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I have an inequality as follows

$$s^T\phi\leq -|s|^TA$$

where $s$, $\phi$ and $A$ are vectors with appropriate dimensions. I want to prove that this inequality holds for the following too

$$s^TM\phi\leq -|s|^TMA$$

where $M$ has positive eigenvalues. Intuitively, it seems to be right. Any ideas?

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    $\begingroup$ I am not clear on your notation. Is $|s|^T$ the norm of $s$ as a vector? What does the ${}^T$ do? $\endgroup$ Commented May 1, 2017 at 10:42
  • $\begingroup$ Hi, @BenMcKay, $T$ is the transpose, and $|.|$ is the absolute value (element-wise). That is $|s|=[|s_1| \ldots |s_n| ]^T$ $\endgroup$ Commented May 1, 2017 at 10:48
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    $\begingroup$ This is the kind of claims that I start believing in only after I have done at least 10.000 random experiments without finding any counterexample... $\endgroup$ Commented May 1, 2017 at 10:50

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Please check my arithmetic. Let $\phi=(1 \ 1)^T$, $s=(1 \ 1)^T$, $A=(-1/2 \ \ {-2})^T$. Then try $$ M=\begin{pmatrix}4 & 0 \\ 0 & 1 \end{pmatrix}. $$ I seem to get $s^T \phi = 2$, $-|s|^T=(-1 \, -1)$, $-|s|^T A = 2 + 1/2$, $s^T M \phi = 5$, $-|s|^T M A =4$.

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Even you let $\phi=A$, the statement still not hold in general.

Let $s=(1,1)$, $\phi=A=(1,-1)$, $M=diag(2,1)$.

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