# Prove that there exists number $k\in \mathbb{N}$ such that $V = \operatorname{Ker}A^{k} \dot{+} \operatorname{Im}A^{k}$

Problem: Let A be linear operator A $$\in L(V)$$. Prove that there exists number $$k\in \mathbb{N}$$ such that $$V = \operatorname{Ker}A^{k} \dot{+} \operatorname{Im}A^{k}$$. Then prove that operator $$\left.A\right|_{\operatorname{Ker}A^{k}}$$ is nilpotent and operator $$\left.A\right|_{\operatorname{Im}A^{k}}$$ is regular.

My attempt: We know that $$A$$ can be written as sum of two operators: $$A = N + S$$ where $$N$$ is nilpotent and $$S$$ diagonalizable operator. Then I construct $$A^k = \sum_{j=0}^{k} {k\choose j}S^{k-j}N^{j}$$ where $$k = \operatorname{ind}(N)$$ but this does not give anything good.

I know that I need to show somehow that for every $$v\in V$$: $$v=a+b, a\in \operatorname{Ker}A^k, b\in \operatorname{Im}A^k$$ but I do not know how. I have no idea how to proceed.

Thanks for any help.

• To prove the $\oplus$, try show that $\mathrm {Ker}(\mathcal A^k) \cap \mathrm {Im}(\mathcal A^k) = 0$. – xbh Dec 3 '18 at 3:57

Maybe an alternative. $$\DeclareMathOperator\Ker{Ker} \DeclareMathOperator\im{Im}$$ Assume $$\dim V =n <\infty$$. Then if $$v \in V$$ that $$\mathcal A^k v = 0$$ for some $$k \in \mathbb N$$, then $$0= \mathcal A(\mathcal A^k v)=\mathcal A^{k+1}v$$, thus $$\Ker (\mathcal A^k) \subseteq \Ker (\mathcal A^{k+1})$$. Also if $$u = \mathcal A^{k+1}v$$, then $$u = \mathcal A^k (\mathcal A v)$$, hence $$\im(\mathcal A^{k+1}) \subseteq \im (\mathcal A^k)$$. Thus we have two chains $$0\subseteq \Ker \mathcal A \subseteq \Ker(\mathcal A^2) \subseteq \dots \subseteq \Ker(\mathcal A^m) \subseteq \cdots \subseteq V$$ and $$V \supseteq \im\mathcal A \supseteq \im(\mathcal A^2) \supseteq \dots \supseteq \im(\mathcal A^m) \supseteq \cdots \supseteq 0.$$ Hence we have the dimension of $$\Ker (\mathcal A^j)$$ increases, the dimension of $$\im(\mathcal A^j)$$ decreases. If either sequence of them is strictly monotonic, then they would be unbounded, contradicting the fact $$\dim V <\infty$$. So there exists some $$k\in \mathbb N$$ s.t. $$\dim(\Ker(\mathcal A^k)) = \dim(\Ker(\mathcal A^{k+1}))$$. Then for each $$p\in \mathbb N^*$$, if $$\mathcal A^{k+p+1} v =0$$, then $$\mathcal A^{k+1} (\mathcal A^p v)$$, i.e. $$\mathcal A^p v \in \Ker(\mathcal A^{k+1}) = \Ker(\mathcal A^k)$$, thus $$\mathcal A^{p+k} v =0$$. Thus $$\Ker(\mathcal A^{p+k+1}) = \Ker(\mathcal A^{p+k})$$. Therefore $$\Ker(\mathcal A^k) = \Ker(\mathcal A^{k+p})$$ for all $$p \in \mathbb N^*$$. Now use the rank-nullity theorem, we have $$\dim(\Ker(\mathcal A^j)) + \dim(\im(\mathcal A^j)) = \dim V$$ for each $$j \in \mathbb N$$. Thus for the same $$k$$, $$\im(\mathcal A^k) = \im (\mathcal A^{k+p})$$ for each $$p \in \mathbb N^*$$.

Now we easy to see that $$\mathcal A\vert_{\Ker(\mathcal A^k)}$$ is nilpotent with index $$k$$: for each $$u \in \Ker(\mathcal A^k)$$, $$\mathcal A^k u =0$$. For $$v \in \im(\mathcal A^k)$$, there is some $$w\in V$$ that $$v = \mathcal A^k w$$. Suppose $$\mathcal Av = 0$$, then $$\mathcal A^{k+1}w =0$$, so $$w \in \Ker (\mathcal A^{k+1}) = \Ker(\mathcal A^k)$$, thus $$v =\mathcal A^k w = 0$$. Therefore $$\mathcal A\vert_{\im(\mathcal A^k)}$$ is invertible.

For the $$\oplus$$, we only need to show that $$\Ker(\mathcal A^k) \cap \im(\mathcal A^k) =0$$. Suppose $$x$$ is in this intersection, then $$x = \mathcal A^k y$$ for some $$y\in V$$. Also $$\mathcal A^k x =0$$, so $$y \in \Ker(\mathcal A^{2k}) = \Ker(\mathcal A^k)$$, i.e. $$\mathcal A^k y = 0$$, hence $$x = 0$$. Now the proof is completed.

• This proof is remarkably beautiful, thank you very much. – Thom Dec 3 '18 at 7:06
• @Thom You are welcome. Glad to help. – xbh Dec 3 '18 at 7:07

Hint: Note that $$A = N \dot+ P$$, where $$N$$ is nilpotent and $$P$$ is invertible. With that, we have $$A^k = 0 \dot + P^k$$.

The problem asks you to decompose $$V$$ as $$V=A^k(V) + A^{-k}(0)$$. If you are familiar with the primary decomposition theorem and its proof, we can prove this statement using the very similar argument. Here's some details.
Let $$m(t)=t^kp(t)$$ be the minimal polynomial of $$A$$ where $$\gcd(t^k, p(t)) = 1$$. In case that $$k=0$$, $$A$$ is bijective and the statement holds trivially for $$k=1$$. So assume $$k\geq 1$$. Then, by Euclidean algorithm, we have for some $$f,g \in \mathbb{F}[t]$$, $$t^kf(t)+p(t)g(t)=1.$$That is, we have $$A^kf(A)+p(A)g(A)=I,$$and $$A^kf(A)v+p(A)g(A)v=v$$for all $$v\in V$$. The first term $$A^kf(A)v$$ belongs to $$A^k(V)$$, and the latter term satisfies $$A^k[p(A)g(A)v]=[A^kp(A)]g(A)v = 0,$$ hence belongs to $$A^{-k}(0)$$. This shows $$V = A^k(V) + A^{-k}(0)$$. Furthermore, dimension theorem says that $$\dim V = \dim A^k(V) + \dim A^{-k}(0)$$, so the sum is direct.
Now notice the fact that $$V = A^k(V) \oplus A^{-k}(0)$$ implies $$A^{2k}x =0 \Rightarrow A^k x=0.$$ We see that $$A^k(V)$$ and $$A^{-k}(0)$$ are both $$A$$-invariant. So $$A$$'s restrictions on both spaces are well-defined. If we write $$T=A|_{A^{-k}(0)}$$, it is straightforward that $$T^k=0$$, that is, $$T$$ is nilpotent. Finally, $$S= A|_{A^k(V)}$$ is regular if and only if $$Sv=0, v=A^kw\in A^k(V) \Rightarrow v=0.$$This is also straightforward since $$Sv = Av = A^{k+1}w = 0 \Rightarrow A^{2k}w=0 \Rightarrow A^kw = 0\Leftrightarrow v=0.$$