# Does every real symmetric matrix with non-positive eigenvalues have an anti-symmetric square root?

Given $$S$$ a real symmetric matrix. The original question is to find an if and only if condition for there to exist an anti-symmetric matrix $$A$$ such that $$S=A^2.$$ My guess is that the condition should be $$S$$ having all non-positive eigencalues.

One direction is trivial: if such an $$A$$ exists, then $$S=-A^TA$$ and we know $$A^TA$$ is positive-definite and thus have non-negative eigenvalues. So $$S$$ has non-positive eigenvalues.

Then I am kind stuck at the other direction. Given such an $$S$$, we can diagonalize $$-S$$ with unitary $$U$$:$$-S=US'U^T$$ with $$S'$$ have positive diagonal elements. Then we can take square root of $$S'$$ i.e. $$-S=U\sqrt{S'}\sqrt{S'}U^T.$$ Then put $$A=\sqrt{S'}U^T$$ we have $$S=-A^TA.$$ But I cannot proceed to show $$-A^TA=A^2$$ i.e. $$A$$ is anti-symmetric. So I am wondering how to proceed from there or is my initial iff condition correct? Any hint is appreciated!

To identify the if and only if condition, denote the order of $$S$$ by $$n$$. If such anti-symmetric matrix $$A$$ exists, suppose $$\pm ib_1, \ldots, \pm ib_s$$ are all non-zero eigenvalues of $$A$$, where $$s \leq n/2$$, then there exists an order $$n$$ orthogonal matrix $$P$$ such that \begin{align*} A = P\mathrm{diag}\left(\begin{pmatrix} 0 & b_1 \\ -b_1 & 0 \end{pmatrix}, \cdots, \begin{pmatrix} 0 & b_s \\ -b_s & 0 \end{pmatrix}, \underbrace{0, \cdots, 0}_{n - 2s}\right)P^T. \tag{*} \end{align*} Hence \begin{align*} A^2 = P\mathrm{diag}(-b_1^2, -b_1^2, \cdots, -b_s^2, -b_s^2, \underbrace{0, \cdots, 0}_{n - 2s})P^T. \end{align*} This shows the eigenvalues of $$S = A^2$$ must satisfy:
2. The number of strictly negative eigenvalues must be an even number, say $$2s$$. And they consist of $$s$$ pairs of identical eigenvalues.
It is easy to verify that any symmetric matrix $$S$$ satisfying 1. and 2. above admits the decomposition as well (just spell out expression $$(*)$$ from $$S$$'s spectral decomposition).
To see just having condition 1. is not sufficient, take \begin{align*} S = \begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix}. \end{align*} If $$S = A^2$$ exists, then as an order $$2$$ anti-symmetric matrix, $$A$$ must have the form $$A = \begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix}$$, where $$a > 0$$, whence \begin{align*} A^2 = \begin{pmatrix} -a^2 & 0 \\ 0 & -a^2 \end{pmatrix} \neq \begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix} = S. \end{align*}