# Why is $(T-{\lambda_i})$ onto?

$$K_{\lambda_i}$$ = $$\{v \in V: (T-{\lambda_i} I)^p(v) = 0$$ 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 $$i. For suppose that $$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. One consequence of this is that for $$i, $$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_{\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.

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?