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Let us consider the class of matrices of the following form: $$ \begin{bmatrix}0&A\\B&0\end{bmatrix}, $$ where $A$ and $B$ are $n\times n$ nonsingular real antisymmetric matrices. After some numerical experiments I noticed that for $n$ even the null space is trivial, but for $n$ odd the null space has dimension 2. I was unable to prove or disprove that, so I am asking this question in the hope that someone else could do it.

EDIT: Any odd dimensional antisymmetric matrix is singular, so the null space has dimension at least one for theses matrices and that is the origin for the 2-dimensional null space I was seeing: one dimension for $A$ and other for $B$.

EDIT2: I was performing my numerical experiments with random matrices, so I did not worry about the matrices been singular because I knew that singular matrices are rare, then a systematic generation of singular matrices was basically impossible unless there is some basic mathematical fact that make them be singular, as it was the case. Anyway, the rank of the odd dimensional antisymmetric matrices $A$ and $B$ generated randomicaly is $n-1$ and not any number. The simple Octave code I was using is the following:

N=20 for n=3:N A=rand(n); A=A-A'; B=rand(n); B=B-B'; M=[zeros(n),A;B,zeros(n)]; disp([n,2*n-rank(M)]) endfor

Anyone can test this code online.

Joca Ramiro
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2 Answers2

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The determinant of this matrix is $(-1)^n \det(A)\det(B)$ (when $n$ is odd you need an odd number of column transpositions to make your matrix block-diagonal), therefore it is not zero since you assumed that the matrices $A$ and $B$ were non-singular. It follows that $M$ can't have a non-trivial nullspace.

Arnaud Mortier
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    You are right. In fact, any odd dimensional antisymmetric matrix must be singular. Each eigenvector for the zero eigenvalue came from a singular matrix, so the dimension 2 for the null space. – Joca Ramiro Feb 08 '18 at 22:30
  • @jobe It could be more than two as well, right? In the extreme case, the zero matrix is anti-symmetric. – Arnaud Mortier Feb 08 '18 at 23:51
  • The matrices were generated randomicaly , so the probability they were systematically singular is zero in practice. That is why I supposed they were nonsingular. I didn't know odd dimensional antisymmetric matrices were singular and ended up discovering in a tortuous way. – Joca Ramiro Feb 09 '18 at 08:58
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If $n$ is odd, then $A$ and $B$ cannot be nonsingular. Indeed, for $n$ odd and $A$ antisymmetric, $$ \det(A)=\det(A^T)=\det(-A)=(-1)^n\det(A)=-\det(A) $$ so $\det(A)=0$ and $A$ is singular.

So if $A$ are to be nonsingular, you conclude $n$ is even and also that $$ M=\begin{bmatrix} 0 & A \\ B & 0 \end{bmatrix} $$ is nonsingular, because with $n$ row exchanges you can bring the matrix in the form $$ \begin{bmatrix} B & 0 \\ 0 & A \end{bmatrix} $$ and this matrix has determinant $\det(B)\det(A)$, by a repeated application of Laplace expansion.

For $n$ odd you can surely say that the kernel has dimension $2$, but you could as well say the kernel has dimension $42$, because there is no matrix of that kind that can falsify $\dim(\ker M)=42$.

egreg
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  • Thanks for your answer. I've made some edits that explain why the rank of the odd dimensional antisymmetric matrices I was using is $n-1$ and why - being ignorant that odd dimensional antisymmetric matrices are singular- I assumed they were not. – Joca Ramiro Feb 09 '18 at 13:34
  • @jobe If you don't assume $A$ and $B$ nonsingular, then the rank of $M$ can be whatever even number between $0$ and $2n$. – egreg Feb 09 '18 at 13:42