# projection - linear alebra

$U$ is a subspace of $\mathbb{C}^n$; $v,w\in\mathbb{C}^n$; $p$ is the projection of $v$ on $U$; $q$ is the projection of $w$ on $U$.

I need to prove that: $$\langle v,w\rangle = \langle p,q\rangle +\langle v-p,w-q\rangle$$

I opened this way the right part: \begin{align} \langle p,q\rangle +\langle v,w-q\rangle-\langle p,w-q\rangle&= \langle p,q\rangle+\langle v,w\rangle+\langle v,-q\rangle-\langle p,w\rangle-\langle p,-q\rangle\\ &=\langle v,w\rangle+\langle v,-q\rangle-\langle p,w\rangle \end{align}

Hope I didn't have any mistakes till here but what can I do now?

• you are not a newbie here, so (-1) for the awfull formating, Mar 3, 2013 at 14:02

Decompose $$v = p + (v-p), \qquad w = q+ (w-q)$$ and use the fact that $v-p$ and $w-q$ aro orthogonal to $U$.

First of all, a projection is a projection onto a subspace $F$ parallel to another subspace $G$ where, if we denote $E$ the space, $E=F\oplus G$.

Now if you have a scalar product on $E$, and $\dim E < +\infty$, you can chose and $F$ and take $G=F^\perp$ which seems to be what you do. So we have an orthogonal projection.

In your case, $E=\mathbb{C}^n$, $F=U$, $G=U^\perp$

You have

$v=p+(v-p)$ where $p\in U, v-p \in U^\perp$

$w=q+(w-q)$ where $q\in U, w-q \in U^\perp$

Note that $\forall x\in U, \forall y\in U^\perp, <x,y>=0$

In particular, $<v-p,q>=0=<p,q-w>$

$<p,q>+<v-p,w-q> = <p,q>+<v-p,w>-<v-p,q>=<p,q>+<v-p,w>=<p,q>-<p,w>+<v,w>=<p,q-w>+<v,w>=<v,w>$