Jacobian Matrix

Suppose \(f: \R^n \to \R^m\), with input \(x \in \R^n\) and output \(y \in \R^m\): \[ f = \begin{cases} y_1 = f_1(x_1, x_2, ..., x_n) \\ y_2 = f_2(x_1, x_2, ..., x_n) \\ ... \\ y_m = f_m(x_1, x_2, ..., x_n) \\ \end{cases} \] Then Jacobian matrix is \(m \times n\): \[ \begin{aligned} J &= \begin{bmatrix} \frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} & \cdots & \frac{\partial f}{\partial x_n} \\ \end{bmatrix} \\ &= \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \cdots & \frac{\partial f_2}{\partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1} & \frac{\partial f_m}{\partial x_2} & \cdots & \frac{\partial f_m}{\partial x_n} \\ \end{bmatrix} \end{aligned} \] When \(f\) is a linear transformation, i.e., \(f\) is a \(m \times n\) matrix \(T\), \(y = Tx\), then, \[ J = T \]

When \(f\) is a linear transformation and \(n = m\), i.e., \(f\) is a \(n \times n\) square matrix \(T\), \[ \begin{bmatrix} dy_1 \\ dy_2 \\ \vdots \\ dy_n \\ \end{bmatrix} = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \cdots & \frac{\partial f_2}{\partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} & \frac{\partial f_n}{\partial x_2} & \cdots & \frac{\partial f_n}{\partial x_n} \\ \end{bmatrix} \begin{bmatrix} dx_1 \\ dx_2 \\ \vdots \\ dx_n \\ \end{bmatrix} \] That is, \[ \underbrace{ \begin{bmatrix} dy_1 & 0 & \cdots & 0 \\ 0 & dy_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & dy_n \end{bmatrix}}_A \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \\ \end{bmatrix} = \underbrace{ \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \cdots & \frac{\partial f_2}{\partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} & \frac{\partial f_n}{\partial x_2} & \cdots & \frac{\partial f_n}{\partial x_n} \\ \end{bmatrix}}_J \underbrace{ \begin{bmatrix} dx_1 & 0 & \cdots & 0 \\ 0 & dx_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & dx_n \end{bmatrix}}_B \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \\ \end{bmatrix} \] \(A\) and \(JB\) are both diagonal. From above equation we can find that \(A = JB\). Therefore, \[ \begin{aligned} |A| &= |JB| \\ dy_1 \dots dy_n &= |J|dx_1 \dots dx_n \\ \end{aligned} \]