Lower bound on smallest eigenvalue of hadamard product of two Hermitian matrices Let A and B be $n \times n$ real symmetric matrix. Suppose A is positive definite and denote its smallest eigenvalue as $\lambda_{\min}(A)>0$. All elements of B are positive and bounded, i.e. $0\le a <B_{ij}<b, B_{ij}=B_{ji}$. Denote the hadamard product of two matrices as $C=A\circ B$.
Here are two questions:

*

*Can we know whether $B$ is positive definite or not? (constant a and b can be chosen arbitrarily to satisfy the requirements)

*Can we lower bound the smallest eigenvalue of C?

P.S.
I'm encountered with this question in an engineering computation problem. The motivation is to control $-v^{\top} C v$ or $\|v^{\top} C\|$ and we can make some restrictions on B, for example, bounding the entries of $B$ or set $n$ as large as we need. I have seen some results in On majorization and Schur products while $B$ has to be positive definite, which may not be guaranteed in my question.
 A: As discussed in the comments, the positivity of the entries of $B$ does not ensure that $B$ is positive definite. I further suspect that there's no good way to take advantage of the positivity of the entries in checking whether $B$ is positive definite.
Here's a proof of a corrected version of the result that I referred to

Claim: If $A$ is positive definite and $B$ is symmetric with at least one negative eigenvalue, then $\lambda_{\min}(A \circ B) \geq \lambda_\max(A)\lambda_\min(B)$.

Proof: We note that $A \circ B$ is a principal submatrix of the Kronecker product $A \otimes B$. The eigenvalues of $A \otimes B$ are of the form $\lambda = \lambda_i(A) \lambda_j(B)$ for $1 \leq i,j \leq n$. By the Cauchy interlacing inequality, it follows that
$$
\lambda_{\max}(A)\lambda_\min(B) = \lambda_{\min}(A \otimes B) \leq \lambda_{\max}(A \circ B). \qquad \square
$$
Alternative proof: Let $\mu = \lambda_{\min}(B)$, so that $B - \mu I$ is positive semidefinite. It follows that $A \circ (B - \mu I)$ is positive semidefinite, so that its minimal eigenvalue is at least $0$. Thus,
$$
\lambda_\min(A \circ B) = \lambda_\min(A \circ (B - \mu I) + \mu A)
\\ 
\geq \lambda_\min(A \circ (B - \mu I) + \lambda_\min(\mu A)
\\ =  \lambda_\min(A \circ (B - \mu I) - |\mu|\lambda_\max(A)
\\ \geq 0- |\mu|\lambda_\min(A) = \lambda_\max(A)\lambda_\min(B).
$$
