Proving a function is an inner product in a complex space

[Adgorn]

Homework Statement

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$

Homework Equations

N/A

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

 

[PeroK]

Homework Statement

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$

Homework Equations

N/A

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

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.

 

[DrClaude]

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

 

[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.

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

 

[Adgorn]

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

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...

 

[PeroK]

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...

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

 

[Adgorn]

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

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

 

[Adgorn]

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

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?...

 

[PeroK]

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?...

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.

 

[Adgorn]

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 $-$?...

 

[PeroK]

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 $-$?...

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.

 

[PeroK]

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
$2(a+bi+a-bi-i(a+bi)+i(a-bi))$ $=$ $2(2a -ai +b +ai +b) = 2(2a+2b)$ $=$ $4(a+b)$

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.

 

[PeroK]

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 $-$?...

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 $\langle u, v \rangle = \overline{\langle v, u \rangle}$

 

[Adgorn]

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 $\langle u, v \rangle = \overline{\langle v, u \rangle}$

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.

 

[PeroK]

Homework Statement

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$

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$

 

[Adgorn]

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$

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.

You can read it here (page 261, problem 7.96) <link to copyrighted material removed>

 

[PeroK]

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.

You can read it here (page 261, problem 7.96) <link to copyrighted material removed>

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.

 

[Adgorn]

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