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Let $V$ be a finite dimensional vector space. If $f$ is any linear transformation from $V$ to $V$. Prove that there is an integer $m$ such that the intersection of the image of $f^m$ and the kernel of $f^m$ is $\{0\}$.

What I've tried:

For any linear transformation $f$ we have $f(0)=0$.

Thus $f\circ f\circ f\circ ....\circ f(0)=0$(m times) and hence $f^m(0)=0$.

Thus $0 \in \ker(f^m) $. And since $f^m(0)=0$ , $0 \in Im(f^m) $.

Then $0$ is in the intersection of $Im(f^m)$ and $\ker(f^m)$

Now for proving that their intersection is only $\{0\}$:

I'm guessing that this implies that either $Im(f^m)=0$ or $\ker(f^m)=0$.

For the first case, this would result in the zero map, but we can't get a zero map from multiple compositions of a non zero map (I think?).

This would imply that $\ker(f^m)=0$ for some $m$, and the function would turn injective after $m$ compositions?

Is my reasoning correct? Can anyone help me solve this or point me in the right direction?

Thanks in advance!

glS
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Raku
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    The guess actually fails. The result is $V = \mathrm {Ker}(f^m) \oplus \mathrm {Im} (f^m)$, but neither of them is $0$. You could easily find counterexamples even for $m = 1$. – xbh Dec 23 '17 at 12:43
  • Hint: try $m = \dim (V)$ – xbh Dec 23 '17 at 12:53
  • Possible duplicate of https://math.stackexchange.com/questions/669417/the-kernel-and-image-of-tn – lhf Dec 23 '17 at 12:54
  • Related: https://math.stackexchange.com/q/919231/173147 – glS Jun 05 '18 at 18:38

2 Answers2

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Since $$\ker(f)\subseteq \ker(f^2)\subseteq\cdots$$ are all subspaces of a finite dimensional space $V$, so there exists integer $m$ such that $\ker(f^m)=\ker(f^{m+1})=\cdots$. Then the claim is that $\ker(f^m)\cap Im(f^m)=\{0\}$.

So if $x\in\ker(f^m)\cap Im(f^m)$, then $f^m(x)=0$. Also there exists $y$ such that $f^m(y)=x$. Therefore $f^{2m}(y)=f^m(x)=0\Longrightarrow y\in\ker(f^{2m})=\ker(f^m)\Longrightarrow f^m(y)=x=0.$ Hence $\ker(f^m)\cap Im(f^m)\subseteq \{0\}$ and so $\ker(f^m)\cap Im(f^m)= \{0\}$ .

Prajwal Kansakar
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Assuming that you are working over $\mathbb C$, take a basis $B$ of $V$ such that the matrix of $f$ with respect to $B$ is in the Jordan normal form. Furthermore, suppose that the first $m$ vectors of $B$ are such that the $m$ first entries of the main diagonal and only those ones are equal to $0$. Then that $m$ is the one that you are interested in: $\ker f^m$ is generatad by the first $m$ vectors of $B$ and $\operatorname{Im}f^m$ is generated be the other vectors of the basis.

José Carlos Santos
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