# Why does $V=K_{\lambda_i}$ for $k=1$?

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 $V=K_{\lambda_i}$?

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 $v\in K_{\lambda_1}$.