Revisit "What is the rank of $Q+Q^T-Q\circ I$ if $Q = qq^T$"

Question

Based on my previous problem: What is the rank of $Q+Q^T-Q\circ I$ if $Q = qq^T$

1. $Q = qq^T$, $q\in \mathbb{R}^n$. Suppose $q^Tq=1$.
2. $A\circ B:$ Hadmard (elementwise) product
3. $Q\circ I$ just takes the diagonal entries of $Q$.

My question: Is $Q+Q^T-Q\circ I$ still positive semidefinite?

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I guess the answer is not fixed by Gershgorin circle theory since the diagonal terms become small. Not sure if I am correct.

Answer 1

It is clearly not always positive semidefinite. E.g. when $q^T=(x,y)$, we have $$ qq^T+qq^T-(qq^T)\circ I=\pmatrix{x^2&2xy\\ 2xy&y^2}, $$ whose determinant $-3x^2y^2$ is nonnegative only when $x=0$ or $y=0$.