# Proof Regarding Projections

• Nov 22nd 2011, 05:39 PM
TaylorM0192
Proof Regarding Projections
Suppose that we have an inner product space $\displaystyle V$ and a finite dimensional subspace $\displaystyle W$.

Let $\displaystyle E(\alpha) = proj_W \space \alpha$. Prove that $\displaystyle <E\alpha, \beta>$ $\displaystyle =$ $\displaystyle <\alpha, E\beta>$ for all $\displaystyle \alpha$ and $\displaystyle \beta$ in $\displaystyle V$.

I've gone about various ways of trying to prove this; my main attack of this problem was to write out a basis for W, and then expand out the projection expression in terms of each basis vector, and then write out the inner products and see what I get (direct proof). This just leads me to a dead end.

I was able to sketch a quick proof in the Euclidean space with the dot product, but of course this is much too restrictive to be a proof (it's easier when you know exactly how the dot product is defined). This led me to the idea of representing the arbitrary dot product as a matrix, and then setting up the equations and trying another direct proof, but again it looks messy, and it seems like this kind of direct proof could get extremely long and bogged down, especially when you start expanding out individual linear combinations and distributing across inner products, etc. etc.

I need a new idea to approach this problem...any help would be appreciated! =)
• Nov 22nd 2011, 07:55 PM
Drexel28
Re: Proof Regarding Projections
Quote:

Originally Posted by TaylorM0192
Suppose that we have an inner product space $\displaystyle V$ and a finite dimensional subspace $\displaystyle W$.

Let $\displaystyle E(\alpha) = proj_W \space \alpha$. Prove that $\displaystyle <E\alpha, \beta>$ $\displaystyle =$ $\displaystyle <\alpha, E\beta>$ for all $\displaystyle \alpha$ and $\displaystyle \beta$ in $\displaystyle V$.

I've gone about various ways of trying to prove this; my main attack of this problem was to write out a basis for W, and then expand out the projection expression in terms of each basis vector, and then write out the inner products and see what I get (direct proof). This just leads me to a dead end.

I was able to sketch a quick proof in the Euclidean space with the dot product, but of course this is much too restrictive to be a proof (it's easier when you know exactly how the dot product is defined). This led me to the idea of representing the arbitrary dot product as a matrix, and then setting up the equations and trying another direct proof, but again it looks messy, and it seems like this kind of direct proof could get extremely long and bogged down, especially when you start expanding out individual linear combinations and distributing across inner products, etc. etc.

I need a new idea to approach this problem...any help would be appreciated! =)

Is this even true? You are stating that all projections onto finite dimensional subspaces are unitary. But, isn't this true if and only if $\displaystyle \ker E=W^\perp$? Am I misunderstanding?
• Nov 22nd 2011, 08:15 PM
TaylorM0192
Re: Proof Regarding Projections
Well, since I haven't proved anything yet, I suppose it could be false;)

In any case, the problem comes from Hoffman and Kunze #8.2.12:

"Let W be a finite dimensional subspace of an inner product space V, and let E be the orthogonal projection of V on W. Prove that (Ea|b) = (a|Eb) for all a and b in V."
• Nov 22nd 2011, 08:19 PM
Drexel28
Re: Proof Regarding Projections
Quote:

Originally Posted by TaylorM0192
Well, since I haven't proved anything yet, I suppose it could be false;)

In any case, the problem comes from Hoffman and Kunze #8.2.12:

"Let W be a finite dimensional subspace of an inner product space V, and let E be the orthogonal projection of V on W. Prove that (Ea|b) = (a|Eb) for all a and b in V."

Oh, ok, fine, so it is an orthogonal projection! In this case you can actually find a nice discussion of this on wiki.
• Nov 22nd 2011, 08:32 PM
TaylorM0192
Re: Proof Regarding Projections
Hmm...is my notation incorrect?

I thought $\displaystyle proj_\alpha \beta$ meant "the orthogonal projection of $\displaystyle \beta$ onto $\displaystyle \alpha$" and $\displaystyle orth_\alpha \beta$ meant "the orthogonal component of $\displaystyle \alpha$ onto $\displaystyle \beta$."

Anyway, this is how I remember the notation from my calculus courses, although I admit I don't see it used much in linear algebra texts.

And in any case, what would be the difference between a projection and an orthogonal projection?
• Nov 22nd 2011, 08:35 PM
Drexel28
Re: Proof Regarding Projections
Quote:

Originally Posted by TaylorM0192
Hmm...is my notation incorrect?

I thought $\displaystyle proj_\alpha \beta$ meant "the orthogonal projection of $\displaystyle \beta$ onto $\displaystyle \alpha$" and $\displaystyle orth_\alpha \beta$ meant "the orthogonal component of $\displaystyle \alpha$ onto $\displaystyle \beta$."

Anyway, this is how I remember the notation from my calculus courses, although I admit I don't see it used much in linear algebra texts.

And in any case, what would be the difference between a projection and an orthogonal projection?

You know that every projection $\displaystyle P:V\to V$ naturally decomposes $\displaystyle V$ as $\displaystyle V=\ker P\oplus \text{im }P$. In an inner product space whenever we have complements (i.e. subspaces whose direct sum is $\displaystyle V$ [implicitly assuming they are disjoint]) we'd love for them to be orthogonal complements. The fact that $\displaystyle \ker P$ and $\displaystyle \text{im }P$ are complements does not imply that they are orthogonal complements. When this nice condition is satisfied we call $\displaystyle P$ an orthogonal projection.
• Nov 22nd 2011, 08:40 PM
TaylorM0192
Re: Proof Regarding Projections
I really dislike my linear algebra professor this quarter...it has been a journey of self-discovery from day one!

Thanks man~