Well, the $ \oplus $ denotes the direct sum which means, that:
$V = \mbox{ker}(\varphi) + \mbox{im}(\varphi)$ and that $ker(\varphi)∩im(\varphi)={0}$ and the linear mapping $\varphi^2=\varphi$ is a projection.
So let $v \in \mbox{im}(\varphi) \cap \mbox{ker}(\varphi)$, how can I continue the proof?
Best Answer
Here is a proof for finite-dimensional spaces: If we have $\varphi^2 = \varphi$, then $\operatorname{Im}(\varphi) = \operatorname{Im}(\varphi^2)$ follows easily (and the same goes for the kernel). Further, we get that the restriction $$\varphi|_{\operatorname{Im}(\varphi)} \colon \operatorname{Im}(\varphi) \to V, v \mapsto \varphi(v) $$ actually must be an automorphism of $\operatorname{Im}(\varphi)$ by a dimension argument (specifically, $\varphi|_{\operatorname{Im}(\varphi)}$ must have full rank to guarantee that the ranks of $\varphi$ and $\varphi^2$ agree). This also yields $\operatorname{Im}(\varphi) \cap \operatorname{Ker}(\varphi) = 0$ as assuming there was a nonzero $v \in \operatorname{Im}(\varphi) \cap \operatorname{Ker}(\varphi)$ would imply $$\operatorname{rank} \varphi = \operatorname{dim} \operatorname{Im}(\varphi) > \operatorname{rank} \varphi|_{\operatorname{Im}(\varphi)} = \operatorname{rank} \varphi.$$ By observing that $\operatorname{Im}(\varphi)$ and $\operatorname{Ker}(\varphi)$ are distinct subspaces of $V$ and using the rank-nullity theorem it follows that $V = \operatorname{Im}(\varphi) + \operatorname{Ker}(\varphi)$ and thus $V = \operatorname{Im}(\varphi) \oplus \operatorname{Ker}(\varphi)$.