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Suppose that $A$ is a real skew-symmetric $n \times n$ matrix and $S = A^2$ (a symmetric matrix). Now we want to find all skew-symmetric square roots of $S$, i.e. all the skew-symmetric matrices $X$ such that $X^2 = S$. We know that the solution exists, namely $X = A$, and it is easy to immediately see one more solution, $X = -A$ (hence, the solution is generally not unique).

The question is: Can we express all the solutions $X$ with the elements of the matrix $S$?

If I'm not mistaken, the answer is affirmative in the

the $n = 2$ case, $$A = \left( \begin{array}{cc} 0 & a \\ -a & 0 \end{array} \right) \ , \quad S = \left( \begin{array}{cc} -a^2 & 0 \\ 0 & -a^2 \end{array} \right) \ , \quad X = \left( \begin{array}{cc} 0 & \pm \sqrt{-S_{11}} \\ \mp \sqrt{-S_{11}} & 0 \end{array} \right)$$

and the $n = 3$ case, $$A = \left( \begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array} \right) \ , \quad S = \left( \begin{array}{ccc} -a^2-b^2 & -bc & ac \\ -bc & -a^2-c^2 & -ab \\ ac & -ab & -b^2-c^2 \end{array} \right)$$ in which case the elements of the matrix $X$ are easily obtained from the linear system, $$X_{12} + X_{13} = \pm \sqrt{-(S_{11} + 2S_{23})}$$ $$X_{12} + X_{23} = \pm \sqrt{-(S_{22} - 2S_{13})}$$ $$X_{13} + X_{23} = \pm \sqrt{-(S_{33} + 2S_{12})}$$

However, I'm stuck with the general $n \ge 4$ case. Does anyone has some useful hint?

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The answer is yes... in a way.

By the spectral theorem, $S$ can be factored as $S = UD_SU^T$ for some orthogonal matrix $U$ and some (real) diagonal matrix $D_S$. If $S$ is the square of some skew-symmetric matrix, then all of $S$'s eigenvalues (i.e. the diagonal entries of $D_S$) are non-positive, and all non-zero eigenvalues have even multiplicity. In particular, we may write $$ D_S = \pmatrix{-a_1^2\\&-a_1^2\\&&-a_2^2\\&&&-a_2^2\\ &&&&\ddots\\ &&&&&-a_k^2\\&&&&&&-a_k^2 \\ &&&&&&&0 \\ &&&&&&&&\ddots \\ &&&&&&&&&0} $$ where the blank entries are zeros, and each of the $a_i$ are non-zero. In other words, we have $$ D_S = \pmatrix{-a_1^2 I\\ & -a_2^2I\\ && \ddots \\ &&& -a_k^2 I \\ &&&& 0_{\ell \times \ell}} $$ Where $I = I_2 = (\begin{smallmatrix}1&0\\0&1 \end{smallmatrix})$ is the identity matrix of size $2$. In the case that none of the $a_i$ repeat, every skew-symmetric square root of $D$ has the form $$ D_A = \pmatrix{\pm a_1^2 J\\ & \pm a_2^2 J\\ && \ddots \\ &&& \pm a_k^2 J \\ &&&& 0_{\ell \times \ell}} $$ Where each $\pm$ can be chosen independently, and $J = (\begin{smallmatrix}0&-1\\1&0 \end{smallmatrix})$. It follows that the skew-symmetric square roots of $S$ can be written in the form $A = UD_AU^T$, with $D_A$ as above.

We may express the entries of $U$ and the entries of $D$ as the roots of polynomials involving the entries of $S$. However, these polynomials are not necessarily "solvable" for $n \geq 10$.


Things become more complicated still if we allow for repeated non-zero eigenvalues. For example, the matrix $S = -I_4$ has infinitely many skew-symmetric square roots of the form $$ A = (V\tilde D V^T) \otimes J $$ where $J$ is as above, $V$ is any orthogonal matrix of size $2$, and $\tilde D = (\begin{smallmatrix}1&0\\0&-1 \end{smallmatrix})$. Here, $\otimes$ denotes the Kronecker product.

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  • $\begingroup$ A notable consequence of the above analysis: a symmetric $S$ with non-positive eigenvalues has finitely many skew-symmetric square roots if and only if each of its non-zero eigenvalues has multiplicity $1$. $\endgroup$ – Omnomnomnom Jan 29 '17 at 21:17
  • $\begingroup$ Thank you for very informative answer! $\endgroup$ – Ivica Smolić Jan 29 '17 at 21:58
  • $\begingroup$ You're welcome! $\endgroup$ – Omnomnomnom Jan 29 '17 at 22:35

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