Square root of matrix that is a square of skew-symmetric matrix

Let's suppose we have a matrix $A$ (dimension $3\times 3$) which is the square of some skew-symmetric matrix $S$ i.e. $A=S^2$. How to obtain from $A$ its skew-symmetric square root $S$?

• This question is related to my previous one StackExchange – Widawensen Apr 28 '16 at 6:53
• Um, $A=S^2$, so you have it already. Or are you looking for the skew-symmetric matrix $S$, given $A$? – Christopher Carl Heckman Apr 28 '16 at 7:35
• I'm looking for $S$ (assuming $S$ is skew-symmetric), given $A$. – Widawensen Apr 28 '16 at 7:46

Firstly, every eigenvalue of the skew-symmatrix $S$ will be purely imaginary. In this specific case where $S$ is a $3\times 3$ matrix, we will have the 3 distinct eigenvalues $i\lambda, -i\lambda, 0$, where $\lambda \in \mathbb R$.

According to this wikipedia article, our real skew-symmetric matrix $S$ can be written in the form $$S = Q\cdot \Sigma\cdot Q^\top,$$ where $Q$ is an orthogonal matrix and $\Sigma$ is of the form $$\Sigma = \begin{bmatrix} 0 &-\lambda &0\\ \lambda & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$$

Matrix $A$ will have the real, non - positive eigenvalues $-\lambda^2, -\lambda^2, 0$. Moreover, since $A$ is symmetric it can be written as: $$A = V \cdot \begin{bmatrix} -\lambda^2 & 0 & 0 \\ 0 & -\lambda^2 & 0 \\ 0 & 0 & 0\end{bmatrix}\cdot V^\top,$$ where $V$ is some orthogonal matrix. So, to answer your question, since you are given the matrix $A$, write it down in the previous form and in order to find $S$, set $Q = V$ and $\Sigma$ as written above.

Notice that alternating the signs in matrix $\Sigma$, you will get either the matrix $S$ or $-S$.

• Great answer, but .. how to find $V$? $S^2$ I suppose is of rank 2. – Widawensen Apr 28 '16 at 8:28
• Ok. So 3 eigenvectors, but one eigenvalue is 0. What to do with that? – Widawensen Apr 28 '16 at 8:49
• $V$ has as its columns $3$ orthogonal vectors. Since you have $V$, you already have $Q$. Since you know the eigenvalues of $A$, you can construct the matrix $\Sigma$ as well. Then finding $S$ is a matter of matrix multiplication, i.e. $S= Q \cdot \Sigma \cdot Q^T.$ – thanasissdr Apr 28 '16 at 8:56
• Eigenvector from 0 eigenvalue will be unigue? $Av_0=0$. I quess it would be perpendicular to the plane defined by $A$. – Widawensen Apr 28 '16 at 9:01
• Yes, it will be unique (up to a scalar). – thanasissdr Apr 28 '16 at 9:07