If $M, N \in \mathbb{R}^{n \times n}$ are symmetric and positive semi-definite then is $M^2 \circ N^2 - (M \circ N)^2 = (MM) \circ(NN) - (M \circ N)^2$ symmetric and positive semi-definite? Here $\circ$ denotes the component-wise product of the matrices. I feel like this should not be true and that there must be a counterexample but I am not completely sure. Even the $2 \times 2$ case seems to get messy quickly when trying to express $M^2 \circ N^2 - (M \circ N)^2$. I believe there may be some trick with expressing $M^2 \circ N^2$ in a different form. The reason I am considering this is that a friend of mine told me she was having a hard time with this problem. It is proving to be quite diifcult.
Any help is welcomed.
Let $M$ and $N$ be symmetric real matrices. Then,
$\left[M^2 \circ N^2\right]_{ij}=\sum_{k,l} M_{ik} M_{kj} N_{il} N_{lj}$
$\left[(M \circ N)^2\right]_{ij}=\sum_{k,l} M_{ik} M_{kj} N_{il} N_{lj}\delta_{kl}$
Let $R={\mathbb 1}-I$, a matrix whose elements $R_{kl}=1- \delta_{kl}\geq 0$. Then
$a^T \left(M^2 \circ N^2 - (M \circ N)^2\right) a=\sum_{i,j}a_i \left[M^2 \circ N^2 - (M \circ N)^2 \right]_{ij} a_j=\sum_{i,j,k,l} R_{kl} a_i M_{ik} M_{kj} N_{il} N_{lj} a_j=\sum_{k,l} R_{kl} (\sum_i a_i M_{ik}N_{il}) (\sum_j M_{kj}N_{lj}a_j)$
Singe-index sums in round brackets are the same real number $x_{kl}$, their product is $x_{kl}^2\geq 0$
Thus $a^T \left(M^2 \circ N^2 - (M \circ N)^2\right) a\geq0$ and the matrix $M^2 \circ N^2 - (M \circ N)^2$ is Positive semi-definite.