Bessel's Inequality: Proving \|v\|^2 \ge c_1^2 + \cdots + c_k^2

[e(ho0n3]

Homework Statement
Let v \in \mathbb{R}^n, let \{u_1, \ldots, u_k\} be an orthonormal subset of \mathbb{R}^n and let c_i be the coefficient of the projection of v to the span of u_i. Show that \|v\|^2 \ge c_1^2 + \cdots + c_k^2.

The attempt at a solution
c_i = v \cdot u_i and \| v \|^2 = v \cdot v so I can write the inequality as

v \cdot (v - (u_1 + \cdots + u_k)) \ge 0

This means the angle between v and v - (u_1 + \cdots + u_k) is less than 90 degrees. This is all I've been able to conjure. I'm trying to reverse-engineer the inequality back to something I know is true. Is this a good approach? Is there a better approach?

 

[Dick]

Couldn't you just complete {u1,...,uk} to an orthonormal basis {u1,...,uk,uk+1,...,un}? So ||v||^2=c1^2+...cn^2.

 

[e(ho0n3]

Ah! Good point. I think that'll work. All that remains to be proved is that v = c_1u_1 + \cdots c_nu_n.

 

[e(ho0n3]

Quick question: the basis doesn't have to be orthonormal does it?

 

[Dick]

Ah! Good point. I think that'll work. All that remains to be proved is that v = c_1u_1 + \cdots c_nu_n.

Do you really have to prove that? v is some linear combination of the basis vectors u_i. So c_i must be v.u_i since they are orthonormal. Isn't that what an orthonormal basis is all about?

 

[e(ho0n3]

Never mind. I understand why it has to be orthonormal: If it isn't, I couldn't write v as a linear combination of the u's using the c's as the coefficients.

 

[Dick]

The only thing that might have to be proved is that you can complete an orthonormal subset to an orthonormal basis. But that's Gram-Schmidt.

 

[e(ho0n3]

I know I can expand the set of u's to a basis, then orthogonalize it using Gram-Schmidt, and then normalize the result. This will yield an orthonormal basis with the original u's.

 

[Dick]

Right. So not much to prove really. That one was easy.

 

[e(ho0n3]

Thank you for your help.

 

[qcokns2008]

Hi
Can you do it these two proof?
I tried but i don't know these proofs...

(b) Prove Parseval’s Indentity: For any w ∈ span(S), we have

||w||^2 = |w · u1 |^2 + |w · u2 |^2+ · · · + |w · uk |^2 .

(c) Prove Bessel’s Inequality: For any x ∈ R^n we have

||x||^2 ≥ |x · u1 |^2 + |x · u2 |^2 + · · · + |x · uk |^2 ,

and this is an equality if and only if x ∈ span(S).