The exterior product
It may be difficult to calculate Jacobians as the determinant of matrices of derivatives, especially when one has many variables. However, there is an alternative method based on the exterior product which may be useful in some cases. This approach is covered in this post.
When one transforms \(x\) to \(y\) according to \(x = x\left( y \right)\), every component of \(x\) is transformed as follows \({x_i} = {x_i}\left( {{y_1},{y_2},...{y_n}} \right)\). So the differential of \({x_i}\) is \[d{x_i} = \frac{{\partial {x_i}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_i}}}{{\partial {y_2}}}d{y_2} + ... + \frac{{\partial {x_i}}}{{\partial {y_n}}}d{y_n}\] for i=1,2,…,n. I now consider the quantities \(d{x_i}\), i=1,2,…,n, and define a product (or more precisely the exterior product or wedge product) by two rules
multilinearity \[d{x_i} \wedge \left( {ad{x_j} + bd{x_k}} \right) = a\left( {d{x_i} \wedge d{x_j}} \right) + b\left( {d{x_i} \wedge d{x_k}} \right)\] and \[\left( {ad{x_j} + bd{x_k}} \right) \wedge d{x_i} = a\left( {d{x_j} \wedge d{x_i}} \right) + b\left( {d{x_k} \wedge d{x_i}} \right)\]
and skew-symmetry \[d{x_i} \wedge d{x_j} = - d{x_j} \wedge d{x_i}\] i.e. if the order of two terms is reversed, the product changes sign.
Notice that when I write \(d{x_1}d{x_2,}....d{x_n}\) I really mean \(d{x_1} \wedge d{x_2} \wedge .... \wedge d{x_n}\). If follows from the property of skew-symmetry that \[d{x_i} \wedge d{x_i} = - d{x_i} \wedge d{x_i}\] i.e.\(d{x_i} \wedge d{x_i} = 0.\)
An important result is the following one. If \[d{x_i} = \frac{{\partial {x_i}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_i}}}{{\partial {y_2}}}d{y_2} + ... + \frac{{\partial {x_i}}}{{\partial {y_n}}}d{y_n}\] for \(i=1,2,…,n\), and I let \[d{x_1} \wedge d{x_2} \wedge d{x_3} \wedge .... \wedge d{x_n} = J\left( {x \to y} \right)\left( {d{y_1} \wedge d{y_2} \wedge d{y_3} \wedge .... \wedge d{y_n}} \right),\] so that this method is equivalent to the one considered before.
I will not prove this, but I will go through an example instead. Suppose that \(n=3\), so that I need to deal with \(d{x_1} \wedge d{x_2} \wedge d{x_3}\) only, and \[\begin{array}{l} d{x_1} = \frac{{\partial {x_1}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_1}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_1}}}{{\partial {y_3}}}d{y_3}\\ d{x_2} = \frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_3}\\ d{x_3} = \frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3} \end{array}\] Then \[\begin{array}{c} d{x_1} \wedge d{x_2} \wedge d{x_3} = \left( {\frac{{\partial {x_1}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_1}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_1}}}{{\partial {y_3}}}d{y_3}} \right) \wedge \\ \left( {\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_3}} \right) \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ = \frac{{\partial {x_1}}}{{\partial {y_1}}}d{y_1} \wedge \left( {\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_3}} \right) \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_2}}}d{y_2} \wedge \left( {\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_3}} \right) \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_3}}}d{y_3} \wedge \left( {\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_3}} \right) \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right) \end{array}\] because of the property of multilinearity. \[\begin{array}{c} d{x_1} \wedge d{x_2} \wedge d{x_3} = \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_1} \wedge d{y_1} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_1} \wedge d{y_2} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_1} \wedge d{y_3} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_2} \wedge d{y_1} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_2} \wedge d{y_2} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_2} \wedge d{y_3} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_3} \wedge d{y_1} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_3} \wedge d{y_2} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_3} \wedge d{y_3} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right) \end{array}\] Again, this follows from multilinearity. Notice that some terms are zero because \(d{y_i} \wedge d{y_i} = 0\), \[\begin{array}{c} d{x_1} \wedge d{x_2} \wedge d{x_3} = \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_1} \wedge d{y_2} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_1} \wedge d{y_3} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_2} \wedge d{y_1} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_2} \wedge d{y_3} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_3} \wedge d{y_1} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ + \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_3} \wedge d{y_2} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right) \end{array}\]
