# Proving a function is an inner product in a complex space

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1. Feb 15, 2017

1. The problem statement, all variables and given/known data
Prove the following form for an inner product in a complex space V:
$\langle u,v \rangle$ $=$ $\frac 1 4$$\left| u+v\right|^2$ $-$ $\frac 1 4$$\left| u-v\right|^2$ $+$ $\frac 1 4$$\left| u+iv\right|^2$ $-$ $\frac 1 4$$\left| u-iv\right|^2$

2. Relevant equations
N/A

3. The attempt at a solution
By opening the expressions and canceling equals I've managed to bring the expression
$\left| u+v\right|^2$ $-$ $\left| u-v\right|^2$ $+$ $\left| u+iv\right|^2$ $-$ $\left| u-iv\right|^2$
into the form of $2(\langle u,v \rangle + \langle v,u \rangle +\langle u,iv \rangle +\langle vi,u \rangle)$. Deviding by 4 means the expression in the question may be written as $\frac 1 2 (\langle u,v \rangle + \langle v,u \rangle +\langle u,iv \rangle +\langle iv,u \rangle)$. This is where I got stuck, I have managed to reach this expression yet I do not know how to show it follows the three axioms or alternatively simplify it further to a point where it is a multiple of $\langle u,v \rangle$.

Any help would be greatly appriciated

2. Feb 15, 2017

### PeroK

I doubt it will simplify to a multiple of $\langle u,v \rangle$. Try one of the axioms and see whether the function you have satisfies it.

3. Feb 15, 2017

### Staff: Mentor

I am not a mathematician, but shouldn't you simply be showing that the definition satisfies all the properties of an inner product?

4. Feb 15, 2017

### PeroK

On reflection, perhaps continue to simplify! Hint: You'll need to think about the properties of the usual inner product and about complex conjugation.

5. Feb 15, 2017

I've been trying every possible way I can think of. The best I've managed is 2$(u\overline v + \overline {u \overline v} - iu\overline v +i \overline{u\overline v})$. I can't help but feel I am missing some key property of complex conjugate that prevents me from solving this...

6. Feb 15, 2017

### PeroK

The property you are missing is that $z + \bar{z} = 2Re(z)$.

7. Feb 15, 2017

Of course, now I feel like an idiot but I guess mistake is the mother of all teachers. Thanks for the help.

8. Feb 15, 2017

No, nevermind, it seems I'm completely helpless on this one. By defining $\langle u,v \rangle = u\overline v$ to be $(a+bi)$ the expression $2(u\overline v + \overline {u \overline v} - iu\overline v +i \overline{u\overline v})$ became
$2(a+bi+a-bi-i(a+bi)+i(a-bi))$ $=$ $2(2a -ai +b +ai +b) = 2(2a+2b)$ $=$ $4(a+b)$

Which can't be write since if I define $u=x+yi$ and $v=z+ti$ where $x,y,z,t\in R$, the expression becomes the real number $4(xz+yt+yz-xt)$, and so the function does not follow the second axiom of $\langle u,v \rangle = \overline {\langle v,u \rangle}$.
Where did I go wrong this time?...

9. Feb 15, 2017

### PeroK

I don't think you did. The original expression is clearly real. And real numbers are always their own complex conjugate. It says nothing in the definition of an inner product that it can't take solely real values.

You can stick with an expression involving $\langle u, v \rangle$. You should get a very simple sum, from which the inner product axioms follow easily.

Note that $u, v$ are vectors. And we are assuming that the norm of a vector is obtained from the usual inner product (*). The new inner product should really have a different notation, like $\langle u, v \rangle_1$

(*) Or, in fact, any inner product.

Last edited: Feb 15, 2017
10. Feb 15, 2017

The expression is real. which means $\langle u,v \rangle = \overline {\langle v,u \rangle} = \langle v,u \rangle$
However $\langle u,v \rangle = xz+yt+yz-xt$ while $\overline {\langle v,u \rangle} = \langle v,u \rangle = zx+ty+tx-zy \neq \langle u,v \rangle$ so the 2nd axiom does not apply
I can't seem to find any mistakes here, perhaps the book replaced some $+$ with a $-$?....

11. Feb 15, 2017

### PeroK

You've gone off track by replacing $u, v$ with complex numbers. These are vectors. It's difficult to spot your mistake, and fairly pointless, since you should be working either with the original expression (which could get messy) or with a simplified expression involving $\langle u,v \rangle$.

Hint: you should be getting Real and Imaginary parts coming out.

12. Feb 15, 2017

### PeroK

This is correct. $a$ and $b$ are the real and imaginary parts of $\langle u, v \rangle$ - the old inner product. So, you can start looking at the inner product axioms for the new inner product from here.

13. Feb 15, 2017

### PeroK

Your mistake is to confuse the new inner product, which is real-valued, with the original inner product, which may complex valued and for which $\langle u, v \rangle = \overline{\langle v, u \rangle}$

14. Feb 15, 2017

So the new inner product, which refers to the product in the original question, is a+b where a,b are the real and imaginary parts of the old inner product $\langle u,v \rangle$ respectively. I suppose I wanted to express it in a more "technical" manner since according to the book it is the complex equivalent of the real inner product function $\langle u,v \rangle$ $=$ $\frac 1 4$ $\left\|u+v\right\|$$^2$ $-$ $\left\|u-v\right\|^2$.
I'll do the axioms from here as you said then.

15. Feb 15, 2017

### PeroK

Are you sure the question isn't:

$\langle u,v \rangle$ $=$ $\frac 1 4$$\left| u+v\right|^2$ $-$ $\frac 1 4$$\left| u-v\right|^2$ $+$ $\frac i 4$$\left| u+iv\right|^2$ $-$ $\frac i 4$$\left| u-iv\right|^2$

16. Feb 15, 2017

Positive, the book is "Schaum's outlines Linear Algebera (fourth addition)". It is full of mistakes and the more I tend to this problem the more I think it is one of them.

Last edited by a moderator: Feb 15, 2017
17. Feb 15, 2017

### PeroK

Yes, you're right. I apologise for some of my posts on this. You can't possibly have a real-valued inner product on a complex vector space. I don't know what I was thinking! It's supposed to be an identity as I suggested in post #15.

Last edited by a moderator: Feb 15, 2017
18. Feb 15, 2017

No need to apologize, I've made my fair share of mistakes as well. Thank you very much for the assitance.