TITLE: The squares of skew-symmetric matrices span all symmetric matrices
QUESTION [7 upvotes]: This is a self-answered question. I post this here, since it wasn't obvious for me at first, and I think it might be helpful for someone at some future time (maybe even future me...).

Claim: Let $n \ge 3$, and let $X$ be the set of all squares of real $n \times n$ skew-symmetric matrices. Then $\text{span}(X)$ is the space of all symmetric matrices.

How to prove this claim?

REPLY [2 votes]: You could also use a result from a previous question of yours. Consider a symmetric matrix $A$ orthogonal to $\operatorname{span}(X)$ with respect to the Frobenius inner product. Let $\lambda_1,\dots,\lambda_n$ be the eigenvalues of $A.$ 
By Characterizing the dual cone of the squares of skew-symmetric matrices applied to $A$ and $-A,$ we get the conditions that for distinct $i,j,$ we have both $\lambda_i+\lambda_j\geq 0$ and $\lambda_i+\lambda_j\leq 0,$ which together give $\lambda_i+\lambda_j=0.$ Hence $\lambda_i=\tfrac12[(\lambda_i+\lambda_j)+(\lambda_i+\lambda_k)-(\lambda_j+\lambda_k)]=0$ for any distinct $i,j,k.$ For $n\geq 3$ the matrix $A$ is forced to be zero. So $X$ spans the space of symmetric matrices.