# Restriction of a diagonalizable linear operator $T$ to any nontrivial $T-$invariant subspace is also diagonalizable using the given lemma.

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 $v_1, v_2, . . . , v_k$ are eigenvectors of $T$ corresponding to distinct eigenvalues. Prove that if $v_1+v_2+· · ·+v_k$ 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 eigen values 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?

Let $w \in W$ Then $w \in V=E_{\lambda_{1}} \bigoplus E_{\lambda_{2}}\bigoplus ...\bigoplus E_{\lambda_{k}}$. So we can write $w=w_1+w_2+...+w_k$ where $w_i \in E_{\lambda_{i}}$ for each $i$. The lemma now tells you that $w_i \in W$ for each $i$. Claim 2 follows immediately from this.