given A linear map $T:V\longrightarrow V$, let $\dim V = n$, prove that for every $k\geq n,\ \operatorname{Im}T^{k}\cap \ker T^{k}=\{0\}$.

Question

Given a linear map $T:V\longrightarrow V$, let $\dim V = n$.

1. prove that for every $k\geq1$ , $\ker T^{k}\subseteq \ker T^{k+1}$

2. prove that for every $k\geq1$ , $ \operatorname{Im}T^{k+1}\subseteq \operatorname{Im}T^{k}$

3. let $\ker T^{k}=\ker T^{k+1}$, prove that $\:\ker T^{k}=\ker T^{k+1}=\ker T^{k+2}=\cdots$

4. let $ \operatorname{Im}T^{k}= \operatorname{Im}T^{k+1}$. Prove: $\:\operatorname{Im}T^{k}= \operatorname{Im}T^{k+1}= \operatorname{Im}T^{k+2}=\cdots$

5. explain why $\:\ker T^{n}=\ker T^{n+1}=\ker T^{n+2}=\cdots$

6. prove that for every $k\geq n, \: \operatorname{Im}T^{k}\cap \ker T^{k}=\{0\}$.

I did the first 5 sections . I got stuck in section No.6

I understand that the max value of $\dim(\ker T^{n})=n $, as $\dim V = n$.

also for the image, $\max(\dim( \operatorname{Im}T^{n})=n$.
but why this means that the intersection could only contain $\{0\}$?

Answer 1

If $x \in \mbox{im } T^k \cap \ker T^k$ for some $k \geq n$ then $x = T^k y$ for some $y$, and $T^k x = T^{2k} y = 0$.

Since $\ker T^{2k} = \ker T^n$ because $2k \geq n$, we get $T^n y = 0$, so $T^k y = T^{k-n} T^ny = 0$.

Consequently $x = T^k y = 0$, so every element in the intersection of $\mbox{im } T^k$ and $\ker T^k$ is zero.