Let $V$ be a finite-dimensional vector space over a field $F$ and let $T:V\to V$ be a linear transformation. Let $W\subseteq V$ be a subspace such that $T(W)\subseteq W$. Suppose T is diagonalizable. Is $T$ restricted to $W$ also diagonalizable?
I have a proof for this which I'm hoping someone could verify, thank you.
Since $T$ is diagonalizable $\exists \beta=\{u_1,u_2,\cdots,u_n\}$ which is an ordered eigenbasis for T with $dim(T)=n$. Let $[T]_\beta$ be the corresponding matrix where the $j$th column is the eigenvalue associated with $u_j$.
We know that $dim(W) \leq dim(V)$, let $\beta_W$ be the basis of $W$ where $\beta_W \subseteq \beta$. Take $v_i \in \beta_W$, we know $T(v_i)=\lambda_iv_i$ since $v_i$ is an eigenvector in $V$. Since every $w \in W$ can be represented as a linear combination of $\beta_W$ by construction, we have that $[T]_{\beta_W}$ is a diagonal matrix with the $i$th diagonal entry equalling $\lambda_i$.
One thing I am not sure about is if I have sufficiently shown that $\beta_W$ can exist for W.
