For a class of matrices $M_{2n}$: $\dim(\ker(M_{2n}))$ is zero if $n$ is even and 2 if $n$ is odd 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.
 A: 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.
A: 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 $\dim(\ker M)=42$.
