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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.

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

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  • $\begingroup$ This proof is remarkably beautiful, thank you very much. $\endgroup$ – Thom Dec 3 '18 at 7:06
  • $\begingroup$ @Thom You are welcome. Glad to help. $\endgroup$ – xbh Dec 3 '18 at 7:07
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Hint: Note that $A = N \dot+ P$, where $N$ is nilpotent and $P$ is invertible. With that, we have $A^k = 0 \dot + P^k$.

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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 $$ Sv = Av = A^{k+1}w = 0 \Rightarrow A^{2k}w=0 \Rightarrow A^kw = 0\Leftrightarrow v=0. $$

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