Jacobian Matrices of Coordinate Transformations

Why Coordinate Maps Need a Matrix of Derivatives

A coordinate map expresses several output coordinates in terms of several input variables. For example, a map might send $(u,v)$ to a point $(x,y)$, with $x$ and $y$ given by formulas involving $u$ and $v$. To describe how all the outputs respond to changes in all the inputs, we need to organize several partial derivatives at once.

The resulting array is the Jacobian matrix. Its rows correspond to output functions, and its columns correspond to input variables. That convention makes the matrix dimensions immediate and gives a consistent way to compute its entries. When the number of outputs equals the number of inputs, the matrix is square and has a determinant.

Here the goal is to construct these matrices and calculate their determinants. The determinant’s geometric interpretation and its role in substitution for multiple integrals are separate matters; a determinant can be computed before making either interpretation.

Definition and Dimension

Definition: Let $F:\mathbb{R}^n\to\mathbb{R}^m$ be a differentiable map, written in components as $F(u_1,\ldots,u_n)=(F_1(u_1,\ldots,u_n),\ldots,F_m(u_1,\ldots,u_n))$. The Jacobian matrix of $F$ at an input point is the $m\times n$ matrix $$ D F = \begin{bmatrix} \frac{\partial F_1}{\partial u_1} & \cdots & \frac{\partial F_1}{\partial u_n}\\ \vdots & \ddots & \vdots\\ \frac{\partial F_m}{\partial u_1} & \cdots & \frac{\partial F_m}{\partial u_n} \end{bmatrix}. $$ There is one row per output function and one column per input variable. A determinant of this matrix is defined only when $m=n$.

For a map $F(u,v)=(x(u,v),y(u,v))$, the Jacobian matrix is $2\times2$. For a map $G(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w))$, it is $3\times3$. A map from two inputs to three outputs instead has a $3\times2$ Jacobian matrix, for which there is no determinant.

Polynomial functions are differentiable everywhere. Rational functions are differentiable at points where their denominators are nonzero. In either setting, the matrix entries are found by differentiating each output function with respect to each input variable, holding the other inputs fixed.

Why the Partial Derivatives Form the Matrix

The Jacobian matrix is the linear part of the map’s change near an input point. Precisely, differentiability at a point $p$ means there is a matrix $A$ such that $F(p+h)=F(p)+Ah+r(h)$, where the remainder satisfies $\lVert r(h)\rVert/\lVert h\rVert\to0$ as $h\to0$. The matrix $A$ gives the best first-order linear approximation to the change in $F$.

To identify its entries, change just one input coordinate at a time. If $h$ has only its $j$-th coordinate nonzero, the $i$-th output’s first-order change is $\frac{\partial F_i}{\partial u_j}h_j$. The $i$-th row of $A$ must therefore contain the partial derivatives of $F_i$ in the order of the inputs. Repeating this for each output gives the stated matrix.

This argument also explains why a list of partial derivatives is not enough by itself to establish differentiability for every possible function: the first-order approximation with a remainder small compared with $\lVert h\rVert$ is required. Polynomial and rational maps on their domains are differentiable, so their partial derivatives do form their Jacobian matrices.

Worked Examples

Worked Example: A Polynomial Map in Two Variables

Let $F(u,v)=(x(u,v),y(u,v))$, where $x=u^2+3v$ and $y=2u-v^2$. Find its Jacobian matrix and determinant.

There are two outputs and two inputs, so the matrix will be $2\times2$. Differentiate each output with respect to each input:

\frac{\partial x}{\partial u}=2u,\qquad \frac{\partial x}{\partial v}=3,\qquad \frac{\partial y}{\partial u}=2,\qquad \frac{\partial y}{\partial v}=-2v.

Place the derivatives of $x$ in the first row and those of $y$ in the second row, keeping the input order $u,v$ in each row:

D F(u,v)= \begin{bmatrix} 2u&3\\ 2&-2v \end{bmatrix}.

For a $2\times2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, its determinant is $ad-bc$. Thus

\det D F(u,v) =(2u)(-2v)-(3)(2) =-4uv-6.

Check: The four entries differentiate the stated output functions in the required input order. Applying $ad-bc$ to the displayed matrix gives $(2u)(-2v)-(3)(2)=-4uv-6$, confirming the determinant.

Worked Example: A Rational Map in Two Variables

Consider $F(u,v)=(x(u,v),y(u,v))$, where $x=u/v$ and $y=u+v$. The map is defined on the domain $v\ne0$. Find its Jacobian matrix and determinant.

Differentiate the quotient $u/v$ with respect to each input. With respect to $u$, $v$ is fixed; with respect to $v$, $u$ is fixed. The derivatives of the second output are both $1$:

\frac{\partial x}{\partial u}=\frac{1}{v},\qquad \frac{\partial x}{\partial v}=-\frac{u}{v^2},\qquad \frac{\partial y}{\partial u}=1,\qquad \frac{\partial y}{\partial v}=1.

Therefore the Jacobian matrix and its determinant are

D F(u,v)= \begin{bmatrix} \frac{1}{v}&-\frac{u}{v^2}\\ 1&1 \end{bmatrix}, \qquad \det D F(u,v) =\frac{1}{v}(1)-\left(-\frac{u}{v^2}\right)(1) =\frac{1}{v}+\frac{u}{v^2} =\frac{u+v}{v^2}.

Check: The denominator $v^2$ is nonzero throughout the map’s domain $v\ne0$. Expanding the determinant directly gives $1/v+u/v^2$; putting these terms over the common denominator $v^2$ gives $(u+v)/v^2$, as stated.

