Pfaffian

By convention, the Pfaffian of the 0×0 matrix is equal to one. The Pfaffian of a skew-symmetric 2n×2n matrix A with n>0 can be computed recursively as

${\displaystyle \operatorname {pf} (A)=\sum _{{j=1} \atop {j\neq i}}^{2n}(-1)^{i+j+1+\theta (i-j)}a_{ij}\operatorname {pf} (A_{{\hat {\imath }}{\hat {\jmath }}}),}$

where index i can be selected arbitrarily, ${\displaystyle \theta (i-j)}$ is the Heaviside step function, and ${\displaystyle A_{{\hat {\imath }}{\hat {\jmath }}}}$ denotes the matrix A with both the i-th and j-th rows and columns removed.[1] Note how for the special choice ${\displaystyle i=1}$ this reduces to the simpler expression:

${\displaystyle \operatorname {pf} (A)=\sum _{j=2}^{2n}(-1)^{j}a_{1j}\operatorname {pf} (A_{{\hat {1}}{\hat {\jmath }}}).}$

One can associate to any skew-symmetric 2n×2n matrix A =(a_ij) a bivector

${\displaystyle \omega =\sum _{i<j}a_{ij}\;e_{i}\wedge e_{j}.}$

where {e₁, e₂, ..., e_2n} is the standard basis of R²ⁿ. The Pfaffian is then defined by the equation

${\displaystyle {\frac {1}{n!}}\omega ^{n}=\operatorname {pf} (A)\;e_{1}\wedge e_{2}\wedge \cdots \wedge e_{2n},}$

here ωⁿ denotes the wedge product of n copies of ω.

Equivalently, we can consider the bivector (which is more convenient when we don't want to impose the summation constraint ${\displaystyle i<j}$):

$${\displaystyle \omega '=2\omega =\sum _{i,j}a_{ij}\;e_{i}\wedge e_{j}.}$$

which gives ${\displaystyle \omega '^{n}=2^{n}n!\operatorname {pf} (A)\;e_{1}\wedge e_{2}\wedge \cdots \wedge e_{2n}}$.

A non-zero generalisation of the Pfaffian to odd dimensional matrices is given in the work of de Bruijn on multiple integrals involving determinants.[2] In particular for any m x m matrix A, we use the formal definition above but set ${\displaystyle n=\lfloor m/2\rfloor }$. For m odd, one can then show that this is equal to the usual Pfaffian of an (m+1) x (m+1) dimensional skew symmetric matrix where we have added an (m+1)th column consisting of m elements 1, an (m+1)th row consisting of m elements -1, and the corner element is zero. The usual properties of Pfaffians, for example the relation to the determinant, then apply to this extended matrix.

For a 2n × 2n skew-symmetric matrix A

${\displaystyle \operatorname {pf} (A^{\text{T}})=(-1)^{n}\operatorname {pf} (A).}$

${\displaystyle \operatorname {pf} (\lambda A)=\lambda ^{n}\operatorname {pf} (A).}$

${\displaystyle \operatorname {pf} (A)^{2}=\det(A).}$

For an arbitrary 2n × 2n matrix B,

${\displaystyle \operatorname {pf} (BAB^{\text{T}})=\det(B)\operatorname {pf} (A).}$

Substituting in this equation B = A^m, one gets for all integer m

${\displaystyle \operatorname {pf} (A^{2m+1})=(-1)^{nm}\operatorname {pf} (A)^{2m+1}.}$

Proof of ${\displaystyle \operatorname {pf} (BAB^{\text{T}})=\det(B)\operatorname {pf} (A)}$

As previously said, ${\displaystyle A\mapsto \sum _{ij}A_{ij}e_{i}\wedge e_{j}\mapsto ^{\wedge n}{2^{n}n!}Pf(A)e_{1}\wedge \cdots \wedge e_{2n}}$. The same with ${\displaystyle BAB^{T}}$:

$${\displaystyle BAB^{T}\mapsto \sum _{ijkl}B_{ik}B_{jl}A_{kl}e_{i}\wedge e_{j}=\sum _{kl}A_{il}f_{k}\wedge f_{l}\mapsto ^{\wedge n}{2^{n}n!}Pf(A)f_{1}\wedge \cdots \wedge f_{2n}={2^{n}n!}Pf(BAB^{T})e_{1}\wedge \cdots \wedge e_{2n}}$$

where we defined ${\displaystyle f_{k}=\sum _{i}B_{ik}e_{i}}$.

Since ${\displaystyle f_{1}\wedge \cdots \wedge f_{2n}=\det(B)e_{1}\wedge \cdots \wedge e_{2n}}$ the proof is finished.

Proof of ${\displaystyle \operatorname {pf} (A)^{2}=\det(A)}$:

Since ${\displaystyle \operatorname {pf} (A)^{2}=\det(A)}$ is an equation of polynomials, it suffices to prove it for real matrices, and it would automatically apply for complex matrices as well.

