# Positive definite matrix bounded below

1. Sep 7, 2011

### Charles49

1. The problem statement, all variables and given/known data

Let $$A$$ be a positive definite $$n\times n$$ real matrix, $$b\in\mathbb{R}^n$$, and consider the quadratic polynomial $$Q(x)=\frac{1}{2}\langle{x, Ax\rangle}-\langle{b, x\rangle}$$. Show that $$Q$$ is bounded below.

2. The attempt at a solution

I have to come up with a constant $$m$$ so that $$Q(x)\ge m$$ for all $$x\in\mathbb{R}^n.$$ I see that $$Q$$ looks a lot like a parabola. I know how to find the lower bound of a parabola opening upward but I don't know how to generalize this to quadratic forms.

Last edited: Sep 7, 2011
2. Sep 7, 2011

What is a??

3. Sep 7, 2011

### Charles49

Sorry, its supposed to be x.

4. Sep 7, 2011

### micromass

Let C be such that $C^2=A$ (why does C exist?) and let $c=C^{-1}b$ (why is C invertible?).

Calculate

$$\frac{1}{2}<Cx-c,Cx-c>$$

5. Sep 7, 2011

### Charles49

This is what I got but I don't know what it does

$$\frac{1}{2}\langle{Cx-C^{-1}b, Cx-C^{-1}b\rangle} = \frac{1}{2}||Cx-C^{-1}b||\\ =\frac{1}{2}||C^{-1}(Ax-b)||$$

$$C$$ exists because A is positive definite. $$C^{-1}[\tex] exists because A has no zero eigenvalues. Last edited: Sep 7, 2011 6. Sep 7, 2011 ### micromass Yes, so that shows that [tex]<Cx-C^{-1}b,Cx-X^{-1}b>\geq 0$$

Now try to evaluate this in another way... Use the linearity of the inproduct.

7. Sep 10, 2011

### Charles49

This suggestion helped a lot. Thanks.