Things simplify even more. I will focus on the first term \[\begin{array}{l} \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_1} \wedge d{y_2} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right)\\ = \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} \wedge d{y_2} \wedge d{y_1}\\ + \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}\frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_1} \wedge d{y_2} \wedge d{y_2}\\ + \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}\frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_1} \wedge d{y_2} \wedge d{y_3} \end{array}\]
Now \[d{y_1} \wedge d{y_2} \wedge d{y_1} = - d{y_1} \wedge d{y_1} \wedge d{y_2}\] because of skew-symmetry. But \(d{y_1} \wedge d{y_1} = 0\), so that \[d{y_1} \wedge d{y_2} \wedge d{y_1} = - d{y_1} \wedge d{y_1} \wedge d{y_2} = 0.\]
In general, if there are two terms which are the same the outer product is zero. So \[\frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_1} \wedge d{y_2} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right) = \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}\frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_1} \wedge d{y_2} \wedge d{y_3}.\]
Using the same reasoning it is easy to show that \[\begin{array}{c} \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_1} \wedge d{y_3} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right) = - \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}\frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_1} \wedge d{y_3} \wedge d{y_3}\\ \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_2} \wedge d{y_1} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right) = \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}\frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_2} \wedge d{y_1} \wedge d{y_3}\\ = - \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}\frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_1} \wedge d{y_2} \wedge d{y_3}\\ \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}d{y_2} \wedge d{y_3} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right) = \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} \wedge d{y_2} \wedge d{y_3}\\ \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}d{y_3} \wedge d{y_1} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right) = \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}\frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_1} \wedge d{y_2} \wedge d{y_3}\\ \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}d{y_3} \wedge d{y_2} \wedge \left( {\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} + \frac{{\partial {x_3}}}{{\partial {y_2}}}d{y_2} + \frac{{\partial {x_3}}}{{\partial {y_3}}}d{y_3}} \right) = \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_3} \wedge d{y_2} \wedge d{y_1}\\ = - \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_3} \wedge d{y_1} \wedge d{y_2}\\ = \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} \wedge d{y_3} \wedge d{y_2}\\ = - \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}\frac{{\partial {x_3}}}{{\partial {y_1}}}d{y_1} \wedge d{y_2} \wedge d{y_3} \end{array}\]
Putting everything together \[\begin{array}{l} d{x_1} \wedge d{x_2} \wedge d{x_3}\\ = \left[ {\frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}\frac{{\partial {x_3}}}{{\partial {y_3}}} - \frac{{\partial {x_1}}}{{\partial {y_1}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}\frac{{\partial {x_3}}}{{\partial {y_3}}} - \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}\frac{{\partial {x_3}}}{{\partial {y_3}}} + \frac{{\partial {x_1}}}{{\partial {y_2}}}\frac{{\partial {x_2}}}{{\partial {y_3}}}\frac{{\partial {x_3}}}{{\partial {y_1}}} + \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_1}}}\frac{{\partial {x_3}}}{{\partial {y_2}}} - \frac{{\partial {x_1}}}{{\partial {y_3}}}\frac{{\partial {x_2}}}{{\partial {y_2}}}\frac{{\partial {x_3}}}{{\partial {y_1}}}} \right]d{y_1} \wedge d{y_2} \wedge d{y_3} \end{array}\]
The complicated thing in square brackets is \(J\left( {x \to y} \right)\). So \[d{x_1} \wedge d{x_2} \wedge d{x_3} = J\left( {x \to y} \right)d{y_1} \wedge d{y_2} \wedge d{y_3}.\] This can be rewritten as \[\left( {dx} \right) = J\left( {x \to y} \right)\left( {dy} \right)\] where \(\left( {dx} \right) = d{x_1} \wedge d{x_2} \wedge d{x_3}\) and \(\left( {dy} \right) = d{y_1} \wedge d{y_2} \wedge d{y_3}\). Sometimes the notation \(d{x_1} \wedge d{x_2} \wedge d{x_3} \wedge ... \wedge d{x_n} = \mathop \wedge \limits_{i = 1}^n d{x_i}\) is used. The quantity \(\left( {dx} \right) = \mathop \wedge \limits_{i = 1}^n d{x_i}\) is called the volume element.
