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

Question

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.

Answer 1

In general, the result is false:

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 $\ker(\varphi)\cap\operatorname{Im}(\varphi)=\{0\}$.