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Let $V$ a vector space and $p:V\to V$ a projector. Can we give to $V$ a structure of inner product vector space ? I would say yes, and the inner product would be the one s.t. $$\langle x,p(x)\rangle =0,$$ but I'm not sure if it possible.

For example, if $\{u,v,w\}$ is a basis (not orthonomal), and $p:V\to V$ is s.t. $\operatorname{im}(p)=\operatorname{span}(w)$, then I think we can defined $\langle .,.\rangle $ s.t. $\langle u,w\rangle =\langle v,w\rangle =0$ and $$\langle v,v\rangle =\|v\|^2,\quad \langle u,u\rangle =\|u\|^2\quad \langle w,w\rangle =\|w\|^2.$$ Therefore, we can make of $V$ an inner space.

Do you think it's correct ?

Alex Provost
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user349449
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  • If $p$ is the identity map, wouldn't that inner product be forbidden? Because every vector would satisfy $<x,x>=0$. – jinawee Jul 06 '16 at 14:31
  • @Surb: In my mind, if $p$ is a projector (and $p$ not the identity). Let say that $V=span(e_1,...,e_n)$ and $p$ project onto $span(e_1,...,e_m)$ with $m<n$. It would make sense to me to defined a bilinear map such that $b(x,y)=0$ if $x\in span(e_1,...,e_m)$ and $y\in span(e_{m+1},...,e_n)$. And thus that $p$ would be an orthogonal projection onto $span(e_1,...,e_m)$. Does it make sense ? – user349449 Jul 06 '16 at 14:36
  • Note that, for finite dimensional $V$, $[x,y]:= \langle x,Ay\rangle$ is an inner product if and only if $A$ is symmetric positive definite. – Surb Jul 06 '16 at 14:37
  • Like I defined $b$, I think its symmetric and defined positive. – user349449 Jul 06 '16 at 14:38
  • @MathBeginner yes you can do that, if $v = \sum \alpha_i e_i$ and $w = \sum \beta_i e_i$ then define $b(v,w)=\sum \alpha_i\beta_i$. – Surb Jul 06 '16 at 14:38
  • So, if $V$ has a projector, then necessarily it's an inner space, right ? @Surb – user349449 Jul 06 '16 at 14:39
  • @MathBeginner well, for any subspace $U,U'\subset V$ with $U\oplus U' = V$, there is an inner product $\langle \cdot,\cdot\rangle$ such that $\langle u,u'\rangle=0$ for every $u\in U, u' \in U'$. – Surb Jul 06 '16 at 14:41

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You can give any vector space over $\mathbb{R}$ or $\mathbb{C}$ an inner product without specific reference to a projector. (See here.) But once you fix an inner product, you may speak of orthogonal projectors. These are by definition the self-adjoint projectors, i.e. the linear operators $P$ satisfying $P^2 = P$ and $$ \langle Pu,v \rangle = \langle u,Pv \rangle,$$and they correspond to orthogonal projections onto closed subspaces (which is what you had in mind).

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Alex Provost
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