$A$ is positive definite, $B$ is positive semidefinite and all the diagonal element is positive, prove that $A\circ B$ is positive definite
It is easy to prove it with Oppenheim Inequality:
Apparently $A\circ B$ is positive semi-definite.
Since $\det(A\circ B)\geq \det A\cdot \prod_{i=1}^n b_{ii} >0 $ (Oppenheim), $A\circ B$ is positive definite(all its eigenvalue is positive).
However, can it be proved without the inequality?
I tried to use mathematical induction:
For $n=1$, apparently $A\circ B > 0$.
If $\forall k<n, k>0, \forall A_k, B_k\in M_k$, $A_k$ is positive definite, $B_k$ is positive semi-defenite and all its eigenvalue is positive $\Rightarrow$ $A\circ B$ is positive definite.
Then we only have to prove that $\det A\circ B >0$. Let $A_{ij}$ represent A without $i$th row and $j$th coloumn, we have:
$\det A\circ B = \sum_{j=1}^na_{1j}\cdot b_{1j}\cdot\det A_{1j}\circ B_{1j} $
But I don't know how to go on.
Here's the proof I prefer: if $A$ has eigendecomposition $A = U\Lambda U^*$ (where $\Lambda$ is diagonal with diagonal entries $\lambda_j$ and $U$ is unitary with columns $u_i$), then we have $$ A = U\Lambda U^* = \sum_{j=1}^n \lambda_j u_ju_j^* $$ Now, note that $$ (u_ju_j^*) \circ B = \operatorname{diag}(u_j)B\operatorname{diag}(u_j)^* $$ Where $\operatorname{diag}(u_j)$ denotes the diagonal matrix whose diagonal entries are the entries of the vector $u_j$. Thus, we see that $(u_ju_j^*) \circ B$ is positive semidefinite. With that, we note that $$ A \circ B = \left( \sum_{j=1}^n \lambda_j u_ju_j^* \right) \circ B = \sum_{j=1}^n(u_ju_j^*) \circ (\lambda_j B) $$ which means that $A \circ B$ is a sum of positive semidefinite, and is therefore itself positive semidefinite.