Is a "indicator matrix" gotten from a symmetrical favorable - semidefinite matrix itself symmetic favorable - semidefinite?

Intend that $A \in {\cal S}_+^n$ is a symmetrical favorable semidefinite matrix. Allow $B = {\rm sign}(A)$, where the indicator is taken elementwise. Is the resulting matrix $B$ constantly favorable semidefinite?

Otherwise, under what problems can we claim that $B \in {\cal S}_+^n$?

2019-05-18 23:07:13
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Answers: 2

An unimportant instance when $B$ is semidef is claim when the initial matrix is elementwise non - adverse (and also you specify indicator (0) = 1). Because instance, $B=ee^T$.

As a whole, it could be tough to offer a non - unimportant set of enough problems on $A$ to make sure that $B$ is semidefinite.

2019-05-31 06:26:43

A tiny symmetrical perturbation of the identification matrix declares precise, yet the equivalent indicator matrix need not declare semidefinite. As an example, allow $A$ be the 3 - by - 3 matrix with 1 on the angled, and also, claim, - 1/ 100 in the various other 6 access.

(I have absolutely nothing to claim concerning your 2nd inquiry.)

2019-05-21 09:38:39