$ be an eigenspace of $T$. Is it true that the restriction of $S$ to $E$ is diagonalizable ?

Answer 1

The restriction of a linear operator $S$ to a subspace only exists (as a linear operator on that subspace) if that subspace is $S$-stable. Moreover the restriction of a diagonalisable linear operator $S$ to any $S$-stable subspace is diagonalisable. So what needs to be proved is that eigenspaces of $T$ are $S$-stable. This follows readily from $TS=ST$: if $v\in\ker(T-\lambda I)$ then $(T-\lambda I)(S(v))=S((T-\lambda I)(v))=S(0)=0$, which proves that $S(v)\in\ker(T-\lambda I)$.