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

$T_W$ is the $T$-invariant subspace.

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.

Now suppose that for some integer $k>1$, the result is established whenever $T$ has fewer than $k$ distinct eigenvalues, and suppose that $T$ has $k$ distinct eigenvalues. Let $m$ be the multiplicity of $\lambda_k$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(t) = (t - \lambda_k)^{m}g(t)$ from some polynomial $g(t)$ not divisible by $(t-\lambda_k)$. Let $W = R((T-\lambda I)^m)$. Clearly $W$ is $T$-invariant. Observe that $(T-\lambda_k I)^m$ maps $K_{\lambda_i}$ onto itself for $i<k$. For suppose that $i<k$. Since $(T - \lambda I)^m$ maps $K_{\lambda_i}$ into itself and $\lambda_k \ne \lambda_i$, the restriction of $T-\lambda_k I$ to $K_{\lambda_i}$ is one-to-one and hence is onto. One consequence of this is that for $i<k$, $K_{\lambda_i}$ is contained in $W$, and hence $\lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_{\lambda_i}$.

So, I'm having some difficulty interpreting the second paragraph.

Let $W = R((T-\lambda I)^m)$. Clearly $W$ is $T$-invariant.

Is $W$ $T$-invariant because $(T-\lambda I)^m(v) = 0$ for some $v \in V$?

Observe that $(T-\lambda_k I)^m$ maps $K_{\lambda_i}$ onto itself for $i<k$

I'm unable to see why $(T-\lambda_k I)^m$ maps $K_{\lambda_i}$ onto itself for $i<k$

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These two facts are both true because the relevant operators commute.

First let's recall that we say that $W$ is $T$-invariant if $T(W) \subseteq W$.

If $x \in W$, then by definition of $W$, $x = (T-\lambda I)^m y$ for some $y \in V$. As a result, $Tx = T(T-\lambda I)^m y = (T-\lambda I)^m Ty \in R((T-\lambda I)^m) = W$. This shows that $T(W) \subseteq W$ as desired.

For the second part, by the justification given in the proof, it is enough to see that $(T-\lambda_k)^m$ maps $K_{\lambda_i}$ into itself.

So suppose $x \in K_{\lambda_i}$. This means that there is a $p$ such that $(T-\lambda_i)^p x = 0$. Then we have $$(T-\lambda_i)^p (T-\lambda_k) x = (T- \lambda_k) (T-\lambda_i)^p x = (T- \lambda_k) 0 = 0$$ and so $(T-\lambda_k)x \in K_{\lambda_i}$ which is what we wanted to show.

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  • $\begingroup$ For the second part, I thought we were trying to show that $(T-\lambda_k)^m$ maps $K_{\lambda}$ into itself, so since $0 \in K_{\lambda_i}$ , $(T-\lambda_k I)^{m}0=0 \in K_{\lambda_i}$? I don't know if I'm misunderstanding your reasoning. $\endgroup$ Commented Aug 5, 2018 at 21:07
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    $\begingroup$ $(T-\lambda_k)^m$ maps $K_{\lambda_i}$ into itself means that for any $x \in K_{\lambda_i}$ we have that $(T-\lambda_k)^m x \in K_{\lambda_i}$. It is not enough to check that this is true for some special $x \in K_{\lambda_i}$ like $x = 0$ since we want it to be true for all such $x$. My argument deals with an arbitrary $x \in K_{\lambda_i}$. $\endgroup$ Commented Aug 5, 2018 at 21:13

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