TITLE: Prove that determinant of an odd dimension anti-symmetric matrix is zero
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.