Projection/Linear subspace/Idempotent/Fact/Proof

Proof

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

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

and hence

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

Suppose now that

$\varphi \colon V\longrightarrow V$

is an endomorphism with

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

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

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

Then

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

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

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

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

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

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

To fact