Show that $V = \mbox{ker}(\varphi) \oplus \mbox{im}(\varphi)$,with $ \varphi : V \rightarrow V$ linear mapping, where $\varphi^2=\varphi$.

Question

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?

does this help?https://math.stackexchange.com/questions/261704/show-that-the-direct-sum-of-a-kernel-of-a-projection-and-its-image-create-the-or?noredirect=1&lq=1 — UnsinkableSam, Jun 09 '20 at 10:27
$v=f(u)$ for some $u$ and $f(v)=f(f(u))=f(u)=v$. Since $f(v)=0$ we get $v=0$. — Kavi Rama Murthy, Jun 09 '20 at 10:28

score 1 · Accepted Answer · answered Jun 09 '20 at 11:37

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 $V = \operatorname{Im}(\varphi) \oplus \operatorname{Ker}(\varphi)$.

Just a short question, what does $ \varphi|_{\operatorname{Im}(\varphi)} $ exactly mean? — Herrpeter, Jun 09 '20 at 11:42
It is $\varphi$ where we restrict the domain to $\operatorname{Im}(\varphi)$, i.e. it is a function with domain $\operatorname{Im}(\varphi)$ rather than $V$. — Watercrystal, Jun 09 '20 at 11:45

Show that $V = \mbox{ker}(\varphi) \oplus \mbox{im}(\varphi)$,with $ \varphi : V \rightarrow V$ linear mapping, where $\varphi^2=\varphi$.

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