Lemma
Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $W$ be a $T$-invariant subspace of $V$. Suppose that $v_1, \dots , v_k$ are eigenvectors of $T$ corresponding to distinct eigenvalues. Prove that if $v_1+\dots+v_k$ is in $W$, then $v_i ∈ W$ for all $i$.
Use the above lemma and prove the given result.
Question
Prove that the restriction of a diagonalizable linear operator $T$ to any nontrivial $T$-invariant subspace is also diagonalizable.
I could prove the above lemma using induction.
My attempt: A linear operator $T$ on a finite-dimensional vector space $V$ is diagonalizable if and only if $V$ is the direct sum of the eigenspaces of $T$. Let $\lambda_1, \lambda_2,..., \lambda_k$ are the eigenvalues of $T$, $V=E_{\lambda_1} \bigoplus E_{\lambda_1}\bigoplus ...\bigoplus E_{\lambda_k}.$ Let $W_{\lambda_j}=E_{\lambda_j}\cap W(1\leq j \leq k)$. Will we get $W=W_{\lambda_1} \bigoplus W_{\lambda_1}\bigoplus ...\bigoplus W_{\lambda_k}?$
Claim 1:
$W=W_{\lambda_1} \bigoplus W_{\lambda_1}\bigoplus ...\bigoplus W_{\lambda_k}.$
$$V=E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_1}.$$ $$V\cap W=W \cap (E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_k}).$$
Claim 2:
$W \cap (E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_k})=(W \cap E_{\lambda_1}) + (W\cap E_{\lambda_1})+ ...+ (W\cap E_{\lambda_k})$
$x\in (W \cap E_{\lambda_1}) + (W\cap E_{\lambda_1})+ ...+ (W\cap E_{\lambda_k})\implies$ $x=x_1+x_2+...x_k \implies x \in W \cap (E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_k})$. How do I prove the converse? Does there exist any counter example to prove that result is false? How do I use the above lemma and prove the result?