Worked Example: A Polynomial Map in Three Variables

Let $G(u,v,w)=(x,y,z)$, where $x=u+v$, $y=v+w$, and $z=uw$. Find its Jacobian matrix and determinant.

There are three inputs and three outputs, so the matrix is $3\times3$. The rows list the derivatives of $x$, $y$, and $z$, each in the input order $u,v,w$:

\frac{\partial x}{\partial(u,v,w)}=(1,1,0),\qquad \frac{\partial y}{\partial(u,v,w)}=(0,1,1),\qquad \frac{\partial z}{\partial(u,v,w)}=(w,0,u).

Hence

D G(u,v,w)= \begin{bmatrix} 1&1&0\\ 0&1&1\\ w&0&u \end{bmatrix}.

Expand along the first row. The first entry contributes $1$ times the determinant of $\begin{bmatrix}1&1\\0&u\end{bmatrix}$; the second entry has a minus sign and contributes the determinant of $\begin{bmatrix}0&1\\w&u\end{bmatrix}$; the third entry contributes zero:

\det D G =1\big((1)(u)-(1)(0)\big) -1\big((0)(u)-(1)(w)\big) +0 =u-(-w) =u+w.

Check: The two $2\times2$ determinants in the expansion are $u$ and $-w$, respectively. Including the minus sign on the second term gives $u-(-w)=u+w$.

Worked Example: A Rational Map in Three Variables

Define $H(u,v,w)=(x,y,z)$ by $x=u/v$, $y=v/w$, and $z=w$, on the domain $v\ne0$ and $w\ne0$. Find its Jacobian matrix and determinant.

The first output has derivatives $1/v$, $-u/v^2$, and $0$. The second has derivatives $0$, $1/w$, and $-v/w^2$. The last output has derivatives $0,0,1$. Thus

D H(u,v,w)= \begin{bmatrix} \frac{1}{v}&-\frac{u}{v^2}&0\\ 0&\frac{1}{w}&-\frac{v}{w^2}\\ 0&0&1 \end{bmatrix}.

Expanding along the third row leaves the determinant of the upper-left $2\times2$ block:

\det D H(u,v,w) =1\left(\frac{1}{v}\frac{1}{w} -\left(-\frac{u}{v^2}\right)(0)\right) =\frac{1}{vw}.

Check: The zero below the first diagonal entry and the zero below the second diagonal entry make the determinant equal to the product of the diagonal entries: $(1/v)(1/w)(1)=1/(vw)$. This is defined on the stated domain.

When a Determinant Is Not Available

For a map $K:\mathbb{R}^2\to\mathbb{R}^3$, the Jacobian has three rows and two columns. For instance, if $K(u,v)=(u+v,u-v,uv)$, then

D K(u,v)= \begin{bmatrix} 1&1\\ 1&-1\\ v&u \end{bmatrix}.

This $3\times2$ matrix is a valid Jacobian matrix, but it has no determinant. The determinant formula applies to square matrices only. One should not delete a row, add an artificial row, or reverse the input and output roles merely to obtain a square array.

Key takeaway: For a map with $n$ inputs and $m$ outputs, the Jacobian matrix has $m$ rows and $n$ columns. Its determinant exists only if $m=n$.

A Common Mistake: Reversing the Input Order

Suppose $F(u,v)=(u^2+v,uv)$. In the fixed input order $u,v$, the first row must contain the derivatives of the first output with respect to $u$ and then $v$; the second row must do the same for the second output. The correct matrix is

D F(u,v)= \begin{bmatrix} 2u&1\\ v&u \end{bmatrix}.

A common error is to write the first row as $(1,2u)$, putting the $v$-derivative before the $u$-derivative, while leaving the second row in the order $(v,u)$. That produces the incorrect matrix $\begin{bmatrix}1&2u\\v&u\end{bmatrix}$ and the incorrect determinant $u-2uv$.

The correction is to keep the same input order in every row. Since the columns are $u,v$, the first row is $(2u,1)$ and the second is $(v,u)$. The correct determinant is therefore

$$ (2u)(u)-(1)(v)=2u^2-v. $$

At $u=1,\ v=2$, the incorrect calculation gives $1-4=-3$, while the corrected calculation gives $2-2=0$. Checking the entry order before calculating the determinant prevents this error.

A Reliable Construction Procedure

List the inputs and outputs.
Write the input variables in a fixed order and list the output functions in their stated order.

Set the matrix dimensions.
Use one row for each output and one column for each input. The size is outputs by inputs.

Differentiate each output.
Within each row, differentiate with respect to the input variables in the fixed column order. For rational functions, retain the restrictions that keep denominators nonzero.

Compute a determinant only if the matrix is square.
For a square matrix, use the appropriate determinant formula and check the signs and arithmetic.

Check Your Understanding

Use the row-by-output and column-by-input convention in each question.

For $F(u,v)=(u+v^2,u^2-v)$, what are the dimensions of the Jacobian matrix, and what are its entries?
For $G(u,v,w)=(uv,v+w,u-w)$, construct the Jacobian matrix and compute its determinant.
For $H(u,v)=(u/v,u+v)$, state the domain restriction and compute the Jacobian determinant.
What are the dimensions of the Jacobian matrix for a map from $\mathbb{R}^2$ to $\mathbb{R}^3$, and why is its determinant undefined?
In a Jacobian matrix, what does a row represent, and what does a column represent?