By the spectral theory of skew-symmetric real matrices, ${\displaystyle A=Q\Sigma Q^{T}}$, where ${\displaystyle Q}$ is orthogonal and

$${\displaystyle \Sigma ={\begin{bmatrix}0&a_{1}&0&0\\-a_{1}&0&0&0\\0&0&0&a_{2}\\0&0&-a_{2}&0&\ddots \\&&&\ddots &\ddots &\\&&&&&0&a_{n}\\&&&&&-a_{n}&0\end{bmatrix}}}$$

for real numbers ${\displaystyle a_{k}}$. Now apply the previous theorem, we have ${\displaystyle pf(A)^{2}=pf(\Sigma )^{2}\det(Q)^{4}=pf(\Sigma )^{2}=\left(\prod a_{i}\right)^{2}=\det(A)}$.

If A depends on some variable x_i, then the gradient of a Pfaffian is given by

${\displaystyle {\frac {1}{\operatorname {pf} (A)}}{\frac {\partial \operatorname {pf} (A)}{\partial x_{i}}}={\frac {1}{2}}\operatorname {tr} \left(A^{-1}{\frac {\partial A}{\partial x_{i}}}\right),}$

and the Hessian of a Pfaffian is given by

${\displaystyle {\frac {1}{\operatorname {pf} (A)}}{\frac {\partial ^{2}\operatorname {pf} (A)}{\partial x_{i}\partial x_{j}}}={\frac {1}{2}}\operatorname {tr} \left(A^{-1}{\frac {\partial ^{2}A}{\partial x_{i}\partial x_{j}}}\right)-{\frac {1}{2}}\operatorname {tr} \left(A^{-1}{\frac {\partial A}{\partial x_{i}}}A^{-1}{\frac {\partial A}{\partial x_{j}}}\right)+{\frac {1}{4}}\operatorname {tr} \left(A^{-1}{\frac {\partial A}{\partial x_{i}}}\right)\operatorname {tr} \left(A^{-1}{\frac {\partial A}{\partial x_{j}}}\right).}$

The product of the Pfaffians of skew-symmetric matrices A and B can be represented in the form of an exponential

${\displaystyle {\textrm {pf}}(A)\,{\textrm {pf}}(B)=\exp({\tfrac {1}{2}}\mathrm {tr} \log(A^{\text{T}}B)).}$

Suppose A and B are 2n × 2n skew-symmetric matrices, then

${\displaystyle \mathrm {pf} (A)\,\mathrm {pf} (B)={\tfrac {1}{n!}}B_{n}(s_{1},s_{2},\ldots ,s_{n}),\qquad \mathrm {where} \qquad s_{l}=-{\tfrac {1}{2}}(l-1)!\,\mathrm {tr} ((AB)^{l})}$

and B_n(s₁,s₂,...,s_n) are Bell polynomials.

For a block-diagonal matrix

${\displaystyle A_{1}\oplus A_{2}={\begin{bmatrix}A_{1}&0\\0&A_{2}\end{bmatrix}},}$

${\displaystyle \operatorname {pf} (A_{1}\oplus A_{2})=\operatorname {pf} (A_{1})\operatorname {pf} (A_{2}).}$

For an arbitrary n × n matrix M:

${\displaystyle \operatorname {pf} {\begin{bmatrix}0&M\\-M^{\text{T}}&0\end{bmatrix}}=(-1)^{n(n-1)/2}\det M.}$

It is often required to compute the pfaffian of a skew-symmetric matrix ${\displaystyle S}$ with the block structure

${\displaystyle S={\begin{pmatrix}M&Q\\-Q^{T}&N\end{pmatrix}}\,}$

where ${\displaystyle M}$ and ${\displaystyle N}$ are skew-symmetric matrices and ${\displaystyle Q}$ is a general rectangular matrix.

When ${\displaystyle M}$ is invertible, one has

${\displaystyle \operatorname {pf} (S)=\operatorname {pf} (M)\operatorname {pf} (N+Q^{T}M^{-1}Q).}$

This can be seen from Aitken block-diagonalization formula,[3][4][5]

${\displaystyle {\begin{pmatrix}M&0\\0&N+Q^{T}M^{-1}Q\end{pmatrix}}={\begin{pmatrix}I&0\\Q^{T}M^{-1}&I\end{pmatrix}}{\begin{pmatrix}M&Q\\-Q^{T}&N\end{pmatrix}}{\begin{pmatrix}I&-M^{-1}Q\\0&I\end{pmatrix}}.}$

This decomposition involves a congruence transformations that allow to use the pfaffian property ${\displaystyle \operatorname {pf} (BAB^{T})=\operatorname {det} (B)\operatorname {pf} (A)}$.

Similarly, when ${\displaystyle N}$ is invertible, one has

${\displaystyle \operatorname {pf} (S)=\operatorname {pf} (N)\operatorname {pf} (M+QN^{-1}Q^{T}),}$

as can be seen by employing the decomposition

${\displaystyle {\begin{pmatrix}M+QN^{-1}Q^{T}&0\\0&N\end{pmatrix}}={\begin{pmatrix}I&-QN^{-1}\\0&I\end{pmatrix}}{\begin{pmatrix}M&Q\\-Q^{T}&N\end{pmatrix}}{\begin{pmatrix}I&0\\N^{-1}Q^{T}&I\end{pmatrix}}.}$