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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.

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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?

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  • $\begingroup$ 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. $\endgroup$
    – mass
    Commented Oct 9, 2014 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
    Commented Oct 10, 2014 at 6:14
  • $\begingroup$ Ah yes you're right. Thanks $\endgroup$
    – mass
    Commented Oct 16, 2014 at 23:22

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