Prove that determinant of an odd dimension anti-symmetric matrix is zero
Suppose $A$ is an $(2n+1) \times (2n+1)$ anti-symmetric matrix $(A=-A^T)$. Show that $\det(A)=0$ using Pfaffian formula.
Well, in the wiki page, the formula is only defined for matrix with even dimension. So I'm not sure how to proceed. Any help is greatly appreciated.
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$\begingroup$Directly from the definition of the determinant:
$\det(M) = \sum_\sigma \operatorname{sgn}(\sigma) \ \prod_i \ M_{i,\sigma(i)} $
Now, there are two kinds of permutations $\sigma$: those such that $\sigma^{-1} = \sigma$, and the others.
For odd dimension, permutations with the property that $\sigma^{-1} = \sigma$ have at least one fixed point, i.e. exists at least an $i$ for which $\sigma(i) = i$. (Otherwise you could divide the $2n+1$ objects into couples with the property $i\to \sigma(i) \to i$). Therefore they do not contribute to the determinant, since $M_{i,i} = 0$.
Let's turn to the other permutations. First note that $\sigma$ and $\sigma^{-1}$ have the same sign: exactly the same number of swaps are required to build them. Next note, because of skew-symmetry and of the odd dimension, $\prod_i M_{i,\sigma(i)} = - \prod_i M_{i,\sigma^{-1}(i)}$
Therefore the contribution of every of these permutations cancels with that of its inverse.
$\endgroup$ 1 $\begingroup$Taking a lead from our OP Idonknow I too checked the linked wiki page on the Pfaffian, but couldn't see how it applies to odd-sized matrices, so here's a pretty simple demonstration based on eigenvalues:
Observation: A real skew-symmetric matrix $A$ of odd size $n$ is always singular; that is, $0$ is an eigenvalue of $A$, and hence $\det A = 0$.
Proof of Observation: The characteristic polynomial of such $A$,
$\chi_A(x) = \det(A - xI) \tag{1}$
is of odd degree $n$; hence it has at least one real root $\mu$, which is an eigenvalue of $A$; thus there exists a unit vector $\vec e \in \Bbb R^n$ such that
$A \vec e = \mu \vec e; \tag{2}$
therefore,
$\mu = \mu \langle \vec e, \vec e \rangle = \langle \vec e, \mu \vec e \rangle = \langle \vec e, A \vec e \rangle = \langle A^T \vec e, \vec e \rangle = \langle -A \vec e, \vec e \rangle$ $= -\langle \mu \vec e, \vec e \rangle = -\mu \langle \vec e, \vec e \rangle = -\mu, \tag{3}$
since
$\langle \vec e, \vec e \rangle = 1; \tag{4}$
thus, since
$\mu = -\mu, \tag{5}$
we conclude that
$\mu = 0, \tag{6}$
whence
$A\vec e = 0, \tag{7}$
that is,
$\ker A \ne \{ 0 \}, \tag{8}$
$A$ is singular, and finally
$\det(A) = 0, \tag{9}$
since it is the product of the eigenvalues of $A$. End: Proof of Observation.
Note: In the above it has been proved that any real eigenvalue of a skew-symmetric matrix must be zero; however, we only need one such $\mu = 0$ to obtain the requisite result. End of Note.
$\endgroup$ $\begingroup$$A$ and $A^t$ have same eigen values. Since $A=-A^t$ and since the characteristic polynomial has at least one real eigen value a being of odd degree we see that $a=-a$ and so $A$ has a $0$ eigen value and so it is singular and $\det A=0$.
$\endgroup$ $\begingroup$From spectral theory, an $m\times m$ skew-symmetric (read, anti-symmetric) matrix $A$ can always be brought to a block diagonal form by a special orthogonal transformation. We write $A=Q\Sigma Q^{T}$, where $Q$ is orthogonal and: $$\Sigma=\begin{pmatrix}0&\lambda_1&0&0&0&0&\cdots\\-\lambda_1&0&0&0&0&0\\0&0&0&\lambda_2&0&0&\\0&0&-\lambda_2&0&0&0\\0&0&0&0&0&\lambda_3\\0&0&0&0&-\lambda_3&0&\ddots\\\vdots&&&&&\ddots&\ddots\end{pmatrix}$$ The non-zero eigenvalues of $\Sigma$ are $\pm i\lambda_k$ and in the odd-dimensional case, $\Sigma$ always has at least one row and column of zeros.
In this form, the Pfaffian of $\Sigma$ is simply calculated to be: $$\text{pf}(\Sigma)=\lambda_1\cdot\lambda_2\cdots\lambda_m$$ And from the identities of the Pfaffian, you know that: $$\text{pf}(A)=\text{pf}(Q\Sigma Q^T)=\det(Q)\cdot\text{pf}(\Sigma)$$ Since $Q$ is orthogonal, its determinant is $\det(Q)=(-1)^m$. And since one of the eigenvalues $\lambda_k$ will be $0$ if $\Sigma$ is odd-dimensional, we finally have: $$\text{pf}(A)=0$$
$\endgroup$ $\begingroup$Use the matrix: $$ \begin{bmatrix} 0 & 1 & 1 & \cdots & 1 \\ -1 & & \\ -1 & & A \\ \,\,\,\, \vdots \\ -1 \end{bmatrix} $$
$\endgroup$ $\begingroup$Skew symmetric matrices are isomorphic to nondirected graphs.
The determinant of such a matrix is zero iff the corresponding graph does not contain a perfect matching. If the graph has an odd number of vertices than it trivially doesn't contain a perfect matching, so the determinant must be zero.
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