0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.