Prove that : $\|x-\sum_{i=1}^{n}\langle x,x_{i}\rangle x_{i}\|^{2}=\|x\|^{2}-\sum_{i=1}^{n}|\langle x,x_{i}\rangle |^{2}$

Question

Let $\mathbb{H}$ be a Hilbert space and $x_{1},x_{2},x_{3},...,x_{n}$ Orthogonal vectors from $\mathbb{H}$ , $x\in\mathbb{H}$ then prove that :

$$\|x-\displaystyle\sum_{i=1}^{n}\langle x,x_{i}\rangle x_{i}\|^{2}=\|x\|^{2}-\displaystyle\sum_{i=1}^{n}|\langle x,x_{i}\rangle |^{2}$$

For all $x\in\mathbb{H}$

I know that for $x,y\in\mathbb{H}$

$\|x-y\|^{2}=\|x\|^{2}+\|y\|^{2}-2\Re\langle x,y\rangle$

But when I applied this here I don't get it ?

$\|x-\sum_{i=1}^{n}\langle x,x_{i}\rangle x_{i}\|^{2}=\|x\|^{2}+\|\sum_{i=1}^{n}\langle x,x_{i}\rangle x_{i}\|^{2}-2\Re\langle x,\sum_{i=1}^{n}\langle x,x_{i}\rangle x_{i}\rangle$

I'm correct or not?

Answer 1

The key is to see the equality as $$\tag1 \|x-\displaystyle\sum_{i=1}^{n}\langle x,x_{i}\rangle x_{i}\|^{2}+\displaystyle\sum_{i=1}^{n}|\langle x,x_{i}\rangle |^{2}=\|x\|^{2}. $$ Using that the $\{x_j\}$ are orthonormal you can get that $$\tag2 \left\|\sum_{i=1}^{n}\langle x,x_{i}\rangle x_{i}\right\|^{2}=\sum_{i=1}^{n}|\langle x,x_{i}\rangle |^{2}. $$ Now if you write $M=\operatorname{span}\{x_1,\ldots,x_n\}$ you have $H=M\oplus M^\perp$. Then $x=x_M+y$, with $x_M\in M$, $y\in M^\perp$. It is easy to check that $$x_M=\sum_{i=1}^{n}\langle x,x_{i}\rangle \,x_i$$ and now $(1)$ is simply $$\tag3 \|x\|^2=\|x_M\|^2+\|y\|^2,$$ together with $(2)$.