Originally Posted by

**Zalren** At least, I think it is hard. I don't know how to start this at all.

Let $\displaystyle V$ be a vector space equipped with a positive definite inner product. Let $\displaystyle W$ be a subspace of $\displaystyle V$. Let $\displaystyle W'$ be its orthogonal complement. Define a map $\displaystyle \pi$: $\displaystyle V$ -> $\displaystyle V$ as $\displaystyle \pi(v) = v_1$ where $\displaystyle v = v_1 + v_2$ with $\displaystyle v_1$ in $\displaystyle W$ and $\displaystyle v_2$ in $\displaystyle W'$. Show that:

a) The map $\displaystyle \pi$ is welldefined, linear with range $\displaystyle W$ and that $\displaystyle \pi^2 = \pi$. The map $\displaystyle \pi$ is called the projectiion map onto the subspace $\displaystyle W$.

b) Define $\displaystyle v := I - \pi$. Show that $\displaystyle v$ is the projection map onto $\displaystyle W'$ i.e. show that $\displaystyle v$ is a linear map with range $\displaystyle W'$ and that $\displaystyle v^2 = v$.