Show Hilbert Space has a pre inner product

**dazed** · October 14th 2010, 07:34 AM

Let V = C^n and A be in Mn(C).
Let

<x,y>=sum i, j, k =1 to n Akj . conjugate Aki . x(j) . conjugate y(i)

Show this is a pre inner product on V.

Do I start by letting say B = Akj . conjugate Aki
Define <x, y> = sum i, j, k =1 to n Bx . conjugate y
& find linearity of V by the linearity of B somehow?

Thanks.

**HappyJoe** · October 14th 2010, 10:49 AM

First of all, I was confused as to why the thread is called "Show Hilbert Space has a pre inner product". Certainly any Hilbert space has a pre-inner product, namely the inner product making the underlying vector space a Hilbert space in the first place.

Second of all, your notation is not really user-friendly, but I think I've deciphered it.

Let me show you how to prove additivity in the first variable. I want to show that $\langle x+z,y\rangle = \langle x,y\rangle + \langle z,y\rangle$ .

The left hand side is just

$\langle x+z,y\rangle = \displaystyle\sum_{i,j,k=1}^n A_{kj}\overline{A_{ki}}\left(x(j)+z(j)\right)\over line{y(i)}.$

On the right hand side of this equality, for each term in this (huge) sum, you just multiply into the parenthesis, giving

$\displaystyle\sum_{i,j,k=1}^n A_{kj}\overline{A_{ki}}\left(x(j)+z(j)\right)\over line{y(i)}=\displaystyle\sum_{i,j,k=1}^n A_{kj}\overline{A_{ki}}x(j)\overline{y(i)}+\displa ystyle\sum_{i,j,k=1}^n A_{kj}\overline{A_{ki}}z(j)\overline{y(i)},$

and hey, this is exactly the right hand side of the equality I was trying to prove.

**dazed** · October 14th 2010, 12:17 PM

Sorry. I realised this mistake after, but did not know how to edit the title of the whole thread and I apologise for the poor notation and thank you for deciphering it!

So to continue:

$\langle y,x\rangle = \displaystyle\sum_{i,j,k=1}^n A_{kj}\overline{A_{ki}}\left(y(i))\right)\overline {x(j)}$

= y.transpose Akj Aki conjugate x
= conjugate(conjugate y. conjugate (Akj Aki conjugate x))
=conjugate <x,y>

Sorry and thanks!

**HappyJoe** · October 14th 2010, 11:28 PM

Yeah. Try comparing this to what you want to end up with, i.e. $\overline{\langle x,y\rangle}$ . Does this suggest which steps to make to come from one to the other?

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