Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Remember that $\sqrt{x^2} = |x|$, not $x$. – Arturo Magidin Dec 9 '11 at 21:21
up vote 3 down vote accepted

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.

share|cite|improve this answer
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?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.