Proposition 1

Let $R$ be an alternating matrix of even order $2m$ over a commutative ring $A$, and let $P$ be a square matrix of order $2m$ over $A$. We have

(7) $\mathrm{Pf}({}^t P . R . P) = (\det P) \mathrm{Pf}(R).$

Indeed, suppose first that $A$ is a field of characteristic 0 and put $R = (\alpha_{ij}), P = (\beta_{st})$. Associate with $R$ the bivector

of $\bigwedge^2 \mathbf{A}^{2m}$, where $(e_i)$ denotes the canonical basis of $\mathbf{A}^{2m}$; consider ${}^t P$ as the matrix, with respect to the basis $(e_i)$, of an endomorphism $f$ of $\mathbf{A}^{2m}$. Then the bivector $(\bigwedge f)(u)$ is associated with the matrix ${}^t P . R . P$ since it is equal to $\frac{1}{2} \sum_{i, j, s, t} \beta_{is} \alpha_{ij} \beta_{jt} e_s \wedge e_t$. Since the extension $\bigwedge f$ of $f$ to the exterior algebra $\bigwedge \mathbf{A}^{2m}$ is an endomorphism of this algebra (chap. III, § 5, no 9), we have $\bigwedge ((\bigwedge f)(u)) = (\bigwedge f)(\bigwedge u)$; since $\bigwedge f$ is the homothety of ratio $\det f$, it follows therefore from (6) and def. 1, that we have $m! \mathrm{Pf}({}^t P . R . P) = m! (\det P) \mathrm{Pf}(R)$, whence (7) in the case under consideration. The general case follows by observing that the two members of (7) are polynomials with integer coefficients in the elements of the matrices $R$ and $P$ (chap. IV, § 2, no 5, Scholium).