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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.