Does this conic combination generate all $n\times n$ real symmetric positive-semidefinite matrices?

Question

Let $e_i$ denote a column-vector of length $n$ whose entries are all zero except for the $i$-th entry that is 1. Now consider the set of $n\times n$ matrices given by $$\mathcal{M}_n=\left\lbrace \left(e_i-e_j\right)\left(e_i-e_j\right)^\mathrm{T}\mid 1\leq j<i\leq n\right\rbrace\cup\left\lbrace \left(e_i+e_j\right)\left(e_i+e_j\right)^\mathrm{T}\mid 1\leq j<i\leq n\right\rbrace.$$

My question is that can we obtain all $n\times n$ real symmetric positive-semidefinite matrices as conic combinations of matrices in $\mathcal{M}_n$?

Answer 1

No. This is not the case for any $n$. For $n=1$, $\mathcal M_n$ is empty, so the cone consists only of the zero matrix, whereas there are non-zero positive-definite matrices. For $n=2$, the diagonal elements of a matrix in the cone are always equal, which need not be the case for a positive-semidefinite matrix. For $n\gt2$, the matrix with all entries $1$ is positive-semidefinite, but it's not in the cone, since for a matrix in the cone the sum of diagonal elements cannot be less than the sum of off-diagonal elements.