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!
 A: 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:

*

*All eigenvalues are non-positive (as you discovered).

*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*}
