Vector subspaces isomorphism?

Question

Let $V$ be a vector space and $φ : V\to V$ a linear transform such that $φ \circφ=φ$. Show that $V \cong kerφ \oplus Imφ$.

I understand I need to find an isomorphism $h:V \to kerφ \oplus Imφ$, but I don't know how to start. Any tips?

Answer 1

Given $v\in V$, write $v=(v-\phi(v))+\phi(v)$. Note that $\phi(v-\phi(v)) = 0$, therefore, $v-\phi(v) \in \ker \phi$. If $w\in \ker \phi \cap \text{im }\phi$, $$ w=\phi(v) \,\,\text{and }\phi(w)=0, $$ which means $\phi(\phi v) = 0$, that is $\phi(v) = w = 0$. Then, the decomposition is unique, showing $V=\ker \phi \oplus \text{im }\phi$.

Answer 2

Since $\ker\varphi$ and $\operatorname{Im}\varphi$ are subspaces of $V$, all it takes is to prove that every $v\in V$ can be written in one and only one way as the sum of an element of $\ker\varphi$ and an element of $\operatorname{Im}\varphi$. And that is easy:$$v=\overbrace{v-\varphi(v)}^{\in\ker\varphi}+\overbrace{\varphi(v)}^{\in\operatorname{Im}\varphi}.$$And if $v=w+\varphi(t)$, with $w\in\ker\varphi$, then $\varphi(v)=\varphi\bigl(\varphi(t)\bigr)=\varphi(t)$ and $w=v-\varphi(t)=v-\varphi(v)$.