Consider a square skew-symmetric $n\times n$ matrix $A$. We know that $\det(A)=\det(A^T)=(-1)^n\det(A)$, so if $n$ is odd, the determinant vanishes.

If $n$ is even, my book claims that the determinant is the square of a polynomial function of the entries, and Wikipedia confirms this. The polynomial in question is called the Pfaffian.

I was wondering if there was an easy (clean, conceptual) way to show that this is the case, without mucking around with the symmetric group.

  • $\begingroup$ See "alternate definitions" in the Wikipedia article. The idea that you can associate to such an $A$ an element in the exterior square $\Lambda^2(V)$ of the vector space $V$ on which $A$ acts and then take exterior powers. $\endgroup$ Jun 8, 2012 at 3:35
  • $\begingroup$ Well, I haven't worked through the details. $\endgroup$ Jun 8, 2012 at 3:41
  • $\begingroup$ Related: sciencedirect.com/science/article/pii/S0001870885710298 $\endgroup$ Jul 21, 2015 at 14:10

4 Answers 4


Here is an elaboration of Qiaochu's comment above:

A $2n\times 2n$ matrix $A$ induces a pairing (say on column vectors), namely $$\langle v,w \rangle := v^T A w.$$ Thus we can think of $A$ as being an element of $(V\otimes V)^*$ (which is the space of all bilinear pairings on $V$), where $V$ is the space of $2n$-dimensional column vectors.

If $A$ is skew-symmetric, then this pairing is anti-symmetric, and so we can actually regard $A$ as an element of $\wedge^2 V^*$. We can then take the $n$th exterior power of $A$, so as to obtain an element of $\wedge^{2n} V^*$. This latter space is $1$-dimensional, and so if we fix some appropriately normalized basis for it, the $n$th exterior power of $A$ can be thought of just as a number. This is the Pfaffian of $A$ (provided we chose the right basis for $\wedge^{2n} V^*$).

How does this compare to the usual description of determinants via exterior powers:

For this, we regard $A$ as an endomorphism $V \to V$, which induces an endomorphism $\wedge^{2n} V \to \wedge^{2n} V$, which is a scalar (being an endomorphism of a $1$-dimensional space); this is $\det A$.

So now we see where the formula $\det(A) =$ Pf$(A)^2$ comes from: computing the determinant involves taking a $2n$th exterior power of $A$, while computing the Pfaffian involves only taking an $n$th exterior power (because we use the skew-symmetry of $A$ to get an exterior square "for free", so to speak).

The sorting out the details of all this should be a fun exercise.

  • 5
    $\begingroup$ It's clear from this argument that $\operatorname{Pf}(A)$ exists and is a polynomial of order $n$ in the elements of $A$, while $\det(A)$ is a polynomial of order $2n$. But to see that $\operatorname{Pf}(A)^2=\det(A)$ using exterior algebra methods, so far I haven't been able. $\endgroup$
    – ziggurism
    Nov 14, 2012 at 4:06
  • $\begingroup$ @ziggurism It seems to me that this argument also shows that they have the same zeroes. So, if you can argue that $Pf(A)$ is irreducible, then $\det(A)$ and $Pf(A)$ are proportional. I agree that I have never been able to figure out a truly clean way to finish this argument, though. $\endgroup$ Sep 17, 2015 at 15:33
  • $\begingroup$ Can you regard $A$ as an endomorphism $V\rightarrow V$ given that it is coordinate dependent? $\endgroup$
    – Alex
    Apr 15 at 13:32
  • $\begingroup$ Have you worked out the details? I am not sure if these two views would help with seeing Pf$(A)^2=$det(A)$ more than the original proof of this relation. At least that's what I find myself ending up doing. I think the two views are separate and not related so to speak. $\endgroup$
    – Alex
    Apr 15 at 15:34

Here is an approach using (possibly complex) Grassmann variables and Berezin integration$^1$ to prove the required relation $${\rm Det}(A)~=~{\rm Pf}(A)^2. \tag{1}$$ This approach isn't purely conceptional, but at least it is easy, we don't fudge the overall sign, we don't muck around much with the symmetric group, and Grassmann variables do implement exterior calculus.

