Proof: are restriction to operators diagnolizable if the operator is? "Let $V$ be a vector space of finite dimension $n\ge1\ \ T:V\Rightarrow V$ a linear operator and $S$ a T-invariant space of $V$. Consider the restriction $$T|_S:S\Rightarrow S$$ and $$(T|_S)(w):=T(w) $$
If $T$ is a diagonal operator in $V$ prove that $T|_S$ will be diagolizable in $S$"
Ok so, as in matrix form the diagonal operator can be divided in "blocks" of  same eigenvalues, each block related to it`s eigenspace I thought about these blocks as T-invariant subspaces, and, as the matrix is diagonal the blocks must be diagonal too so the operator restricted to this eigenspaces would be diagonalizable.
My problem is basically how to work this ideas into the proof(Are them right?)
Thanks in advance.
 A: One way to turn that idea into a proof is to work in terms of a specific choice of eigenvectors which act as bases of the eigenspaces you mention.  To get there, I would use the characterization of a diagonalizable matrix as (taken from Diagonalizable Matrix)
A matrix is diagonalizable if there exists a basis of V consisting of 
eigenvectors of T. With respect to such a basis, T will be represented by a 
diagonal matrix. 

Now you can appeal to the definition of an eigenvector as well:
$$ Tv = \lambda v$$
to make statements about whether those eigenvectors are still eigenvectors of the restricted operator.
A: Ok, I came up with this trial now, and I'd like to know if it's acceptable in your opinion - my teacher is really strict when it comes to prove something:
Let $w\in S$ and $S=V(\lambda_k)$ a eigenspace of $T$ and $B=${$v_1,v_2,...,v_n $} a bases of eigenvectors of $V$ where $[T]_B$ is diagonal 
As $S$ is a subspace of $V$ let $B^S=${$v_1,...,v_k $} be a bases of  $S$  $k\le n$.
By the definition of a eigenvector we have $$T(v)=\lambda v $$
And by the defiinition of a application of a restrict operator $$(T|_S)(w):=T(w)$$
If we take $w=\alpha_1v_1+...+\alpha_k v_k \rightarrow$ 
$T(w)=\alpha_1T(v_1)+...+\alpha_k T(v_k)$ and as all this $v_i$ are associated with the same eigenvalue $\lambda_k$ thus we have:
$T(w)=\lambda_k(\alpha_1v_1+...+\alpha_k v_k$) which in fact is in the subspace $S$ holding the definition of the strict operator
So $(T|_S)(w)=\lambda_k(\alpha_1v_1+...+\alpha_k v_k$)
Now at least, for a convenient choice  $w=v_i(1\le i \le k)$ as the  vectors of the bases $B^S$ we can see that:
$$(T|_S)(v_i)=\lambda_kv_i $$
definig eigenvectors of $T|_S$, hence it must be diagonalizable since it holds for every choice of $1\le i \le k$, the whole bases of $S$.
I think I may be missing something about algebraic-geometric multiplicity but at the same time I have the sensation that if $T(v)=\lambda v $ holds already it`d be kinda redundant.
