Some "Product" of Positive Definite Matrices I could remember that if $A,B$ are two positive definite matrices, then $(a_{ij}b_{ij})$ is positive definite also. But I could not see how to prove it then.

Edit: Following the remark of user126154, I suppose here that the two matrix $A$ and $B$ are symmetric.
Let $x=(x_1,\cdots,x_n)\in \mathbb{R}^n=E$ and put $q_A(x)=\sum_{i,j}a_{i,j}x_ix_j$. As $A$ is positive definite, there exists $n$ independant linear forms $\displaystyle T_k(x)=\sum_{l=1}^n \alpha_{k,l}x_l$ such that $\displaystyle q_A(x)=\sum_{k=1}^n (T_k(x))^2$. Hence you get that $a_{i,j}=\sum \alpha_{k,i}\alpha_{k,j}$.
Hence, if you put $y_k=(\alpha_{k,i}x_i, i=1,\cdots,n)$:
$$q(x)=\sum_{i,j}b_{i,j}a_{i,j}x_ix_j=\sum_{i,j,k}b_{i,j}\alpha_{k,i}\alpha_{k,j}x_ix_j
=\sum_{i,j,k}b_{i,j}(\alpha_{k,i}x_i)(\alpha_{k,j}x_j)=\sum_{k}q_B(y_k)$$
You have hence $q(x)\geq 0$ for all $x$; if $q(x)=0$, you get $q_B(y_k)=0$ for all $k$, hence $y_k=0$ for all $k$, and $T_l(x)=0$ for all $l$; as the $T_l$, $1\leq l\leq n$ are independant, this show that $x=0$.
