0
$\begingroup$

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 $S$?

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

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 $-S$.

$\endgroup$
  • $\begingroup$ Great answer, but .. how to find $V$? $S^2$ I suppose is of rank 2. $\endgroup$ – Widawensen Apr 28 '16 at 8:28
  • $\begingroup$ Ok. So 3 eigenvectors, but one eigenvalue is 0. What to do with that? $\endgroup$ – Widawensen Apr 28 '16 at 8:49
  • $\begingroup$ $V$ 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. $S= Q \cdot \Sigma \cdot Q^T.$ $\endgroup$ – thanasissdr Apr 28 '16 at 8:56
  • $\begingroup$ Eigenvector from 0 eigenvalue will be unigue? $Av_0=0$. I quess it would be perpendicular to the plane defined by $A$. $\endgroup$ – Widawensen Apr 28 '16 at 9:01
  • $\begingroup$ Yes, it will be unique (up to a scalar). $\endgroup$ – thanasissdr Apr 28 '16 at 9:07

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.