$V$ is a vector space of dim $= n$. $T$ is a diagonalizable linear operator on $V$.
If $U \subset V$ is any invariant subspace, i.e. for $u \in U$, $T(u) \in U$, then SHOW that the following induced map is also diagonalizable:
$T|_U : U \to U$
the above is just $T$ restricted to $U$.
I guess we also need to assume dim U $\geq 1$.
My thoughts: (feel free to ignore).
We know there exists an eigenvector basis of $V$ , $\beta = \{v_1, ..., v_n \}$ so that $[T]_\beta$ is a diagonal matrix.
But for any invariant subspace $U$ with dim U = $m \leq n$, I cannot conclude a subset $\{v_{n_1}, ..., v_{n_m} \}$ is a basis for $U$, where $1 \leq n_i \leq n$ for $i = 1, ..., m$. Can I? If not, how else to tackle this problem.
I tried using induction on the dimension of $U$, but in the induction step, I still need to express a $(m+1)-$dimensional invariant subspace as a direct sum of $m-$dimensional invariant subspace plus a 1-dimensional invariant subspace, which is not obvious to me.
First factor out the kernel $\ker T\vert_U = \ker T \cap U $ since this portion is the zero operator, wich is diagonalizable.
Suppose then $T\vert_U$ invertible. Take a base $\alpha$ of $U$ and then complete it to $\gamma$ a base of $V$.
$$ [T]_\gamma = \begin{bmatrix} [T\vert_U]_\alpha & 0\\ 0 & B \end{bmatrix} $$
for some matrix $B$, now we can diagonalize it, take $S = \begin{bmatrix} S_{11} & S_{22} \\ S_{12} & S_{22} \end{bmatrix}$ the coordinate transformation with $S[T]_\gamma = D S$ where $D = \begin{bmatrix} D_{11} & 0 \\ 0 & D_{22} \end{bmatrix}$ is diagonal, then
$$ S[T]_\gamma = \begin{bmatrix} S_{11}[T\vert_U]_\alpha & 0\\ 0 & S_{22} B \end{bmatrix} = \begin{bmatrix} D_{11}S_{11} & 0\\ 0 & D_{22} S_{22} \end{bmatrix} = DS $$
Then $S_{11}[T\vert_U]_\alpha = D_{11}S_{11}$, we have then to prove that $S_{11}$ is not singular. If it is, if $S_{11}v = 0$ then $S[T]_\gamma \begin{bmatrix} v \\ 0\end{bmatrix} = 0$, but we assumed $T$ invertible and $S$ coordinate transformation.