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Theorem 7.1. Let $T$ be a linear operator on a vector space $V$, and let $\lambda$ be an eigenvalue of $T$. Define $K_\lambda=\{x\in V\mid (T-\lambda I)^p(x)=0\; \text{for some}\; p\in\mathbb{N}\}$. Then

(a) $K_{\lambda}$ is a $T$-invariant subspace of $V$ containing $E_{\lambda}$

(b) For any scalar $u \ne \lambda$, the restriction of $T-uI$ to $K_{\lambda}$ is one-to-one.

Theorem 7.2. Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits. Suppose that $\lambda$ is an eigenvalue of $T$ with multiplicity $m$. Then

(a) dim$(K_{\lambda}) \le m$

Proof. Let $W = K_{\lambda}$, and let $h(t)$ be the characteristic polynomial of $T_W$. By Theorem 5.21 (p.314), $h(t)$ divides the characteristic polynomial of $T$, and by Theorem 7.1(b), $\lambda$ is the only eigenvalue of $T_W$. Hence $h(t) = (-1)^d(t-\lambda)^d$, where $d$ = dim$(W)$, and $d \le m$.

I wanted to know why $\lambda$ is the only eigenvalue of $T_W$ by Theorem 7.1(b)?

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  • $\begingroup$ What is denoted $K_\lambda$? $\endgroup$ – Bernard Aug 4 '18 at 19:55
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    $\begingroup$ $K_{\lambda}$ is the eigenspace of $T$ associated with $\lambda$, I assume. If $(T - uI)|_{K_{\lambda}}$ is one-to-one, then $\mathrm{null}(T-uI) \cap K_{\lambda} = \{0\}$ and so $u$ is not an eigenvalue. 7.1 states this is true for all $u \neq \lambda$. $\endgroup$ – Morgan Rodgers Aug 4 '18 at 19:58
  • $\begingroup$ $K_{\lambda} = \{x \in V: (T-\lambda I)^p(x) = 0$ for some positive integer $p$\} $\endgroup$ – K.M Aug 4 '18 at 19:59
  • $\begingroup$ What does $T_W$ indicate? What does $E_\lambda$ indicate? It's rather difficult to answer questions like this, unless someone happens to own the text from which it is drawn, recall having seen it, and feel like referencing said text so that they can answer the question. $\endgroup$ – Cameron Buie Aug 4 '18 at 20:26
  • $\begingroup$ $T_W$ is the $T$-invariant subspace, and $E_{\lambda}$ is the eigenspace corresponding to $\lambda$ $\endgroup$ – K.M Aug 4 '18 at 20:46
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If $\mu$ is another eigenvalue of $T_W$ on $K_\lambda$ with the corresponding eigenvector $v\neq 0$ then $(T-\mu I)(v)=0$. However $T-\mu I$ is one-to-one on $K_\lambda$.

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