Prove: if $A\not=0$ is a nilpotent matrix $\implies\dim(\ker A\cap\operatorname{Im} A)\geq 1$

Question

$A$ is nilpotent $\implies 0$ is it's only eigenvalue. I intuitively understand that there's probebly a connection because $Av=\lambda v=0v=0$ but that's true for specific vectors $v$ (the eigenvectors) so i'm stuck.

Answer 1

Let $k$ be the smallest number with $A^k=0$. Because of $A\ne 0$ we have $k\ge 2$.

Let $x$ be a vector with $A^{k-1}x\ne 0$ Then, we have

$A(A^{k-1}x)=A^kx=0$ , so $A^{k-1}x$ is in the kernel of $A$.

Furthermore, we have $A(A^{k-2}x)=A^{k-1}x$, so $A^{k-1}x$ is also in the image of $A$ (If $k=2$, we simply have $A^{k-1}x=Ax$).

This proves the claim.

Answer 2

Suppose we're working on an $\;n\,- $ dimensional space $\;V\;$ and $\;A^k=0\;$ but $\;A^{k-1}\neq0\;$ (observe that it must be $\;k>1\;$ , otherwise $\;A=0\;$).

So let $\;v\in V\;$ be such that $\;A^{k-1}v=w\neq 0\;$ . Observe that

$$\;Aw=A(A^{k-1}v)=A^kv=0\implies w\in\ker A\;,\;\;\text{and also}\;\;w=A(A^{k-2}v)\in\text{Im}\,A$$

so there: we've shown that $\;0\neq w\in\ker A\cap\text{Im}\,A\;\implies \dim(\ker A\cap\text{Im}\,A)\ge1$

Answer 3

The important observation is that if $T \colon V \rightarrow V$ is a nilpotent operator on $V$ where $0 < \dim V < \infty$ then $0$ is an eigenvalue of $T$. If $A \neq 0$ is nilpotent then $\operatorname{Im}(A)$ is an $A$-invariant subspace, $0 < \dim \operatorname{Im}(A) < \infty$ and $A|_{\operatorname{Im}(A)}$ is also nilpotent and so $0$ is an eigenvalue $A|_{\operatorname{Im}(A)}$ and thus we can find $0 \neq v \in \operatorname{Im}(A)$ with $Av = 0$ which shows that $v \in \ker(A) \cap \operatorname{Im}(A)$ and so $\dim \ker(A) \cap \operatorname{Im}(A) \geq 1$.