Example 1. Let \({q_1}\sim{\chi ^2}\left( {{v_1}} \right)\) and \({q_2}\sim{\chi ^2}\left( {{v_2}} \right)\) be independent random variables. Suppose I transform variables as \({q_1} = fq\) and \({q_2} = q\) (i.e. \(f = {q_1}/{q_2}\)). What is the Jacobian of this transformation? To find it, I need to calculate the differentials \[\begin{array}{l} d{q_1} = qdf + fdq\\ d{q_2} = dq \end{array}\] so \[d{q_1} \wedge d{q_2} = \left( {qdf + fdq} \right) \wedge dq = q\left( {df \wedge dq} \right).\]
Exercise 2. Let \(x\) be an \(n \times 1\) vector then I denote \[\left( {dx} \right) = d{x_1} \wedge d{x_2} \wedge d{x_3} \wedge ... \wedge d{x_n} = \mathop \wedge \limits_{i = 1}^n d{x_i}\] (note the order of the elements of the product). Suppose I set \(x = Ay\) where \(A\) is an \(n \times n\) non-singular matrix. Then, I have \(\left( {dx} \right) = \left| A \right|\left( {dy} \right)\).
Example 3. Let \(S > 0\) be a positve definite matrix of dimension \((k + 1 \times k + 1)\). Partition is as \[S = \left( {\begin{array}{*{20}{c}} {{S_{11}}}&{{S_{21}}'}\\ {{S_{21}}}&{{S_{22}}} \end{array}} \right)\] where \({S_{11}}\) is \(1 \times 1\), \({S_{22}}\) is \(k \times k\) and \({S_{21}}\) is \(k \times 1\). Let \[\begin{array}{l} q = {S_{11}} - {S_{21}}'S_{22}^{ - 1}{S_{21}}\,\,\,\,\left( {1 \times 1} \right)\\ r = S_{22}^{ - 1}{S_{21}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \times 1} \right)\\ R = {S_{22}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \times k} \right) \end{array}\] Then \[\begin{array}{l} {S_{22}} = R\\ S{{\kern 1pt} _{12}} = Rr\\ {S_{11}} = q + r'Rr \end{array}\] and \[\begin{array}{l} d{S_{11}} = dq + {\text{terms in }}d{r_i}{\text{ and }}d{R_{ij}}{\text{'s}}\\ dS{{\kern 1pt} _{12}} = \left| R \right|dr + {\text{terms in }}d{R_{ij}}{\text{'s}}\\ d{S_{22}} = \left\{ {d{R_{ij}}} \right\} \end{array}\] Putting things together, \[\begin{array}{c} d{S_{11}} \wedge \left( {d{S_{12}}} \right) \wedge \left( {d{S_{22}}} \right) = dq \wedge \left| R \right|\left( {dr} \right) \wedge \left( {dR} \right) + {\text{all other terms that vanish}}\\ = \left| R \right|dq \wedge \left( {dr} \right) \wedge \left( {dR} \right). \end{array}\] Here I have used \[\begin{array}{l} \left( {dr} \right) = d{r_1} \wedge d{r_2} \wedge ... \wedge d{r_k} = \mathop \wedge \limits_{i = 1}^k d{r_i}\\ \left( {d{S_{12}}} \right) = d{S_{12\left( 1 \right)}} \wedge d{S_{12\left( 2 \right)}} \wedge ... \wedge d{S_{12\left( k \right)}} = \mathop \wedge \limits_{i = 1}^k d{S_{12\left( i \right)}}\\ \left( {dR} \right) = \left( {d{R_1}} \right) \wedge \left( {d{R_2}} \right) \wedge ... \wedge \left( {d{R_k}} \right)\, = \mathop \wedge \limits_{i = 1}^k \mathop \wedge \limits_{j = 1}^k d{R_{ij}}\,\,\,\,{R_j} = j{\text{ - th column of }}R\\ \left( {d{S_{22}}} \right) = \left( {d{S_{22\left( 1 \right)}}} \right) \wedge \left( {d{S_{22\left( 2 \right)}}} \right) \wedge ... \wedge \left( {d{S_{22\left( k \right)}}} \right) = \mathop \wedge \limits_{i = 1}^k \mathop \wedge \limits_{j = 1}^k d{S_{22\left( {i,j} \right)}}\,\,\,\,\,{S_{22\left( j \right)}} = j{\text{ - th column of }}{S_{22}} \end{array}\]