[proofplan] We encode a skew-symmetric matrix $A$ as the alternating $2$-form $\omega_A$ on $\mathbb R^{2m}$ and use the coefficient of the standard volume form in $\omega_A^m/m!$ as the Pfaffian convention. The congruence formula follows by pulling back $\omega_A$ along the [linear map](/page/Linear%20Map) with matrix $B^\top$ and comparing the induced action on the top exterior power. The determinant identity is then reduced, using the real orthogonal normal form for skew-symmetric matrices, to the [direct sum](/page/Direct%20Sum) of standard $2\times 2$ skew blocks, where both sides can be computed explicitly. [/proofplan]

[step:Relate congruence of skew matrices to pullback of alternating forms] Let $V:=\mathbb R^{2m}$, and let $(e_1,\dots,e_{2m})$ be its standard ordered basis. Let $(e_1^*,\dots,e_{2m}^*)$ be the [dual basis](/theorems/414) of $V^*$. For a real skew-symmetric matrix $A=(A_{ij})\in M_{2m}(\mathbb R)$, define \begin{align*} \omega_A:=\sum_{1\le i<j\le 2m}A_{ij}e_i^*\wedge e_j^*\in \Lambda^2V^*. \end{align*} For vectors $u,v\in V$, written as coordinate columns in the standard basis, skew-symmetry gives \begin{align*} \omega_A(u,v)=u^\top A v. \end{align*} Let $C:V\to V$ be the linear map whose standard matrix is $C\in M_{2m}(\mathbb R)$. For each integer $k\in\{0,\dots,2m\}$, its pullback on alternating $k$-forms is the map \begin{align*} C^*:\Lambda^kV^*\to \Lambda^kV^* \end{align*} defined by \begin{align*} (C^*\eta)(v_1,\dots,v_k)=\eta(Cv_1,\dots,Cv_k) \end{align*} for every $\eta\in\Lambda^kV^*$ and $v_1,\dots,v_k\in V$. Applying this definition with $k=2$ and $\eta=\omega_A$ gives \begin{align*} (C^*\omega_A)(u,v)=\omega_A(Cu,Cv)=u^\top C^\top A C v. \end{align*} Therefore \begin{align*} C^*\omega_A=\omega_{C^\top A C}. \end{align*} Taking $C=B^\top$ gives \begin{align*} (B^\top)^*\omega_A=\omega_{BAB^\top}. \end{align*} [/step]

[step:Compare top exterior coefficients to prove the congruence law]By the defining normalization of the Pfaffian, \begin{align*} \frac{1}{m!}\omega_A^m=\operatorname{Pf}(A)e_1^*\wedge\cdots\wedge e_{2m}^*. \end{align*} Apply $(B^\top)^*$ to both sides. Pullback commutes with wedge products, so \begin{align*} \frac{1}{m!}\bigl((B^\top)^*\omega_A\bigr)^m=\operatorname{Pf}(A)(B^\top)^*(e_1^*\wedge\cdots\wedge e_{2m}^*). \end{align*} Since $(B^\top)^*\omega_A=\omega_{BAB^\top}$, the left-hand side is \begin{align*} \operatorname{Pf}(BAB^\top)e_1^*\wedge\cdots\wedge e_{2m}^*. \end{align*} On the top exterior power, the pullback by a linear map with matrix $B^\top$ multiplies the standard volume covector by $\det(B^\top)=\det(B)$. Hence \begin{align*} (B^\top)^*(e_1^*\wedge\cdots\wedge e_{2m}^*)=\det(B)e_1^*\wedge\cdots\wedge e_{2m}^*. \end{align*} Comparing the coefficient of the nonzero covector $e_1^*\wedge\cdots\wedge e_{2m}^*$ gives \begin{align*} \operatorname{Pf}(BAB^\top)=\det(B)\operatorname{Pf}(A). \end{align*}[/step]

