Question: Prove or disprove: if $A$ is an $n \times n$ complex matrix, and $\ker(A) \cap \operatorname{range}(A) \neq \{0\}$, is $A^n = 0_{n \times n}$?

My attempt: I've shown that the converse of this statement is true: that if $A^{n \times n} = 0$, then $\ker(A) \cap \operatorname{range}(A) \neq \{0\}$.

However, I haven't made any headway on this problem. Instinctively, I think that the statement is false. but I can't seem to come up with a counterexample - I've tried projection matrices, and "shifting" matrices, but both of those are nilpotent.

I've also tried to prove that the statement is true, but haven't made any progress. I've most been messing around with the Rank-Nullity theorem: we know that $\dim(C^n) = n$. Now, $\ker(A) \cap \operatorname{range}(A)$ is a subspace of $C^n$, since $\ker(A)$ and $\operatorname{range}(A)$ are subspaces of $C^n$. Let $\beta$ be a basis for $\ker(A) \cap \operatorname{range}(A)$. $\beta$ must contain at least on non-zero vector, by hypothesis. We can extend $\beta$ to a basis $\beta_r$ of $\operatorname{range}(A)$, and to a basis $\beta_k$ for $\ker(A)$. My idea from here was to extend either $\beta_r$ or $\beta_k$ to a basis for $C^n$, and then true to apply $L_A$ to vectors written in this basis, but that doesn't seem to be getting me anywhere.

If anyone knows of a good counterexample, I would appreciate seeing it. If the statement is true, a hint for a proof would be awesome. Thanks!

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    $\begingroup$ How about $A(e_1)=0, A(e_2)=e_1, A(e_3)=e_3$. We have $\ker(A)=span(e_1)$ and $range(A)=span(e_1,e_3)$ but $A^n(e_3)=e_3$ for all $n$. Now, what is the matrix that describes that operator? $\endgroup$ – JMoravitz Jun 17 '16 at 21:50
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    $\begingroup$ As an aside, for $A$ the zero matrix, the converse is still false. $range(0_{n\times n})=\{0\}=range(0_{n\times n})\cap \ker(0_{n\times n})$ despite $0^n = 0$ $\endgroup$ – JMoravitz Jun 17 '16 at 21:54
  • $\begingroup$ @JMoravitz ah yes, thank you, I had neglected to mention that an additional assumption on the converse proof was that $A$ was not the zero matrix. $\endgroup$ – poppy3345 Jun 17 '16 at 21:55

Hint: Consider $Vec(e_1,e_2,e_3), f(e_1)=0, f(e_2)=e_1, f(e_3)=e_3$.

  • $\begingroup$ Ahh, thank you! I will stash this away in my toolbox of handy counterexamples. $\endgroup$ – poppy3345 Jun 17 '16 at 22:04

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