$T: V \to V$ is a linear map. If $\dim(\ker T \cap \text{Im}T)\neq0$, prove $\dim\ker T^2\geq2$

Question

$T: V \to V$ is a linear map. If $\dim(\ker T \cap \text{Im}T)\neq0$, prove $\dim\ker T^2\geq2$

so I can deduce $\dim\ker T\geq 1$ Because the intersection of the image and the kernel, is contained in the kernel.

And it is known that $\dim\ker T^2\geq \dim\ker T$

So $\dim\ker T^2\geq 1$. But I have no idea how to prove that it can't be $1$.

Answer 1

Let $v \in\ker T \cap \text{im } T$ be such that $v \neq 0$. Then there is some $u$ such that $Tu = v$. Clearly $u, v \in \ker T^2$.

We wish to show that $u, v$ are linearly independent, i.e. $u \neq kv$ for any $k \in \mathbb{F}$.

If $u = kv$, then $Tu = T(kv) = k(Tv) = k(0) = 0$, so $v = 0$, a contradiction.