1
$\begingroup$

Is $\frac{1}{Tr(X^TPX)}$ concave, if $P$ is p.s.d? All matrices have real-valued entries. Is there a property that can be generalized to check here. I was thinking of composition with decreasing function- power fn with negative powers. I know if the power is positive, it would be convex. But does it mean that for negative power it would be concave? or would it be non-convex?. second derivative tes is suggesting non convex in general..

I think required condition has to be $[λ f ( y)+ (1− λ) f ( x)][ f (λ x+ (1− λ) y)]≤ f ( x) f ( y)$ to be true for $1/f(x)$ to be concave when $f(x)$ is convex. If so, does $\frac{1}{Tr(X^TPX)}$ satisfy this and if not under what condition could it be satisfied?

$\endgroup$

1 Answer 1

1
$\begingroup$

Notice that in the simplest case when $P$ is a positive number, and $X$ is a real number, we should consider $\frac{1}{x^2}$, which is convex in each piece of its domain.

$\endgroup$
1
  • $\begingroup$ yeah had posted in a bit of haste. when i compute the second derivatives, I got that answer already :) $\endgroup$
    – hearse
    Apr 29, 2020 at 15:50

You must log in to answer this question.

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