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Inner product in n-dimensional vector space

gucci

New member
Sep 1, 2013
13
So, I have an equivalence I need to prove, but I think I'm having trouble understanding the problem at a basic level.

The problem is to prove that the inner product of a and b equals 1/4[|a+b|^2 - |a-b|^2] (a, b in C^n or an n-dimensional vector space with complex elements).

I don't understand how to write out |a+b|^2 in other terms. If anyone has any guidance here, that would be awesome. :-/
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
So, I have an equivalence I need to prove, but I think I'm having trouble understanding the problem at a basic level.

The problem is to prove that the inner product of a and b equals 1/4[|a+b|^2 - |a-b|^2] (a, b in C^n or an n-dimensional vector space with complex elements).

I don't understand how to write out |a+b|^2 in other terms. If anyone has any guidance here, that would be awesome. :-/
Welcome to MHB, gucci! :)

By definition the norm is given by $|x|^2 = \langle x, x \rangle$.

So:
$$|a+b|^2 = \langle a+b, a+b \rangle = \langle a, a+b \rangle + \langle b, a+b \rangle = ...$$

Can you continue using the axioms of an inner product?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Since you are working with a complex inner product, it might be helpful to know that:

\(\displaystyle \langle a,a+b \rangle = \overline{\langle a+b,a \rangle} = \overline{\langle a,a \rangle + \langle b,a \rangle}= \overline{\langle a,a\rangle} + \overline{\langle b,a\rangle} = \langle a,a\rangle + \langle a,b\rangle\)

and that:

\(\displaystyle \langle a,a-b\rangle = \langle a,a+(-b)\rangle = \langle a,a\rangle + \langle a,-b\rangle = \langle a,a\rangle + (\overline{-1})\langle a,b\rangle = \langle a,a\rangle - \langle a,b\rangle\)
 

gucci

New member
Sep 1, 2013
13
So, with your help I am much closer to the answer, but I'm making a mistake somewhere I guess. This is what I'm coming up with:

1/4 [|a+b|^2 - |a-b|^2] = 1/4 [⟨a+b,a+b⟩ - ⟨a+(-1)b,a+(-1)b⟩]
= 1/4 [⟨a,a+b⟩ + ⟨b,a+b⟩ - ⟨a,a+(-1)b⟩ + ⟨b,a+(-1)b⟩]
= 1/4 [⟨a,a⟩ + ⟨a,b⟩ + ⟨b,a⟩ + ⟨b,b⟩ - ⟨a,a⟩ + ⟨a,b⟩ + ⟨b,a⟩ - ⟨b,b⟩]
= 1/4 [2⟨a,b⟩ + 2⟨b,a⟩]

Can anyone spot where I slipped up? Thanks for all your help
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
So, with your help I am much closer to the answer, but I'm making a mistake somewhere I guess. This is what I'm coming up with:

1/4 [|a+b|^2 - |a-b|^2] = 1/4 [⟨a+b,a+b⟩ - ⟨a+(-1)b,a+(-1)b⟩]
= 1/4 [⟨a,a+b⟩ + ⟨b,a+b⟩ - ⟨a,a+(-1)b⟩ + ⟨b,a+(-1)b⟩]
= 1/4 [⟨a,a⟩ + ⟨a,b⟩ + ⟨b,a⟩ + ⟨b,b⟩ - ⟨a,a⟩ + ⟨a,b⟩ + ⟨b,a⟩ - ⟨b,b⟩]
= 1/4 [2⟨a,b⟩ + 2⟨b,a⟩]

Can anyone spot where I slipped up? Thanks for all your help
I don't think you have made a mistake, in a complex inner-product space what you have is:

\(\displaystyle \frac{1}{4}(2\langle a,b\rangle + 2\langle b,a\rangle) = \frac{1}{2}(\langle a,b\rangle + \overline{\langle a,b\rangle}) = \mathfrak{Re}(\langle a,b\rangle)\)

What you are being asked to PROVE is incorrect, in a complex inner product space the polarization identity is actually:

\(\displaystyle \langle a,b\rangle = \frac{|a+b|^2 - |a-b|^2 + i|a+ib|^2 - i|a-ib|^2}{4}\)
 

gucci

New member
Sep 1, 2013
13
Thank you so much! I love this forum, everyone here is so helpful :D