Prove that determinant of an odd dimension anti-symmetric matrix is zero

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

$A$ 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 $\det A=0$.

Answer 4

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: $$\text{pf}(A)=0$$

Answer 5

Use the matrix: $$ \begin{bmatrix} 0 & 1 & 1 & \cdots & 1 \\ -1 & & \\ -1 & & A \\ \,\,\,\, \vdots \\ -1 \end{bmatrix} $$

Answer 6

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.