Projection/Linear subspace/Idempotent/Fact/Proof

Let ${\displaystyle {}\pi _{U}}$ be the projection onto ${\displaystyle {}U}$. Write ${\displaystyle {}v=u+u'}$ with ${\displaystyle {}u\in U,\,u'\in U'}$. Then we have

${\displaystyle {}\pi _{U}(\pi _{U}(u+u'))=\pi _{U}(u)=u=\pi _{U}(u+u')\,,}$

and hence

${\displaystyle {}\pi _{U}^{2}=\pi _{U}\,.}$

Suppose now that

${\displaystyle \varphi \colon V\longrightarrow V}$

is an endomorphism with

${\displaystyle {}\varphi ^{2}=\varphi \,.}$

Let ${\displaystyle {}x\in \operatorname {kern} \varphi \cap \operatorname {Im} \varphi }$. Then there exists some ${\displaystyle {}y\in V}$ such that

${\displaystyle {}\varphi (y)=x\,.}$

Then

${\displaystyle {}x=\varphi (y)=\varphi (\varphi (y))=\varphi (x)=0\,.}$

This means that the intersection of these linear subspaces is the zero space. For an arbitrary ${\displaystyle {}v\in V}$, we write

${\displaystyle {}v=\varphi (v)+(v-\varphi (v))\,.}$

Here, the first summand belongs to the image and, because of

${\displaystyle {}\varphi (v-\varphi (v))=\varphi (v)-\varphi (\varphi (v))=\varphi (v)-\varphi (v)=0\,,}$

the second summand belongs to the kernel. Therefore, we have a direct sum decomposition.