  1. Define the Pfaffian of a (possibly complex) antisymmetric matrix $A^{jk}=-A^{kj}$ (in $n$ dimensions) as$^2$ $$ \begin{align} {\rm Pf}(A)&~:=~\int \!d\theta_n \ldots d\theta_1~ e^{\frac{1}{2}\theta_j A^{jk}\theta_k} \cr &~=~(-1)^{\lfloor\frac{n}{2}\rfloor} \int \!d\theta_1 \ldots d\theta_n~ e^{\frac{1}{2}\theta_j A^{jk}\theta_k}\cr & \cr &~=~(-1)^{\frac{n}{2}} \int \!d\theta_1 \ldots d\theta_n~ e^{\frac{1}{2}\theta_j A^{jk}\theta_k}\cr \cr &~=~i^n \int \!d\theta_1 \ldots d\theta_n~ e^{\frac{1}{2}\theta_j A^{jk}\theta_k}\cr &~=~ \int \!d\theta_1 \ldots d\theta_n~ e^{-\frac{1}{2}\theta_j A^{jk}\theta_k}.\end{align} \tag{2}$$ In the last equality of eq. (2), we rotated the Grassmann variables $\theta_k\to i\theta_k$ with the imaginary unit.

  2. Define the determinant as $$ {\rm Det}(A)~:=~\int \!d\theta_1 ~d\widetilde{\theta}_1 \ldots d\theta_n ~d\widetilde{\theta}_n~ e^{\widetilde{\theta}_j A^{jk}\theta_k} . \tag{3}$$
    It is not hard to prove via coordinate substitution that eq. (3) indeed reproduces the standard definition of the determinant.

  3. If we make a change of coordinates $$ \theta^{\pm}_k~=~ \frac{\theta_k\pm \widetilde{\theta}_k}{\sqrt{2}}, \qquad k~\in~\{1,\ldots,n\},\tag{4} $$ in eq. (3), the super-Jacobian becomes $(-1)^n$.

  4. Therefore we calculate $$\begin{align} {\rm Det}(A)&\stackrel{(3)+(4)}{=}~(-1)^n\int \!d\theta^+_1 ~d\theta^-_1 \ldots d\theta^+_n ~d\theta^-_n~ e^{\frac{1}{2}\theta^+_j A^{jk}\theta^+_k -\frac{1}{2}\theta^-_j A^{jk}\theta^-_k}\cr &~~=~\int \!d\theta^-_1 \ldots d\theta^-_n~d\theta^+_n\ldots d\theta^+_1 ~~e^{\frac{1}{2}\theta^+_j A^{jk}\theta^+_k} e^{-\frac{1}{2}\theta^-_j A^{jk}\theta^-_k}\cr &~~\stackrel{(2)}{=}~{\rm Pf}(A)^2, \end{align}\tag{5}$$
    which proves eq. (1).$\Box$


$^1$ We use the sign convention that Berezin integration $$\int d\theta_i~\equiv~\frac{\partial}{\partial \theta_i}\tag{6} $$ is the same as differentiation wrt. $\theta_i$ acting from left. See e.g. this Phys.SE post and this Math.SE post.

$^2$ The sign of the permutation $(1, \ldots, n)\mapsto(n, \ldots, 1)$ is given by $(-1)^{\frac{n(n-1)}{2}}=(-1)^{\lfloor\frac{n}{2}\rfloor}$, where $\lfloor\frac{n}{2}\rfloor$ denotes the integer part of $\frac{n}{2}$. One may show that the Pfaffian (2) vanishes in odd dimensions $n$.

  • $\begingroup$ Notes for later: Transposed matrix: $\quad {\rm Det}(M^T):=\int \!d\theta_1 ~d\widetilde{\theta}_1 \ldots d\theta_n ~d\widetilde{\theta}_n~ e^{\widetilde{\theta}_j (M^T)^{jk}\theta_k}$ $=(-1)^n\int \!d\widetilde{\theta}_1~d\theta_1 \ldots d\widetilde{\theta}_n~d\theta_n ~ e^{-\theta_kM^{kj}\widetilde{\theta}_j} $ $=\int \!d\widetilde{\theta}_1~d\theta_1 \ldots d\widetilde{\theta}_n~d\theta_n ~ e^{\theta_kA^{kj}\widetilde{\theta}_j}={\rm Det}(M). \quad $ Here we rotated the Grassmann variables with the imaginary unit. $\endgroup$
    – Qmechanic
    Sep 4 at 10:59

Let $(V,\Omega)$ be a vector space of dimension $2n$ with a nonzero $\Omega\in \bigwedge^{2n} V^*$.

