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

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 $\mathcal{M}_n$?

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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.
I made a mistake when I wrote the second set in $\mathcal{M}_n$. For the second set $i=j$ is permitted. Would that make any difference? – S.B. Sep 27 '12 at 1:59
@S.B.: It certainly makes a difference for $n=1$, where you now have the multiple of the identity matrix that you need. I'm not sure for $n=2$, but for $n\gt2$ it doesn't make a difference, since this new matrix can only be used to increase, not to decrease the diagonal elements. – joriki Sep 27 '12 at 2:02
@S.B.: I don't think so. For $n=2$, the matrix $$\pmatrix{a&b\\b&c}$$ is positive-semidefinite iff $a\ge0$, $c\ge0$ and $ac\ge b^2$. I don't think a curved region like that can be the conical hull of a finite set of points, which I would expect to be bounded by hyperplanes. – joriki Sep 27 '12 at 2:25