I've been working on a couple of algebra identities' demonstrations, among which the Bessel inequality. Here is the problem set:
We consider a vector space $V$, a vector $u$ in $V$, a subspace $E$ of $V$ of dimension $n$ and a basis $B$ for $E$ such that $B=\{u_1, u_2, ..., u_n\}$.
The orthogonal projection of $u$ on $E$, named $v$, is given by
$v=\sum_{i=1}^n {\lt u,u_i \gt\over \lt u_i,u_i \gt}\cdot u_i$
We define the vector $w$ as $u-v$ (hence $w$ is orthogonal to $E$).
We want to prove that
$||u||^2\geq \sum_{i=1}^n ({\lt u,u_i \gt^2\over \lt u_i, u_i \gt})\cdot u_i$ (Bessel Inequality)
By the generalization of the Pythagorean Theorem, we have:
$ \begin{align} {\|u\|}^2 & = {\|w\|}^2 + {\|v\|}^2 \\ & = {\|w\|}^2 + {\|\sum_{i=1}^n {\lt u,u_i \gt\over \lt u_i,u_i \gt}\cdot u_i \|}^2 \\ \end{align}$
Hence,
$ {\|u\|}^2 \geq {\|\sum_{i=1}^n {\lt u,u_i \gt\over \lt u_i,u_i \gt}\cdot u_i \|}^2 $
But I'm stuck here. I have to show that
$ {\|\sum_{i=1}^n {\lt u,u_i \gt\over \lt u_i,u_i \gt}\cdot u_i \|}^2 = {\sum_{i=1}^n ({\lt u,u_i \gt^2\over \lt u_i, u_i \gt})\cdot u_i}$
And I don't know how. I have tried the inner product properties, but I just do not understand how the norm squared gives this sum. I hope you can help me on this.
By the way, as this is my first time using MathJax, sorry for any errors in my format style. Thank you.
In order to see that $ {\|\sum_{i=1}^n {\lt u,u_i \gt\over \lt u_i,u_i \gt}\cdot u_i \|}^2 = {\sum_{i=1}^n ({\lt u,u_i \gt^2\over \lt u_i, u_i \gt})\cdot u_i}, $ it suffices to (inductively) use the generalization of the Pythagorean theorem. In particular, $$ \left\|\sum_{i=1}^n {\langle u,u_i \rangle\over \langle u_i,u_i \rangle}\cdot u_i \right\|^2 = \left\|\left(\sum_{i=1}^{n-1} {\langle u,u_i \rangle\over \langle u_i,u_i \rangle}\cdot u_i \right) + {\langle u,u_n \rangle\over \langle u_n,u_n \rangle}\right\|^2 \\= \left\|\sum_{i=1}^{n-1} {\langle u,u_i \rangle\over \langle u_i,u_i \rangle}\cdot u_i\right\|^2 + \left\|{\langle u,u_n \rangle\over \langle u_n,u_n \rangle} \cdot u_n\right\|^2 $$
From there, note that (by the homogeneity of norms) $$ \left\|{\langle u,u_n \rangle\over \langle u_n,u_n \rangle} \cdot u_n\right\|^2 = \left| {\langle u,u_n \rangle\over \langle u_n,u_n \rangle} \right|^2 \cdot \|u_n\|^2 = \frac{|\langle u,u_n \rangle|^2}{\langle u_n,u_n \rangle} \cdot \langle u_n,u_n \rangle = |\langle u,u_n\rangle|^2. $$