TITLE: Diagonalizable transformation restricted to an invariant subspace is diagonalizable
QUESTION [43 upvotes]: Suppose $V$ is a vector space over $\mathbb{C}$, and $A$ is a linear transformation on $V$ which is diagonalizable. I.e. there is a basis of $V$ consisting of eigenvectors of $A$. If $W\subseteq V$ is an invariant subspace of $A$ (so $A(W)\subseteq W$), show that $A|_W$ is also diagonalizable.

I tried supposing $A$ has distinct eigenvalues $\lambda_1,\ldots,\lambda_m$, with $V_i=\{v\in V: Av=\lambda_i v\}$. Then we can write $V=V_1\oplus\cdots\oplus V_m,$ but I'm not sure whether it is true that
$$W=(W\cap V_1)\oplus\cdots\oplus (W\cap V_m),.$$
If it is true, then we're done, but it may be wrong.

REPLY [16 votes]: Here is a minor variation of Zorn's very nice argument. I'll use Zorn's notation: 
Let $w=v_1 + v_2 + \cdots + v_k$ be in $W$, each $v_i$ being a $\lambda_i$-eigenvector of $A$, and the $\lambda_i$ being distinct. 
It suffices to check that each $v_i$ is in $W$. 
But this is clear since 
$$v_i=\left(\prod_{j\neq i}\ \frac{A-\lambda_j\,I}{\lambda_i-\lambda_j}\right)(w)\quad.$$