$x^TAx-b^Tx\leq 0\Rightarrow\|x\|_2\leq\|A^{-1}b\|_2$?

Question

Let $x$ be a vector, $A\succ 0$ an inverse matrix and $b$ a vector with proper dimensions. If $$x^TAx-b^Tx\leq 0,$$ do we have $$\|x\|_2\leq\|A^{-1}b\|_2?$$

I don't think it's a hard problem, but I cannot think of a proof or a counterexample right now. Thank you.

Robert Israel · Accepted Answer · 2014-12-29T01:40:31.710

1

To get a counterexample, start with any vector $x$ with $\|x\|_2 > \|A^{-1} b\|_2$. If $x^T A x - b^T x$ happens to be positive, replace $A$ by $-A$ and $b$ by $-b$.

EDIT: I presume you're working over the real numbers, and the added requirement $A \succ 0$ means $A$ is positive definite (and in particular symmetric).

For a counterexample, take $n=2$, $$ A = \pmatrix{1 & 0\cr 0 & 3\cr},\ b = \pmatrix{2\cr 8\cr},\ x = \pmatrix{3.4 \cr 1.7\cr} $$

edited Dec 29 '14 at 01:40

answered Dec 28 '14 at 22:07

Robert Israel

448,999

Thanks. I was to mean $A\succ 0$, so basically it's a strongly convex function in $x$. – ZhuShY Dec 28 '14 at 22:21
This is good.Thanks. – ZhuShY Dec 29 '14 at 01:56
BTW, if considering it as an ellipse, the length of the long radius should be less than $||A^{-1}b||_2$? – ZhuShY Dec 29 '14 at 02:10

$x^TAx-b^Tx\leq 0\Rightarrow\|x\|_2\leq\|A^{-1}b\|_2$?

1 Answers1