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This is part of Problem 11.1.6 of Dummit and Foote. The problem reads

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

On my homework, I wrote

If $x\in Im(\phi^n)$, then for some $a\in V$, $x=\phi^n(a)=\phi^{n-1}(\phi(a))$, which implies $x\in Im(\phi^{n-1})$. Thus we have a sequence of vector subspaces ordered by inclusion

$$ \cdots\subseteq Im(\phi^n)\subseteq\cdots\subseteq Im(\phi^2)\subseteq Im(\phi) $$

My professor said my argument is wrong, and I agreed after he explained why. The problem is that this was a few months ago, and I've forgotten his reasoning and can't locate my hardcopy of the homework.

If this argument isn't good enough, I would've just said $\phi(V)\subseteq V$, therefore $\phi^m(V)\subseteq\phi^{m-1}(V)$ and have the equivalent conclusion of the existence of a chain (which stabilizes due to the finite dimensionality of the vector space) and continued the rest of the argument (which I'm omitting). Is this approach okay?

Maybe it's something really obvious and 4:30AM is not the right time to think about this. Thanks a lot.

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1 Answer 1

up vote 3 down vote accepted

Your argument that the subspace $\mathrm{Im}(\phi^n)=\phi^n(V)$ weakly decreases with $n$ is correct, so it stabilises to some subspace $W$. However one need not have $W=\{0\}$; indeed this would mean that $\phi$ is nilpotent which is certainly not always the case. Nonetheless you can complete the argument: $W$ is manifestly $\phi$-stable, and the restriction of $\phi$ to $W$ has zero kernel (otherwise the chain wouldn't have stabilised at $W$), so it is invertible $W\to W$. But then $(\phi|_W)^n:W\to W$ is invertible, and $W\cap\ker\phi^n=\{0\}$

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