I Finding the orthogonal projection of a vector without an orthogonal basis

[AimaneSN]

Hi there,

I am currently reading a course on euclidian spaces and I came across this result that I am struggling to prove :

Let $F$ be a subspace of $E$ (of finite dimension) such that $F=span(e_1, e_2, ..., e_p)$ (not necessarily an orthogonal family of vectors), let $x \in E$

Then we have:

$\forall y \in F$ : $y=p_F(x) \Leftrightarrow \forall i= 1,...,p : (x-y,e_i) = 0$

where $(.,.)$ denotes an inner product
and the linear map $p_F$ is the orthogonal projection onto $F$.

I managed to prove the equivalence only when the family of the vectors $(e_i)$ is orthogonal but the result is more general.

Thank you and I would appreciate any hint or help.

 

[Infrared]

The inner product $(x-y,e_i)$ is zero for all $i$ if and only if $x-y\in F^\perp$ if and only if $x$ is of the form $x=y+z$ where $z\in F^{\perp}$ if and only if $y$ is the orthogonal projection of $x$ onto $F$.

Your post was a bit of effort to read- it would be better to use latex.

 

[AimaneSN]

Thank you for your reply, it's much clearer now. I just modified my post using Latex code.

 

[WWGD]

OP: Notice you can always use Gram-Schmidt to orthogonalize your basis vectors, and this won't affect the span .