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If I have a positive semidefinite matrix $A$ and a negative definite matrix $B$, is it true that their Hadamard product $A\circ B$ is negative semidefinite? Ideally I am looking for a proof / a complete argument for why it is true / false that I can replicate.

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Schur product theorem states that Hadamard product of two positive semidefinite matrices is positive semidefinite.

$B$ is negative definite $\implies -B$ is positive definite.

Since $$A \circ B = -(A \circ (-B)),$$ and $A \circ (-B)$ is positive semidefinite by Schur product theorem.

We conclude that $A \circ B$ is negative semidefinite.

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  • $\begingroup$ This is only true if the dimension of the matrices $A$ and $B$ is odd, correct? Else multiplying them by the negative identity leaves their determinant unchanged. $\endgroup$ – jdizzle Jan 27 '18 at 15:52
  • $\begingroup$ a negative definite matrix can have a positive determinant. For example $-I_2$, the negative of the identity matrix of size $2$. $\endgroup$ – Siong Thye Goh Jan 27 '18 at 16:00
  • $\begingroup$ Ah sorry, i think I am using a different definition. Ive also found a counter example to this arguement. $$A=\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad B=\begin{pmatrix}1&3\\3&1\end{pmatrix}.$$ Here $A$ and $A\circ B$ are positive definite, $B$ is not. Any positive definite diagonal matrix will do in place of $A$, and it even still works if the off-diagonal entries of $A$ are nonzero but small. $\endgroup$ – jdizzle Jan 27 '18 at 16:03
  • $\begingroup$ Note that your $B$ is not negative definite. $\endgroup$ – Siong Thye Goh Jan 27 '18 at 16:10

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