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!
Best Answer
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 $\ker(f^m)\cap Im(f^m)= \{0\}$ .