# Existence of the Pfaffian?

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.

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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. – Qiaochu Yuan Jun 8 '12 at 3:35
Well, I haven't worked through the details. – Qiaochu Yuan Jun 8 '12 at 3:41
– darij grinberg Jul 21 '15 at 14:10

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.

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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. – ziggurism Nov 14 '12 at 4:06
@JoeHannon 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. – David Speyer Sep 17 '15 at 15:33

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.

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