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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?

Rodrigo de Azevedo
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Asaf Shachar
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3 Answers3

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Setting $B = E_{ij} - E_{ji}$ for $i \neq j$, we get $B^2 = -(E_{ii} + E_{jj})$.

Thus $E_{ii} + E_{jj} \in \text{span}(X)$ for all $i \neq j$. Now, fix $i \neq j$. Since $n \ge 3$, there is a $1 \le k \le n$, $k\neq i,j$. Thus $E_{ii} + E_{jj},E_{ii} + E_{kk},E_{kk} + E_{jj} \in \text{span}(X)$. Writing $E_{ii}=a,E_{kk}=b,E_{jj}=c$, we have

$$ a+b,a+c,b+c \in \text{span}(X).$$

By substracting, we obtain $b-c,a-c \in \text{span}(X)$, hence $2a=(a-c)+(a+c) \in \text{span}(X)$. Thus $E_{ii} \in \text{span}(X)$ for every $i$, so $\text{span}(X)$ contains all diagonal matrices.

Now, let $Q \in \text{O(n)}$. Since $Q^T X Q=X$, we get $Q^T \text{span}(X) Q=\text{span}(X)$.

So, $\text{span}(X)$ contains all orthogonal conjugations of diagonal matrices, i.e. all symmetric matrices, as required. (The last sentence is the spectral theorem).

Asaf Shachar
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    May I ask why this answer is community wiki? It is (a good example of) a self-answer question, quite the opposite of a community work. – Taladris Jan 02 '20 at 16:49
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More directly, for every orthonormal subset of vectors $\{u,v,w\}$, \begin{aligned} 2uu^T &=(-vv^T-ww^T)-(-uu^T-vv^T)-(-uu^T-ww^T)\\ &=(vw^T-wv^T)^2-(uv^T-vu^T)^2-(wu^T-uw^T)^2. \end{aligned} Therefore the set of all symmetric matrices lies inside the span of the squares of all skew symmetric matrices. The reverse inclusion is obvious.

user1551
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  • Thank you. The fact that every symmetric matrix can be represented as a sum of matrices of the form $uu^T$ follows from the spectral theorem, right? (we can write every symmetric matrix $A$ as a sum $\sum_i \lambda_i u_iu_i^T$ where the $u_i$ is a set of orthonormal eigenvalues of $A$). – Asaf Shachar Jan 08 '20 at 13:14
  • @AsafShachar Yes, it can be proved by using spectral decomposition. – user1551 Jan 08 '20 at 13:24
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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.

Dap
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