Positive definite matrix bounded below

[Charles49]

Homework Statement

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.

 

[micromass]

What is a??

 

[Charles49]

Sorry, its supposed to be x.

 

[micromass]

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

Calculate

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

 

[Charles49]

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

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

$$C$$ exists because A is positive definite. $$C^{-1}[\tex] exists because A has no zero eigenvalues.$$

 

[micromass]

Yes, so that shows that

$$<Cx-C^{-1}b,Cx-X^{-1}b>\geq 0$$

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

 

[Charles49]

This suggestion helped a lot. Thanks.