Inner Product Space - Pythagorean?

Sorry.In summary, the conversation discusses how to show that if ##V## is an inner product space and ##V_0## is a finite dimensional subspace of ##V##, then for a vector ##v \in V## with ##v_0 = proj_{V_0}(v)##, the expression ||v - vo||^2 = ||v||^2 - ||vo||^2 can be derived using general inner product space properties. The conversation also includes a discussion on working backwards from this expression and using the definition of projection and the property of orthogonal vectors to show that <v - vo, vo> = 0.
  • #1
ElijahRockers
Gold Member
270
10

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.

Any advice is appreciated.
 
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  • #2
ElijahRockers said:

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 [itex]\|v - v_0\|^2 = \langle v - v_0, v - v_0 \rangle[/itex]. You should get four terms. Why do the two terms you don't want vanish?
 
  • #3
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?
 
  • #4
ElijahRockers said:

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.

Any advice is appreciated.

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?
 
  • #5
Dick said:
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
 
  • #6
ElijahRockers said:
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||.
 
  • #7
Dick said:
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.
 

What is an Inner Product Space?

An Inner Product Space is a mathematical concept in linear algebra that refers to a vector space with an inner product operation defined on it. This inner product operation takes two vectors as input and produces a scalar value as output, which represents the "angle" between the two vectors. This concept is important in many areas of mathematics and physics, including geometry, quantum mechanics, and signal processing.

What is the Pythagorean Theorem in an Inner Product Space?

The Pythagorean Theorem in an Inner Product Space is a generalization of the classical Pythagorean Theorem from Euclidean geometry. In an Inner Product Space, the Pythagorean Theorem states that the squared length of the sum of two vectors is equal to the sum of the squares of the individual vector lengths, minus twice the inner product of the two vectors. This theorem is useful for calculating the distance between vectors and for proving other theorems in linear algebra.

How is the Pythagorean Theorem proven in an Inner Product Space?

The Pythagorean Theorem can be proven in an Inner Product Space using the properties of the inner product operation. By expanding the inner product of the sum of two vectors, we can show that it is equal to the sum of the squares of the individual vector lengths, minus twice the inner product of the two vectors. This proof is similar to the proof in Euclidean geometry, but it uses the properties of the inner product instead of the Pythagorean Theorem.

What are some examples of Inner Product Spaces?

There are many examples of Inner Product Spaces, including the Euclidean space R^n, where n is any positive integer. This is the most commonly used Inner Product Space, and it is the space that we are most familiar with in everyday life. Other examples include complex vector spaces, function spaces, and sequence spaces. In fact, any vector space that has an inner product operation defined on it can be considered an Inner Product Space.

How is the Pythagorean Theorem used in real-world applications?

The Pythagorean Theorem in an Inner Product Space has many real-world applications. In geometry and engineering, it is used to calculate distances between points and to prove theorems about vector spaces. In physics, it is used to calculate the energy of a particle in quantum mechanics and to analyze signals in signal processing. It is also used in computer graphics to calculate the distance between points in three-dimensional space. The applications of the Pythagorean Theorem in an Inner Product Space are diverse and far-reaching.

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