Let $A$ be an alternating bilinear form on $V$. Then $A\in \bigwedge^2 V^*$ and $\Omega$ spans $\bigwedge^{2n} V^*$.

Define the Pfaffian $Pf(A)$ of $A$ w.r.t. $\Omega$ by $$Pf(A)\Omega:=\frac{A^{\wedge n}}{n!}\in \bigwedge { }^{2n} V^*$$

If a basis is chosen and $A$ is represented by a matrix, then clearly $Pf(A)$ is a homogeneous polynomial of degree $n$ in the coefficients of $A$.

Let now $\omega$ be a symplectic form on $V$. Then there exists $B\in \text{End}(V)$ such that $B^*\omega=A$. If $A$ is nondegenerate (thus also symplectic), this amounts to the fact that any two linear symplectic forms on $V$ are equivalent.

Since $A^{\wedge n}=(B^*\omega)^{\wedge n}=B^*\omega^{\wedge n}=\det(B)\omega^{\wedge n}$ we have $Pf(A)=\det(B)$, if $\omega$ is chosen such that $\omega^{\wedge n}/n!=\Omega$.

Fix an inner product on $V$. Then $\omega$ can be viewed as an endomorphism of $V$. Since $B^T\omega B=A$, we have $\det(A)=\det(B)^2\det(\omega)$ and thus $$ \det(A)=Pf(A)^2\det(\omega).$$ If the inner product is such that $\det (\omega)=1$, this reduces to $\det(A)=Pf(A)^2$.


By continuity, we can assume that $A$ can always be reduced to block diagonal form with blocks $$ \left(\begin{matrix} 0 &\lambda_i\cr -\lambda_i &0 \end{matrix} \right) $$ on the diagonal. In this case computing the determinant gives $\prod \lambda^2_i$ and computing the Pfaffian gives $\prod \lambda_i$, so the determinant is the square of the Pfaffian.

  • $\begingroup$ you did not understand one word about the asked question. We have to find a polynomial $P_n\in \mathbb{Z}[x]$, that depends on $n$ and not on $A\in skew_n(K)$, s.t. $\det(A)=P_n((a_{i,j})_{i<j})^2$. When you block diagonalize $A$, you kill the $(a_{i,j})$ and you replace them with the $(\lambda_i)$ and with the coefficients of some orthogonal matrix of change of basis. Unfortunately (for you) these new variables are in an algebraic extension of the underlying field (say $K$)! How are you going to link these variables to the $(a_{i,j})$ ? -to be continued- $\endgroup$
    – user91684
    Mar 24, 2018 at 10:32
  • $\begingroup$ In fact, we can (we must) work in the field $K$ . In particular, this result is valid for every commutative ring... Then the best is to remove your post that, since $3$ years, looks out of place. $\endgroup$
    – user91684
    Mar 24, 2018 at 10:32
  • $\begingroup$ The question askes us to show that $Pf(A)= \epsilon^{a_1a_2\cdots a_{2n}}A_{a_1a_2}\cdots A_{a_{2n-1} a_{2n}}/(2^n n!)$ squares to the determinant. This is a a polynomial in the entries of $A$. No field extension is required to contruct it. To prove that it squares to det can be done by tedious combinatorics, or by showing it true for diagonalizable matrices. $\endgroup$
    – mike stone
    Mar 25, 2018 at 12:43
  • $\begingroup$ You write anything; there exists no compact formula for $Pf$; in fact, it can only be recursively defined. On the other hand, to show the result for the block diagonalized matrix does not prove anything. $\endgroup$
    – user91684
    Mar 26, 2018 at 18:16
  • $\begingroup$ @loup blanc. The formula for the Pfaffian is standard: en.wikipedia.org/wiki/Pfaffian. Also recall that any real skew symmetric matrix can be reduced into skew 2-by-2 blocks by a real orthogonal transformation: en.wikipedia.org/wiki/Skew-symmetric_matrix#Spectral_theory. Thus, if the formula holds for the block diagonal matrices(and it does) it holds for all matrices. The OP does not request for working in an arbitrary field, and will probably be satisfied with the result for real or complexes which is what I give. $\endgroup$
    – mike stone
    Mar 26, 2018 at 19:01

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