Under what conditions on a square matrix $A$ of size $n$ do we have

$|x^TAy| \le |x^Ty|$ for all $x,y \in \mathbb R^n$ ?


The above inequalities hold for $A \in \{0, I\}$, and so by simple continuity arguments, ought to be true for many more matrices...

Rough guesses

What if

  • $\|A\| \le 1$, where $\|A\|_2 := \sup_{\|x\|_2 \le 1}\|Ax\|_2$ ?, or
  • $A$ is row stochastic, i.e $a_{ij} \ge 0$ and $\sum_{k=1}^na_{ik} = 1$ for all $i, j$ ?
  • $\begingroup$ Oops, yes of course. Thanks very much. $\endgroup$ – dohmatob Apr 16 at 8:22

This is true iff $A=cI$ for some scalar $A$ with $|c| \leq 1$. Proof: let $\{e_1,e_2,...,e_n\}$ be an orthonormal basis. Then $e_1$ is orthogonal to $e_j$ for all $j >1$. From the hypothesis this implies that $Ae_1$ is also orthogonal to $e_j$ for all $j >1$. This means $Ae_1$is a multiple of $e_1$. Similar argument holds for $e_2,e_3,...,e_n$.

  • $\begingroup$ Thanks. What if the constraint on $x$ is strengthened to $x \in \Delta_n := \{x \in \mathbb R^n \mid x \ge 0, \; \sum_i x_i = 1\}$ ? $\endgroup$ – dohmatob Apr 16 at 8:27
  • $\begingroup$ I am using the hypothesis only for $x=e_i$ so the conclusion is still valid. $\endgroup$ – Kavi Rama Murthy Apr 16 at 8:29
  • $\begingroup$ Oops!, sure. Thanks. $\endgroup$ – dohmatob Apr 16 at 8:33

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.