[guided]The point of this step is to turn the matrix identity into a statement about the single coefficient of a top-degree form. The Pfaffian convention says exactly that this coefficient is $\operatorname{Pf}(A)$: \begin{align*} \frac{1}{m!}\omega_A^m=\operatorname{Pf}(A)e_1^*\wedge\cdots\wedge e_{2m}^*. \end{align*} We apply the pullback $(B^\top)^*$ to this identity. From the preceding pullback computation, applied with $C=B^\top$, we have \begin{align*} (B^\top)^*\omega_A=\omega_{BAB^\top}. \end{align*} This is the bridge between the matrix congruence $A\mapsto BAB^\top$ and the exterior-algebra calculation. The pullback $(B^\top)^*$ is an algebra homomorphism on the exterior algebra: it preserves wedge products and scalar multiplication. Thus \begin{align*} (B^\top)^*\left(\frac{1}{m!}\omega_A^m\right)=\frac{1}{m!}\bigl((B^\top)^*\omega_A\bigr)^m. \end{align*} From the previous step, $(B^\top)^*\omega_A=\omega_{BAB^\top}$, so the left-hand side becomes \begin{align*} \frac{1}{m!}\omega_{BAB^\top}^m=\operatorname{Pf}(BAB^\top)e_1^*\wedge\cdots\wedge e_{2m}^*. \end{align*} Now we compute the right-hand side. The space $\Lambda^{2m}V^*$ is one-dimensional and is spanned by the standard volume covector \begin{align*} e_1^*\wedge\cdots\wedge e_{2m}^*. \end{align*} The top exterior power change-of-basis determinant law says that pullback by a linear map multiplies this volume covector by the determinant of the map. Here the map has matrix $B^\top$, so the multiplier is \begin{align*} \det(B^\top)=\det(B). \end{align*} Therefore \begin{align*} (B^\top)^*(e_1^*\wedge\cdots\wedge e_{2m}^*)=\det(B)e_1^*\wedge\cdots\wedge e_{2m}^*. \end{align*} Putting the two computations together gives \begin{align*} \operatorname{Pf}(BAB^\top)e_1^*\wedge\cdots\wedge e_{2m}^*=\det(B)\operatorname{Pf}(A)e_1^*\wedge\cdots\wedge e_{2m}^*. \end{align*} Because $e_1^*\wedge\cdots\wedge e_{2m}^*$ is nonzero, equality of these top-degree forms is equality of their scalar coefficients. Hence \begin{align*} \operatorname{Pf}(BAB^\top)=\det(B)\operatorname{Pf}(A). \end{align*}[/guided]

[step:Compute the Pfaffian and determinant on skew-symmetric block normal form] We use the standard real skew-symmetric orthogonal block diagonal normal form. It states that there exist an [orthogonal matrix](/page/Orthogonal%20Matrix) $Q\in O(2m)$ and [real numbers](/page/Real%20Numbers) $\lambda_1,\dots,\lambda_m\in\mathbb R$ such that \begin{align*} A=QJQ^\top, \end{align*} where $J\in M_{2m}(\mathbb R)$ is the block diagonal matrix with $2\times 2$ diagonal blocks $J_j\in M_2(\mathbb R)$ determined by \begin{align*} (J_j)_{11}=0,\quad (J_j)_{12}=\lambda_j,\quad (J_j)_{21}=-\lambda_j,\quad (J_j)_{22}=0 \end{align*} for $j\in\{1,\dots,m\}$. For this block diagonal matrix, \begin{align*} \omega_J=\sum_{j=1}^m \lambda_j e_{2j-1}^*\wedge e_{2j}^*. \end{align*} In the expansion of $\omega_J^m$, the only nonzero terms are those in which each of the $m$ two-forms $\lambda_j e_{2j-1}^*\wedge e_{2j}^*$ appears exactly once. Since two-forms commute under the wedge product, the $m!$ such terms have the same sign. Hence \begin{align*} \frac{1}{m!}\omega_J^m=\lambda_1\cdots\lambda_m e_1^*\wedge\cdots\wedge e_{2m}^*. \end{align*} Thus \begin{align*} \operatorname{Pf}(J)=\lambda_1\cdots\lambda_m. \end{align*} The determinant of a block diagonal matrix is the product of the determinants of its diagonal blocks, and \begin{align*} \det J_j=\lambda_j^2. \end{align*} Therefore \begin{align*} \det J=\prod_{j=1}^m \lambda_j^2=(\lambda_1\cdots\lambda_m)^2=\operatorname{Pf}(J)^2. \end{align*} [/step]

[step:Transfer the block computation back to the original matrix] Since $A=QJQ^\top$ and $Q\in O(2m)\subset GL(2m,\mathbb R)$, the congruence formula gives \begin{align*} \operatorname{Pf}(A)=\operatorname{Pf}(QJQ^\top)=\det(Q)\operatorname{Pf}(J). \end{align*} Squaring and using $\det(Q)^2=1$, we obtain \begin{align*} \operatorname{Pf}(A)^2=\operatorname{Pf}(J)^2. \end{align*} From the block computation, \begin{align*} \operatorname{Pf}(J)^2=\det J. \end{align*} Finally, \begin{align*} \det A=\det(QJQ^\top)=\det(Q)\det(J)\det(Q^\top)=\det(Q)^2\det(J)=\det J. \end{align*} Combining these equalities yields \begin{align*} \operatorname{Pf}(A)^2=\det A. \end{align*} Together with the congruence law already proved, this completes the proof. [/step]