# Bound $|x^TAy|$ in terms of $\|A\|$ and $|x^Ty|$

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 $$i, j$$ ?
• Oops, yes of course. Thanks very much. – 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 $$e_2,e_3,...,e_n$$.
• Thanks. What if the constraint on $x$ is strengthened to $x \in \Delta_n := \{x \in \mathbb R^n \mid x \ge 0, \; \sum_i x_i = 1\}$ ? – dohmatob Apr 16 at 8:27
• I am using the hypothesis only for $x=e_i$ so the conclusion is still valid. – Kavi Rama Murthy Apr 16 at 8:29