Quadratic and billinear form inequality

57 Views Asked by At

Working in reals, consider a positive semidefinite matrix $A$, and two vectors $x,z$ that contain strictly positive elements only. Assume that we know

$$x'Ax < z'Az$$

Is it true that $$x'Ax < x'Az < z'Az\;\;\; ??$$

It "feels" true, but I haven't been able to prove it, so it may not be the case after all.

1

There are 1 best solutions below

0
On BEST ANSWER

Even assuming that the elements of $x$ and $z$ are strictly positive, the statement is false. Take for example $A$ the identity matrix, $x = (1, a)$, $z = (2a, 1)$ - in the limit $a \to 0$, $x$ and $z$ are orthogonal. We have

$$x^TAx = 1 + a^2, \quad z^TAz = 1 + 4a^2, \quad \text{but} \quad x^TAz = 3a,$$

so the inequality to be disproved is not satisfied when $a$ is small enough.