Bessel's Inequality

1. Jul 26, 2008

e(ho0n3

The problem statement, all variables and given/known data
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?

2. Jul 26, 2008

Dick

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

3. Jul 26, 2008

e(ho0n3

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

4. Jul 27, 2008

e(ho0n3

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

5. Jul 27, 2008

Dick

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?

6. Jul 27, 2008

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.

7. Jul 27, 2008

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.

8. Jul 27, 2008

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.

9. Jul 27, 2008

Dick

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

10. Jul 27, 2008

e(ho0n3

Thank you for your help.

11. Mar 7, 2011

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).