TITLE: Range of linear operator restricted in generalized eigenspace
QUESTION [0 upvotes]: Let $T:V\to V$ be a linear operator on a finite-dimensional complex vector
space $V$. Let $\lambda_1,\lambda_2,...,\lambda_k$ be the distinct eigenvalues of $T$ and $K_{\lambda_i}$ be the generalized eigenspace of $\lambda_i$ for all $i$. Let $h(t)=\prod_{i=1}^{k}(t-\lambda_i)$.
Show that $$R(h(T))=\oplus_{i=1}^{k}R(h(T)|_{K_{\lambda_i}}) $$.
I know how to prove the existance, but got stuck at the uniqueness part (i.e. for any $x\in R(h(T))$ , the expression $x=v_1+v_2+...+v_k$ where $v_i\in R(h(T)|_{K_{\lambda_i}})$ is unique). Could you please give me some hints? Thank you.

REPLY [1 votes]: I assume you have shown $R(h(T))=R(h(T)|_{K_{\lambda_1}})+\cdots+R(h(T)|_{K_{\lambda_n}})$, and attack the uniqueness part.
Note that $K_i$ is invariant under $h(T)$ as $h(T)w_i=p(T-\lambda_i)w_i\in K_i, \forall w_i\in K_i$, where $p(x)$ is the polynomial defined by $p(x)=h(x+\lambda_i)$. Thus $R(h(T)|_{K_{_i}}\leq K_i$, and $V=\oplus K_i$ implies the uniqueness.
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