0
$\begingroup$

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?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ I am not clear on your notation. Is $|s|^T$ the norm of $s$ as a vector? What does the ${}^T$ do? $\endgroup$
    – Ben McKay
    Commented May 1, 2017 at 10:42
  • $\begingroup$ Hi, @BenMcKay, $T$ is the transpose, and $|.|$ is the absolute value (element-wise). That is $|s|=[|s_1| \ldots |s_n| ]^T$ $\endgroup$
    – Has
    Commented May 1, 2017 at 10:48
  • 2
    $\begingroup$ 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... $\endgroup$
    – Federico Poloni
    Commented May 1, 2017 at 10:50

2 Answers 2

Reset to default
1
$\begingroup$

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 $s^T \phi = 2$, $-|s|^T=(-1 \, -1)$, $-|s|^T A = 2 + 1/2$, $s^T M \phi = 5$, $-|s|^T M A =4$.

Improve this answer
$\endgroup$
1
$\begingroup$

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

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

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .