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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.