# Inner Product Space - Pythagorean?

• ElijahRockers

Gold Member

## Homework Statement

Let ##V## be an inner product space and let ##V_0## be a finite dimensional subspace of ##V##. Show that if ##v ∈ V## has ##v_0 = proj_{V_0}(v)##:

||v - vo||^2 = ||v||^2 - ||vo||^2

## Homework Equations

General inner product space properties, I believe.

## The Attempt at a Solution

So Vo is finite dimensional and has an orthonormal basis {v1, v2,..., vn} for finite n. I can say that right? If so,

##v_0 = \sum_{i=1}^{n}<v,v_i>v_i## (Not sure if that helps really)

Also <v - vo , vo> = 0 since they are orthogonal.

That's about where I get stuck. I see that the expression I'm supposed to arrive at is basically pythagorean theorem for inner product spaces, and I can see it in my head quite easily if n is 2 dimensional.

So I tried working backwards:

||v - vo||^2 = ||v||^2 - ||vo||^2

||v - vo||^2 + ||vo||^2 = ||v||^2

And using inner product properties I can get:

<v - vo , v - vo> + <vo, vo> = <v , v>

I don't know if that really helps though.

## Homework Statement

Let ##V## be an inner product space and let ##V_0## be a finite dimensional subspace of ##V##. Show that if ##v ∈ V## has ##v_0 = proj_{V_0}(v)##:

||v - vo||^2 = ||v||^2 - ||vo||^2

## Homework Equations

General inner product space properties, I believe.

## The Attempt at a Solution

So Vo is finite dimensional and has an orthonormal basis {v1, v2,..., vn} for finite n. I can say that right? If so,

##v_0 = \sum_{i=1}^{n}<v,v_i>v_i## (Not sure if that helps really)

Also <v - vo , vo> = 0 since they are orthogonal.

That's about where I get stuck. I see that the expression I'm supposed to arrive at is basically pythagorean theorem for inner product spaces, and I can see it in my head quite easily if n is 2 dimensional.

So I tried working backwards:

||v - vo||^2 = ||v||^2 - ||vo||^2

||v - vo||^2 + ||vo||^2 = ||v||^2

And using inner product properties I can get:

<v - vo , v - vo> + <vo, vo> = <v , v>

I don't know if that really helps though.

Expand $\|v - v_0\|^2 = \langle v - v_0, v - v_0 \rangle$. You should get four terms. Why do the two terms you don't want vanish?

Didn't think I could...

So would it be:
##<v,v> -2<v,v_0 > + 2<v_0 , v_0> = <v, v>##
##<v,v_0 > = <v_0, v_0>##
##v = v_0##
##v - v_0 = 0##

so ##<v - v_0 , v_0> = 0##

Is that right?

## Homework Statement

Let ##V## be an inner product space and let ##V_0## be a finite dimensional subspace of ##V##. Show that if ##v ∈ V## has ##v_0 = proj_{V_0}(v)##:

||v - vo||^2 = ||v||^2 - ||vo||^2

## Homework Equations

General inner product space properties, I believe.

## The Attempt at a Solution

So Vo is finite dimensional and has an orthonormal basis {v1, v2,..., vn} for finite n. I can say that right? If so,

##v_0 = \sum_{i=1}^{n}<v,v_i>v_i## (Not sure if that helps really)

Also <v - vo , vo> = 0 since they are orthogonal.

That's about where I get stuck. I see that the expression I'm supposed to arrive at is basically pythagorean theorem for inner product spaces, and I can see it in my head quite easily if n is 2 dimensional.

So I tried working backwards:

||v - vo||^2 = ||v||^2 - ||vo||^2

||v - vo||^2 + ||vo||^2 = ||v||^2

And using inner product properties I can get:

<v - vo , v - vo> + <vo, vo> = <v , v>

I don't know if that really helps though.

You say you can show <v - vo , v - vo> + <vo, vo> = <v , v>? Isn't that the same thing as ||v - vo||^2 + ||vo||^2 = ||v||^2?

You say you can show <v - vo , v - vo> + <vo, vo> = <v , v>? Isn't that the same thing as ||v - vo||^2 + ||vo||^2 = ||v||^2?

Yea. but as far as I am able to tell I'm supposed to be able to show that given vo = proj_Vo (v). I was working backwards from ||v - vo||^2 etc

Yea. but as far as I am able to tell I'm supposed to be able to show that given vo = proj_Vo (v). I was working backwards from ||v - vo||^2 etc

How did you define projection? Have you shown <v-vo,vo>=0? That's all you really need. Expand ||v|| = ||(v-vo)+vo||.

How did you define projection? Have you shown <v-vo,vo>=0? That's all you really need. Expand ||v|| = ||(v-vo)+vo||.
Yes, I defined that in my original post. That's what I was thinking! Thanks!

EDIT: Shown, not defined.