Is there any sufficient condition under which the hadamard product of two square matrices is invertible?

Question

Is there any sufficient condition under which the hadamard product of two square matrices is invertible ?

Answer 1

Schur product theorem states that for symmetric positive semidefinite matrices $A$, $B$, we have

$$\det(A \circ B) \geq \det(A) \det(B).$$

Hence a sufficient condition cound be both $A$ and $B$ are symmetric positive definite.

Hence $\det(A)\det(B) > 0$

Edit:

Using Oppenheim's inequality, $$\det(A \circ B) \geq \det(A) \prod_{i=1}^nB_{ii}$$

for symmetic positive semidefinite matrices. One sufficient condition is that $A$ is symmetric positive definite, $B$ is symmetric positive semidefnite and all diagonal entries of $B$ is non-zero.

Similarly, we can have $B$ is symmetric positive definite, $A$ is symmetric positive semidefinite and all diagonal entries of $A$ is non-zero.