# Does every negative semidefinite matrix lie in the convex cone generated by the squares of skew-symmetric matrices?

Let $$C$$ be the convex cone generated by all the squares of real $$n \times n$$ skew-symmetric matrices.

Does every negative-semidefinite matrix lie in $$C$$?

I know that every square of a skew-symmetric matrix can be written in the form $$-\big(a_1 \, (x_1x_1^T + y_1y_1^T) + \cdots + a_k \, (x_kx_k^T + y_ky_k^T)\big),$$

where $$a_i \ge 0$$.

This can be seen easily for $$n=1$$ (trivially) and $$n=2$$ (squares of skew-symmetric matrices are multiples of the identity).
Let $$n\ge2$$. Then the following diagonal matrix $$A:=\pmatrix{ -1 \\ &0\\&&\ddots\\&&&0}$$ is not in the cone. Assume otherwise. Then there are skew-symmetric matrices $$A_k$$ such that $$A = \sum_k A_k^2$$.
Multiplying this equation from the left with $$e_i^T$$ and from the right with $$e_i$$ yields $$- \delta_{1,i} = \sum_k e_i^TA_k^2e_i = -\sum_k e_i^TA_k^TA_ke_i =-\sum_k \|A_ke_i\|^2.$$ For $$i>1$$ this implies that the $$i$$-th column of all matrices $$A_k$$ is zero. By skew-symmetry, the first column of all $$A_k$$ has to be zero as well. So $$A_k=0$$ for all $$k$$, which implies $$A=0$$, a contradiction.