1
$\begingroup$

$A$ is an adjacency matrix for a graph so $A$ is symmetric. $x$ is a unit vector.

Why is $\langle Ax,x\rangle^4\leq \langle A^4x,x\rangle$?

So far I have that $\langle Ax,x\rangle^4\leq ||Ax||^2||x||^2= ||Ax||^2$ By the Cauchy Schwartz inequality.

$\endgroup$
1
  • $\begingroup$ Shouldn't the first inequality be $\langle Ax, x \rangle^4 \leq ||Ax||^4 ||x||^4$? $\endgroup$ Mar 30 '20 at 4:10
2
$\begingroup$

Hint: $\langle A^4x, x \rangle = \langle A^2x, (A^T)^2x \rangle$.

$\endgroup$
2
$\begingroup$

This is true for any Hermitian matrix $A$. By a change of orthonormal basis, we may assume that $A=\operatorname{diag}(a_1,a_2,\ldots,a_n)$ is a real diagonal matrix. Let $c_i=|x_i|^2$. Then $c_i\ge0,\,\sum_ic_i=1$ and the inequality in question is equivalent to $\left(\sum_i c_ia_i\right)^4\le\sum_ic_ia_i^4$, which is true because $f(t)=t^4$ is a convex function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.