I'm having troubles with this problem
Let $T:V\rightarrow V$ be a linear transformation. Show that: $$Ker(T^2)=Ker(T)\Rightarrow V=Ker(T)\oplus Im(T)$$
So far, using the rank-nullity theorem (and the fact that $Im(T^2)\subseteq Im(T)$) I've showed that $Im(T^2)= Im(T)$. Then, I want to show that $V=Ker(T)+Im(T)$, so I write any $v \in V$ as $v=T(v)+(v-T(v))$, with the first term obviously in $Im(T)$. To conclude this part, I'd like to show that $v-T(v) \in Ker(T)$, I'm inclined to think that $T^2\equiv T$, but I'm not sure if I have enough information to conclude that.
Best Answer
Hint: By the rank-nullity theorem, you know that the image and kernel have dimensions that add to $n$. What you need to show (using the given information) is that $$ Ker(T) \cap Im(T)=\{0\} $$