# Finding conditions so that something is an inner product

I'm having trouble with a type of exercise where you have to find conditions so that something is an inner product. The most general of these exercises is the following one:

Find $a, b \in \mathbb{R}$ so that the following function, $\phi : \mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}$ is an inner product: $$\phi(x,y) = ax_1 y_1 + b x_1y_2 + bx_2y_1 + b x_2y_2+(1+b)x_3y_3.$$

It's easy to see that $\phi(v+w,z)=\phi(v,z)+\phi(w,z)$, that $\phi(\alpha v, w)=\alpha \phi(v,w)$ and that $\phi(v,w)=\phi(w,v)$ without running into any limitations for a and b.

So the only thing that's left is cheking for which a's and b's $\phi(v,v) \gt 0$ with $v \neq (0,0,0)$ .

$\phi(x,x)= ax_1^2 +2 b x_1x_2 +b x_2^2 + (1+b)x_3^2$

This is where I get pretty lost. I tried the following with not much of a result

$\phi(x,x)= (x_1 + x_2)^2 + (a-1)x_1^2 + (b-1)x_2^2 + (2b - 1)x_1x_2 +(1+b)x_3^2$

Then since $(2b-1)x_1x_2 = b x_1x_2 + (b-1)x_1x_2$

$\phi(x,x)= (x_1+x_2)^2 +(a-1)x_1^2+(b-1)x_2 (x_2+x_1)+bx_1x_2+(1+b)x_3^2$
$\phi(x,x)= (x_1+x_2)(x_1+x_2+(b-1)x_2) +x_1((a-1)x_1+bx_2)+(1+b)x_3^2$

Then since I want to find the a's and b's so that it's only 0 is $x_1=x_2=x_3=0$ I'd want the a's and b's so that the only solution to this is the trivial one

$(x_1+x_2)(x_1+x_2+(b-1)x_2) +x_1((a-1)x_1+bx_2)+(1+b)x_3^2 = 0$
$(x_1+x_2)(x_1+x_2+(b-1)x_2) = - x_1((a-1)x_1+bx_2)-(1+b)x_3^2$

These are all pretty ugly things, and most likely laughable, but I can't really find a way to make this work. In some other cases where I only had to find one value I ended up with a square root so I'd find the values so that the argument is negative and that's it, but here I can't seem to do that.

Any help would be greatly appreciated.

EDIT: Sometimes writings things down slowly like this really helps, and I think I may have found the solution.

$(x_1+x_2)^2 +(a-1)x_1^2+(b-1)x_2 (x_2+x_1)+bx_1x_2+(1+b)x_3^2 = 0$

$x_1 + x_2 = \sqrt{-1}\sqrt{(a-1)x_1^2+(b-1)x_2 (x_2+x_1)+bx_1x_2+(1+b)x_3^2}$

So I have to exclude a and b so that
$a -1 < 0$
$b-1 < 0$
$b < 0$
$b +1 < 0$

So that leaves me with $a \gt 1$ and $b > -1$. I think that pretty much solves it.

Sorry for polluting the board like this.

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Remember that $\sqrt{x^2} = |x|$, not $x$. – Arturo Magidin Dec 9 '11 at 21:21

The matrix of $\phi$ with respect to the standard basis is $\begin{pmatrix} a & b & 0 \\ b & b & 0 \\ 0 & 0 & 1 + b \end{pmatrix}$ and there are lots of criteria for checking that a matrix (and hence its corresponding form) is positive definite. Sylvester's criterion, for example, seems to give me the same conditions as in Prof Magidin's answer.

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That's a great way of thinking the problem. Really simple and elegant, I love it. I'll read a little about it and start using it! Thanks! – Bananas Dec 9 '11 at 21:36

You have $$\phi\Bigl( (x_1,x_2,x_3),(x_1,x_2,x_3)\Bigr) = ax_1^2 + 2bx_1x_2 + bx_2^2 + (b+1)x_3^2;$$ you want to see what conditions on $a$ and $b$ will guarantee that this is nonnegative, and positive if $x_1^2+x_2^2+x_3^2\neq 0$.

The key is to recognize that the first three terms are almost $b$ times the square of $(x_1+x_2)$. So we rewrite it so that it is that: $$\phi\Bigl( (x_1,x_2,x_3),(x_1,x_2,x_3)\Bigr) = (a-b)x_1^2 + b(x_1+x_2)^2 + (b+1)x_3^2.$$

Evaluating at $(1,0,0)$ we get $a$, so we can see that we need $a\gt 0$.

Evaluating at $(0,1,0)$ we get $b$, so we also need $b\gt 0$.

Evaluating at $(1,-1,0)$ shows that we need $a-b\gt 0$, or $a\gt b$.

Are the conditions $a\gt b \gt 0$ enough to guarantee that $\phi$ is an inner product?

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It never occur to me to actually try it with a few values to find conditions, that's an awesome tip that I'm sure will help me with a lot of stuff. I edited my main post just before reading yours, and we got to pretty similar results, although my way is a lot less elegant. I think that it still works if b> -1 instead of 0, but I'm not completely sure yet. Thanks for your help. – Bananas Dec 9 '11 at 21:10
@Collman: Take any $b$ with $-1\lt b\lt 0$, and plug in $(0,1,0)$. You get $b$, so if $b\lt 0$, you contradict positive definiteness. – Arturo Magidin Dec 9 '11 at 21:20
I was checking that, you're right. Thanks a ton for your help! – Bananas Dec 9 '11 at 21:27