Let $n$ be an even integers. Let $r\in \mathbb R^n$ and $e=[1,1,\dots,1]^T$. If $$A = re^T - er^T,$$ then $A\in \mathbb{R}^{n\times n}$ is of rank 2 and skew-symmetric, i.e., $$A = -A^T.$$

This does not represent all rank skew-symmetric matrices, but a useful subset which appears in ranking problems. Clearly there are no more than $n$ degrees of freedom in choosing $A$.

Does there exist a similar method of generating an arbitrary (even) rank skew-symmetric matrix? Skew-symmetric matrices of any rank in general have $\frac{n(n-1)}{2}$ degrees of freedom, representing $\binom{n}{2}$ row/column pairs. I'm looking for skew-symmetric matrices that somehow represent requiring much fewer comparisons than this such that the space of possible $A$ has fewer than $\binom{n}{2}$ degrees of freedom.


Let $A= -A^T$. Let $e_i$ denote the $i-th$ unit vector. Then it holds $$ A = \sum_{i=2}^n\sum_{j=1}^{i-1} a_{ij}( e_ie_j^T-e_je_i^T). $$ In that way you represent any skew-symmetric matrix as a sum of rank-2 matrices.

By truncating the sum, you obtain matrices with a specified even rank. Is this what you are after?

  • $\begingroup$ Nice. It's not exactly what I want because your matrices are somewhat "coherent," i.e., they are low rank because of missing rows and columns (I realize I didn't specify). However, you gave me the idea to use a different sum of rank-2 matrices, $A = \sum_{i=1}^{k/2} \sigma_i( r_i e^T - e^T r_i)$. I just wonder how far this set of matrices is from the set of all rank-k skew-symmetric matrices. $\endgroup$ – SCarm Oct 9 '14 at 22:40
  • $\begingroup$ Such a matrix has only rank $k/2+1$ at most, its image is spanned by the $r_i$'s and $e$. $\endgroup$ – daw Oct 10 '14 at 6:14
  • $\begingroup$ Ah yes you're right. Thanks $\endgroup$ – SCarm Oct 16 '14 at 23:22

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.