0
$\begingroup$

Theorem 7.3. Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and lett $\lambda_1,\lambda_2,...,\lambda_k$ be the distinct eigenvalues of $T$. Then, for every $x\in V$, there exist vectors $v_i \in K_{\lambda_i}$, $1\le i \le k$, such that

$x = v_1 + v_2 + \cdots +v_k$.

Proof. The proof is by mathematical induction on the number $k$ of distinct eigenvalues of $T$. First suppose that $k = 1$, and let $m$ be the multiplicity of $\lambda_1$. Then $(\lambda_1 - t)^m$is the characteristic polynomial of $T$, and hence $(\lambda_1 I - T)^m = T_0$ by the Cayley-Hamilton theorem(p.317). Thus $V=K_{\lambda_i}$, and the result follows.

where $K_{\lambda_i}$ = $\{v \in V: (T-\lambda_1 I)^p(v) = 0$ for some positive integer $p$}.

I wanted to know why it follows that $V=K_{\lambda_i}$?

$\endgroup$
1
$\begingroup$

Cayley-Hamilton tells you that the characteristic polynomial of your linear operator annihilates it. Hence, what you must write is $(\lambda_1I-T)^m=0_{\operatorname{End}(V)}$. Now, consider any $v\in V$. We have $$0=0_{\operatorname{End}(V)}(v)=(\lambda_1I-T)^m(v)$$ Hence, by the definition you gave of $K_{\lambda_1}$, we may conclude that $v\in K_{\lambda_1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.