I am having trouble with a part of my assignment. I need to show that
$||x – P_Ex||^2$ = $||x||^2 – \sum_{j=1}^n |\langle e_j,x \rangle|^2$
where $e_j$ is in a hilbert space and an orthonormal set, and $P_Ex$ is the orthogonal projection of x onto E and is defined as
$P_Ex = \sum_{i=1}^n \langle e_i,x \rangle e_i$,
and
$ \langle e_j,e_i \rangle = 1$ if $i=j$ and $0$ if $i\neq j$
So far I have started by expanding the lefthand side and got this.
$||x – P_Ex||^2$ = $ \langle x, x\rangle – \langle x,P_Ex \rangle – \langle P_Ex,x \rangle + \langle P_Ex,P_Ex \rangle$
Since $P_Ex$ is orthogonal to x (I believe), the above line reduces to
$||x – P_Ex||^2$ = $ \langle x, x\rangle + \langle P_Ex,P_Ex \rangle$
From here I would like to expand the last part
$\langle P_Ex,P_Ex \rangle = \left\langle \sum_{i=1}^n \langle e_i,x \rangle e_i, \quad \sum_{j=1}^n \langle e_j,x \rangle e_j \right\rangle$
I feel like this would result in a vector of vectors but I don't really understand how it would work exactly.
Best Answer
By sesquilinearity$$\langle P_Ex,\,P_Ex\rangle=\sum_{ij}\underbrace{\langle e_i,\,x\rangle}_{\overline{\langle x,\,e_i\rangle}}\langle x,\,e_j\rangle\underbrace{\langle e_i,\,e_j}_{\delta_{ij}}\rangle=\sum_j|\langle x,\,e_j\rangle|^2.$$Now use the same technique to show $\langle x,\,P_Ex\rangle=\sum_j|\langle x,\,e_j\rangle|^2$, so the final coefficient of $\sum_j|\langle x,\,e_j\rangle|^2$ is $-2+1=-1$.