Kernel of skew-symmetric matrix of rank $n-1$ with $n$ odd: is this a known result?

Question

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?

Answer 1

$\def\pf{\mathop{\rm pf}}$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.

Answer 2

Another proof:

Let $\omega\in\Lambda^2(V^*)$ correspond to the skew-symmetric matrix $M$, i.e. $\omega = \sum_{i<j} M_{ij} f_i\wedge f_j$ where $f_i$ is the canonical basis of $(\mathbb{R}^n)^*$. Let $\eta = f_1\wedge\cdots\wedge f_n$ be the canonical volume form on $\mathbb{R}^n$. Then $v$ is uniquely defined by $\iota_{v}\eta = \omega\wedge\cdots\wedge\omega$ ($\frac{n-1}{2}$ terms). Now that we have defined everything in a coordinate-free way, we can choose a basis where

$M=\left(\begin{array}{cccccc}0&\lambda_1&&&&\\-\lambda_1&0&&&&\\&&\ddots&&&\\&&&0&\lambda_\frac{n-1}{2}&\\ &&&-\lambda_\frac{n-1}{2}&0&\\&&&&& 0\end{array}\right)$

The change of basis can be taken to be in $SO(n)$, in particular the expression for $\eta$ does not change.

In this case, deleting anything but the last row replaces one of the blocks by a $0$, and the resulting Pfaffian vanishes by inspection. Thus only the last component of $v$ does not vanish, so $v$ is in the kernel of $M$.

Answer 3

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.

Answer 4

I think this is covered by I. Kaplansky. A contribution to von Neumann’s theory of games. II. Linear Algebra and its Applications, 226-228:371–373, 1995.