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> Let $\varphi,\psi$ be endomorphisms on a finite-dimensional vector space over a field $K$ such that $\varphi\circ\psi=\psi\circ\varphi$. Assume $V$ has a basis of eigenvectors of $\psi$ and a basis of eigenvectors of $\varphi$. Prove that $V$ has a basis consisting of vectors that are eigenvectors of both $\varphi$ and $\psi$.
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**My work so far:**

Let $\lambda\_1,\ldots,\lambda\_n$ be the distinct eigenvalues of $\varphi$. For each $\lambda\_i$, let $c\_{i,1},\ldots,c\_{i,m\_i}$ be the eigenvectors of $\varphi$ corresponding to $\lambda\_i$. By assumption, $\varphi$ is diagonalizable, so we can write
$$
V\cong V\_1\oplus\cdots\oplus V\_n
$$
where each $V\_i=\text{eig}(\varphi,\lambda\_i)$. Using the fact that $\varphi$ and $\psi$ commute, it is easy to show that for all $v\_i\in V\_i$, $\psi(v\_i)$ is an eigenvector of $\varphi$ corresponding to $\lambda\_i$, so $\psi(v\_i)\in V\_i$. Apparently, I need this fact in order to show that since $\psi$ is also diagonalizable by assumption, we can find a basis $\{d\_{i,1},\ldots,d\_{i,m\_i}\}$ for each $V\_i$ consisting of eigenvectors of $\psi$. From here it follows that each $d\_{i,j}$ is an eigenvector of $\varphi$, since $\varphi(d\_{i,j})=\lambda\_id\_{i,j}$ ($\varphi$ acts as multiplication by $\lambda\_i$ on $V\_i$).

**My question:** I know that $\psi$ being diagonalizable implies I can decompose $V$ into eigenspaces corresponding to $\psi$, but how do I know I can find a basis of eigenvectors of $\psi$ for each $V\_i$? Like I said before, I believe this comes from the fact that $\psi(V\_i)\subset V\_i$, but I am not sure. Any ideas?