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Let $A : V \rightarrow V$ be an endomorphism. Then we have that $$V = \operatorname{im}(A-\lambda \operatorname{Id})^n \oplus \ker(A-\lambda\operatorname{Id})^n$$ for some $n \in \mathbb{N}$.

I tried to prove this by using the JNF, but apparently I do not get there somehow.

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For every endomorphism $T\colon V \to V$, you have $\dim V = \dim \ker T + \dim \operatorname{im} T$. That holds in particular if $T = S^n$ for an endomorphism $S$ and an $n\in\mathbb{N}$. Now, $\ker S^0 \subset \ker S^1 \subset \ker S^2 \subset \dotsb$, and $\DeclareMathOperator{\im}{im} \im S^0 \supset \im S^1 \supset \im S^2 \supset \dotsb$. What does that tell you? – Daniel Fischer Aug 29 '13 at 20:05
Thank you. this is actually great. – user66906 Aug 29 '13 at 20:15
I don't think you mean $\otimes$. – Dan Petersen Aug 29 '13 at 20:17
Probably $\oplus$ – Cortizol Aug 29 '13 at 20:33

1 Answer 1

up vote 3 down vote accepted

For brevity, I shall denote $A - \lambda\operatorname{Id}$ by $T$.

Aside remark: we always have the trivial decomposition $\DeclareMathOperator{\im}{im} V = \ker T^0 \oplus \im T^0$, but that is uninteresting, so let's assume $0 \notin \mathbb{N}$ here.

Further remark: if $\dim V = \infty$, such a decomposition need not exist, consider the space $V$ of all sequences, and $T(x_1,\,x_2,\,x_3,\,\dotsc) = (x_2,\,x_3,\,x_4,\,\dotsc)$. Then $\im T^n = V$ for all $n$ ($T$ has the right inverse $S\colon (x_1,\,x_2,\,\dotsc) \mapsto (0,\,x_1,\,x_2,\,\dotsc$), so $T^n \circ S^n = \operatorname{Id}$), but $\ker T^n \neq \{0\}$ for $n > 0$.

So we assume $\dim V < \infty$.

$\ker T \subset \ker T^2 \subset \ker T^3 \subset \dotsc$ cannot be a strictly increasing sequence of subspaces, since $\dim V < \infty$, so there is a smallest $n \geqslant 1$ with $\ker T^n = \ker T^{n+1}$. Then $\ker T^{n+k} = \ker T^n$ for all $k > 0$: if $v \in \ker T^{n+k}$, then $T^{k-1}v \in \ker T^{n+1} = \ker T^n$, hence $T^{n+k-1}v = 0$, i.e. $v \in \ker T^{n+k-1}$, thus $\ker T^{n+k} \subset \ker T^{n+k-1}$, and the reverse inclusion is generally true, so $\ker T^{n+k} = \ker T^{n+k-1} = \dotsb = \ker T^n$.

Since $\dim V = \dim \ker T^k + \dim \im T^k$ for all $k$, the decreasing sequence of subspaces $\im T \supset \im T^2 \supset \im T^3 \supset \dotsc$ becomes stationary at the same $n$. $\im T^{n+1} = \im T^n$ implies $\im T^n \cap \ker T = \varnothing$, hence a fortiori $\im T^n \cap \ker T^n = \varnothing$, and by the dimension formula, that means

$$V = \im T^n \oplus \ker T^n.$$

As pointed out by YACP in the comment, for the decompositon, you can always choose $n = \dim V$, since that is the latest point at which the $\ker T^k$ sequence can possibly become stationary.

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This is a particular case of Fitting Decomposition Theorem. Unfortunately, most proofs don't mention that one can choose $n=\dim V$. – user26857 Aug 31 '13 at 7:49

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