# Boundedness of a Quadratic Form

1. Jul 20, 2011

### thegreenlaser

Given:
$$H_0 = x^2 + y^2 + 2axy$$

How does one go about finding the bounds on x and y, based on a? The author of a book I'm reading says that bounds are simple to show based on the different conditions |a|<1, |a|>1, or |a|=1. Unfortunately, I'm not finding it so simple, except for the |a|=1 case. Could someone at least point me in the right direction?

Thanks

2. Jul 20, 2011

### pmsrw3

Hmm. Does it help to write H0 as

$$\alpha(x+y)^2 + \beta(x-y)^2$$

?

3. Jul 20, 2011

### thegreenlaser

I'm trying that now. It turns out to be:

$$\frac{1}{2}(1+a)(x+y)^2 + \frac{1}{2} (1-a)(x-y)^2$$

I have yet to see if that helps me in any way.

Also, just in case it was a little ambiguous in my first post, the goal is to find out whether if H0 is bounded then x and y are bounded as well. If they are bounded, I need to know how they're bounded in terms of H0

4. Jul 20, 2011

### pmsrw3

In your OP, you said you were trying to find bounds on x and y. Does that mean that H0 is a fixed number (presumably >0)?

5. Jul 20, 2011

### thegreenlaser

^Just edited that in my last post. Yes, sorry, that was a little unclear.

6. Jul 20, 2011

### pmsrw3

Ah! In that case, I think the answer is clear. If |a|<1, then you have a sum of two positive squares. Obviously those are both going to be bounded, and that will bound x and y. If |a|>1, you have a positive and a negative square, and you can make one as big as you like as long as you compensate by making the other big, too.

7. Jul 20, 2011

### thegreenlaser

That does make sense. Thanks!

Now is there any way to do that quantitatively? i.e. Can we bound x and y directly in terms of H0?

EDIT:
Breaking it into it's $$H_0 =\frac{1}{2}(1+a)(x+y)^2 + \frac{1}{2} (1-a)(x-y)^2$$ form did give me a lot of useful information about where the zeroes are and where it's positive/negative, but I still haven't coaxed any boundedness info out of it.

Last edited: Jul 20, 2011