# Prove that if $\varphi \in \mathcal L(V)$, then $V=\ker(\varphi )\oplus \text{Im}(\varphi )$.

Let $$V$$ a vector space of finite dimension and $$\varphi \in \mathcal L(V)$$ an endomorphism. Prove that $$V=\ker(\varphi )\oplus \text{Im}(\varphi ).$$

• Using rank theorem yield $$\dim(V)\geq \dim(\ker(\varphi )+\text{Im}(\varphi ))=\dim(\ker(\varphi ))+\dim(\text{Im}(\varphi ))-\dim(\ker(\varphi )\cap \text{Im}(\varphi ))$$ $$=\dim(V)-\dim(\ker(\varphi )\cap \text{Im}(\varphi ))$$

and thus $$\dim(\ker(\varphi )\cap \text{Im}(\varphi ))\geq 0$$, and thus irrelevant.

• Then I tried to get a contradiction as follow : Let $$x\in \ker(\varphi )\cap\text{Im}(\varphi )$$, i.e. $$\varphi (x)=0$$ and $$x=\varphi (y)$$ for some $$y$$. Therefore $$\varphi ^2(y)=0$$, but I can't get a contradiction. So maybe it's wrong.

• So, is there $$\varphi \in \mathcal L(V)$$ s.t. $$\ker(\varphi )\cap \text{Im}(\varphi )\neq \{0\}$$ ? Because I can't find one.

Consider $$\varphi:\mathbb{R}^{2}\to\mathbb{R}^{2}$$ given by $$\varphi(x,y)=(y,0)$$. You can check that $$\varphi\in \mathcal{L}(\mathbb{R}^{2})$$ and $$\ker(\varphi)=\langle (1,0) \rangle=\operatorname{Im}(\varphi)$$, so it is not true that $$\ker(\varphi)\cap\operatorname{Im}(\varphi)=\{0\}$$.