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

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

Notes

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 $i, j$ ?
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  • $\begingroup$ Oops, yes of course. Thanks very much. $\endgroup$ – dohmatob Apr 16 at 8:22
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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 $e_2,e_3,...,e_n$.

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  • $\begingroup$ 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

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