How frequently do $5 \times 5$ dimensional skew-symmetric real-valued matrices of rank $2$ occur?

Question

Since a $5\times 5$ skew-symmetric, real-valued matrix can only take a rank of either $0$, or $2$, or $4$, how much probable is coming across such a rank $2$ matrix at random? In other words, is there any general form the entries of the matrix should satisfy in order for it to be of rank $2$?

For example, a general (any rank out of $0, 2$ or $4$) $5\times5$ skew-symm. matrix may be written as, say,

$\begin{bmatrix} 0 & c_1 & c_2 & c_3 & c_4 \\ -c_1 & 0 & c_5 & c_6 & c_7 \\ -c_2 & -c_5 & 0 & c_8 & c_9 \\ -c_3 & -c_6 & -c_8 & 0 & c_{10} \\ -c_4 & -c_7 & -c_9 & -c_{10} & 0 \end{bmatrix}.$

Any relation the $c_i$'s should satisfy so that the above matrix is of rank $2$?

Answer 1

One has $$\det(1-\lambda A) = 1 + \lambda^2 {\mbox{tr}} \wedge^2 A + \lambda^4 {\mbox{tr}} \wedge^4 A= 1 + c_2(A)\lambda^2 +c_4(A) \lambda^4$$ $A$ has rank 2 iff there are only 2 roots, i.e. if $c_4(A)=0$. Let me write $$A = \left( \begin{matrix} 0 & a_{12} & .. & a_{15} \\ a_{21} & 0 & .. & a_{25} \\ . & . & .. & .\\ a_{51} & a_{52} & .. & 0 \end{matrix} \right)$$ with the understanding that $a_{ij}+a_{ji}=0$ (in particular $a_{ii}=0$). The coefficient $c_4(A)$ comes from taking one diagonal element from the identity and 4 elements from the $A$ matrix. We have $$ c_4(A)= \frac{1}{2}\sum (a_{i_1 i_2} a_{i_2 i_1}) (a_{i_3 i_4} a_{i_4 i_3}) - \sum (a_{i_1 i_2} a_{i_2 i_3} a_{i_3 i_4} a_{i_4 i_1}) $$ The first sum is over all (disjoint) quadruples $(i_1,i_2,i_3,i_4)$ for which $i_1<i_2$ and $i_3<i_4$. The second sum is over all distinct cycles $(i_1 i_2 i_3 i_4)$. With your notation it is less easy to write it down. Anyway, it is a polynomial of degree 4 in the coefficients. The zero set has co-dimension 1 in the set of all such matrices. Another way to put it: In a generic 1 parameter family of such matrices the rank 2 will occur as isolated points that persist under perturbation.

Answer 2

Let $p(x)=\det(x I_5-A)=x^5+\sigma_2x^3+\sigma_4x$. Since $A$ (skew-symmetric) is diagonalizable over $\mathbb{C}$, $rank(A)=2$ iff $\sigma_4=0,\sigma_2\not= 0$. Now $4\sigma_4=\sigma_3S_1-\sigma_2S_2+\sigma_1S_3-S_4=-\sigma_2S_2-S_4=0$, where $S_k=\sum_i \lambda_i^k$ (the $(\lambda_i)$ are the eigenvalues of $A$). Since $2\sigma_2=\sigma_1^2-S_2=-S_2$, $\sigma_2S_2+S_4=0$ is equivalent to $2S_4=S_2^2$, that is $2tr(A^4)=(tr(A^2))^2$. Note that, when $A$ is skew-symmetric, $\sigma_2$ is always $\geq 0$.

Finally, we give a NS condition about the entries of $A$:

$rank(A)=2$ IFF $tr(A^2)\not= 0,2tr(A^4)=(tr(A^2))^2$.