TITLE: Kernel of skew-symmetric matrix of rank $n-1$ with $n$ odd: is this a known result?
QUESTION [14 upvotes]: 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?

REPLY [2 votes]: There is a proceeding by D. Knuth about that. You can see the relevant formula in a) of Exercise 93 in my list of exercises on matrices. 
Let $n$ be odd and $1\le i,j\le n$ be given. Then the minor $M[\hat i|\hat j]$ equals the product of the Pfaffians ${\rm Pf}M[\hat i]$ and  ${\rm Pf}M[\hat j]$ ; I have used the notation that $\hat i$ is the complement of $i$ in the set of indices. Application: because the rank of $M$ is $n-1$, we can find a index $j$ such that  ${\rm Pf}M[\hat j]\ne0$. Then the kernel is spanned that the vector of coordinates $(-1)^i M[\hat i|\hat j]$. Simplifying by ${\rm Pf}M[\hat j]$, we obtain your claim.
I apologize that I could not find the reference of Knuth's paper. If I remember well, he claimed in it there is no determinant, there are only Pfaffians.