I am not good with vector spaces so I would be grateful for any help.
As I've been told I need to take $v \in \mathrm{Im}(T)$, $v\neq 0$, and show that if $T(v) = \mathbf{0}$ then $T^2 = 0$. But if $T(v) \neq \mathbf{0}$, then need to show that if $\{e_1,\ldots,e_{n-1}\}$ is a basis of $\mathrm{Ker}(T)$, then $\{e_1,\dots,e_{n-1},v\}$ is a basis of $V$. Then deduce that $T$ is diagonalisable. I don't really know how to do that though...
Let $V$ be a $k$-vector space of dimension $n$ and $T: V \to V$ a linear map of rank 1. Show that either $T^2 = 0$ or that $T$ is diagonalisable?
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linear-algebra
vector-spaces