Schur's product theorem says that for positive definite matrices $A$,$B$ also their Hadamard (entry-wise) product $A \circ B$ is positive definite. I am trying to prove a converse result of the following type:
Let $A$ be positive (semi-)definite and $B$ Hermitian. If $A \circ B$ is positive (semi-)definite, then also $B$ is positive (semi-)definite.
My first observation is that it is sufficient to prove that $\det(B) \ge 0$, by induction on the dimension $n$ of the matrices. Indeed, for $n=1$ the result is trivial. For $n > 1$ we can use the fact that $B$ is positive definite if all its leading principal minors are positive. The $k-th$ leading principal minors for all $k \le n-1$ are positive by induction hypothesis and the $n$-th leading principal minor is $\det(B)$.
Using this observation I can prove the result under the additional assumption that $A$ has rank one and no zero elements: In this case there is a complex vector $c$ without zeroes such that $A = cc^*$. Then we can write:
$$ 0 \le \det(A \circ B) = \det(D_c B D_\bar{c}) = \prod_i c_i \bar{c_i} \det(B) = \prod_i |c_i|^2 \det(B)$$
where $D_c$ is the diagonal matrix with the vector $c$ on its diagonal and $\bar{c}$ denotes complex conjugation.
But I am stuck on proving the theorem in general, without assumptions on the rank of $A$. Any ideas?
The stated converse of Schur's product theorem does not hold true in general.
A simple counterexample is given by $$ A= \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 1 \\ 1 & 3/4 \end{pmatrix}. $$ Clearly, both matrices are symmetric, $A$ is positive definite, $B$ is not positive definite. However, their Hadamard product $$A \circ B = \begin{pmatrix} 1 & 1 \\ 1 & 3/2 \end{pmatrix}$$ is again positive definite.