Prove: Let S be any subspace of X. Then, for any x in X and s in S, we have <P_S x - x, s> = 0?

Question

Okay... instead of that other one I listed... lets try the simpler one...

and the hint for this one is: Since S is a subspace, s + P_S x is again in S, for all s, as is cs, for every scalar c.

I know it's something simple that i'm not understanding... and it's frustrating me.

P_S x is the orthogonal projection of x onto S (which means it's perpendicular)...

so P_S x - x leaves you with Px (projection of x to S), which is perpendicular to S... so that is 0... but then why is s in the original vector ? maybe that is what threw me off... because everything i just said, i knew... and i kept drawing it over and over again...

Answer 1

This looks like something where P is a projection. Is that right? There must be something known about P because if P is the zero operator then
P_sx = 0
and so
=<-x,s> which is not necessarily equal to 0.

I do know that for P_v being the projection onto subspace V of X, then
for any x e X
x-P_v x is perpendicular to any v e V, thus the zero inner product.
A simple sketch helps to see this.

So, it sure looks like P_s is the projection onto S.

You now I can help.
Let W be the orthogonal complement of S
then X is a direct sum of S + W
then any x e X can be written:
x = s + w
P_s x = P_s s + 0 = s
so
x = P_s x + w
x-P_s x = w
= 0 because they are from orthogonal subspaces:
= 0


A 2 dim sketch does help because when you subtract the component of x along S from x, you get the component of x that is perpendicular to S, thus perpendicular to any vector in S.

Answer 2

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