When $n$ is odd, the kernel of a skew-symmetric matrix $M$ of size $n\times n$ and rank $n-1$ is the span of $v$, where $v$ is a vector whose $i$-th component is the Pfaffian of the matrix obtained by removing the $i$-th line and $i$-th column of $M$, multiplied by $(-1)^i$. 


Example with $n=3$:
$$\begin{bmatrix} 0 & a&b \\ -a&0&c \\ -b&-c&0\end{bmatrix} \begin{bmatrix} c\\-b\\a\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$$


Example with $n=5$:
$$\begin{bmatrix}
 0 & a & b & c & d \\
 -a & 0 & e & f & g \\
 -b & -e & 0 & h & i \\
 -c & -f & -h & 0 & j \\
 -d & -g & -i & -j & 0 \\
\end{bmatrix}\begin{bmatrix} -e j+f i-g h \\ b j-c i+d h \\ -a j+c g-d f \\a i-b g+d e \\ -a h+b f-c e\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$


I suppose this is a known result: would you know a reference which mentions it?



Kernel of skew-symmetric matrix of rank $n-1$ with $n$ odd: is this a known result?
$\def\pf{\mathop{\rm pf}}$I do not know the reference, and it seems that it is easier to prove the claim than to find a reference ;). 


The proof follows the lines of the proof of a similar formula for determinants. Let $M=[m\_{ij}]\_{i,j=1}^n$. Denote by $M\_i$ the matrix obtained from $M$ by deleting the $i$th row and the $i$th column. Let $\overline M\_i$ be the $(n+1)\times(n+1)$ skew-symmetric matrix obtained from $M$ by doubling the $i$th row and the $i$th column and putting the copies to the top and to the left, respectively. We have $\det \overline M\_i=0$ and hence $\pf \overline M\_i=0$. 


On the other hand, in view of the formula
$$
 \pf A=\sum\_{j=2}^{2k}(-1)^j a\_{1j}\pf A\_{\hat 1\hat j}
$$
(for a skew-symmetric $2k\times 2k$ matrix $A=[a\_{ij}]$), we have
$$
 0=\pf \overline M\_i=\sum\_{j=1}^nm\_{ij}\cdot (-1)^{j+1}\pf M\_j,
$$
which shows that the $i$th element in the product of $M$ by your column is zero.


