# Prove $T|_{V_\lambda}$ is diagonalizable

Let $V$, an $n$-dimensional vector space and let $T, S:V\to V$, two diagonalizable linear operators. Show that if $TS=ST$ then every $V_\lambda$ of $S$ is $T$-invariant and the restriction, $T|_{V_\lambda}$ is diagonalizable.

So first it easy to show that $V_\lambda$ (an eigensapce of $S$) is $T$-invariant. Let $v\in V_\lambda$. Then

$$S(T(v)) = T(S(v)) = T(\lambda v) = \lambda T(v)$$

So indeed, $T(v)$ is an eigenvector of $V_\lambda$.

Now, I need to show that $T|_{V_\lambda}$ is diagonalizable.

Since $ST=TS$ and $S,T$ are diagonalizable then we know that there's a basis $B$ such that $$[T]_B = \text{Diag}(\lambda_1, \ldots ,\lambda_n) \\ [S]_B = \text{Diag}(\mu_1, \ldots ,\mu_n)$$.

What should I do next? (Or is there another approach, not using the fact that $S, T$ are simultaneously diagonalizable?)

Now suppose $m(x)$ is the minimal polynomial for $T$. Then, $m(T|_{V_\lambda})=0$ so that the minimal polynomial $m_1(x)$ of $T|_{V_\lambda}$ divides $m(x)$. Since $m$ has no multiple roots, $m_1$ doesn't have multiple roots, too, so that $T|_{V_\lambda}$ is diagonalizable.