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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 $K_{\lambda_i}$.

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.

I'm unable to see why $(T-\lambda_k)$ being restricted to $K_{\lambda_i}$ and being one-to-one makes $(T-{\lambda_i})$ 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 $K_{\lambda_i}$.

I was wondering if someone could also further elaborate why this is so.

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You write

I'm unable to see why $(T−\lambda_i)$ being restricted to $K_{\lambda_i}$ and being one-to-one makes $(T−\lambda_i)$ onto?

This is because the vector space is finite-dimensional, and a one-to-one linear map from a finite-dimensional vector space to itself is onto. (Rank-nullity formula).

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  • $\begingroup$ from the rank-nullity formula, I know that dim V = dim null(T) + dim range(T), but how do I apply that to the linear map being onto? $\endgroup$
    – Skm
    Aug 6, 2018 at 3:45

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