What is Jacobian matrix?

Last updated: 2026-04-12

The Jacobian, or rather its modulus, is a measure of how a general mapping stretches space locally, near a particular point, even when this stretching effect varies from point to point. The Jacobian takes its name from the German mathematician Carl Jacobi (1804-1851).

Definition

Given two co-ordinates

u(x,y)

and

v(x,y)

which depend on variables

x

and

y

, we define the Jacobian

\begin{equation*}\frac{\partial(u,v)}{\partial(x,y)}\end{equation*}

to be the determinant

\begin{equation*}\begin{vmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{vmatrix}.\end{equation*}

Motivation and Examples

Illustration how area changes under mapping

x = r\cos\theta

and

y = r\sin\theta

Let

x = r\cos\theta

and

y = r\sin\theta

where

r

and

\theta

are polar co-ordinates. Then

\begin{align*}\frac{\partial(x,y)}{\partial(r,\theta)} &= \det\begin{pmatrix}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{pmatrix} \\&= \det\begin{pmatrix}\cos\theta & -r\sin\theta \\\sin\theta & r\cos\theta\end{pmatrix} \\&= r(\cos^2\theta + \sin^2\theta) = r.\end{align*}

Mathematical justification for Jacobian

Theorem

Let

\mathbf{f}: R \to S

be a bijection between two regions of

\mathbb{R}^2

, which is differentiable and has differentiable inverse with

\begin{equation*}\frac{\partial(u,v)}{\partial(x,y)}, \quad \frac{\partial(x,y)}{\partial(u,v)},\end{equation*}

defined and non-zero everywhere. Further, write

(u,v) = \mathbf{f}(x,y)

and let

\psi(x,y) = \Psi(u,v)

. Then

\begin{equation*}\iint_{(u,v)\in S} \Psi(u,v)\,du\,dv = \iint_{(x,y)\in R} \psi(x,y)\left|\frac{\partial(u,v)}{\partial(x,y)}\right|\,dx\,dy,\end{equation*}

\begin{equation*}\iint_{(x,y)\in R} \psi(x,y)\,dx\,dy = \iint_{(u,v)\in S} \Psi(u,v)\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv.\end{equation*}

\end{theorem}

\quad

Proof. It is sufficient to prove the first integral identity.

Split the region into squares

Divide the region

R

into

N

square elements of equal area

\delta x\delta y

as previously with

\psi_i

; the scalar field at the centre of the

i^{th}

element,

i \in \{1,\ldots N\}

Consider the mapping of the

i^{th}

element, which is bounded by the co-ordinate lines

x = x_i

and

x = x_i + \delta x

and

y = y_i

and

y = y_i + \delta y

. The value of the scalar field at the mapped element centre is

\begin{equation*}\Psi_i = \Psi(\mathbf{f}(x_i + \delta x/2, y_i + \delta y/2)) = \psi_i = \psi(x_i + \delta x/2, y_i + \delta y/2).\end{equation*}

Change the area of squares under the mapping

Also, given sufficiently small

\delta x

\delta y

the

i^{th}

element maps to an image region, denoted

IM_i

, which is a deformed parallelogram spanned by the vectors

\begin{align*}\mathbf{a} &= \mathbf{f}(x_i + \delta x, y_i) - \mathbf{f}(x_i, y_i) \approx \frac{\partial\mathbf{f}}{\partial x}(x_i, y_i)\,\delta x, \\\mathbf{b} &= \mathbf{f}(x_i, y_i + \delta y) - \mathbf{f}(x_i, y_i) \approx \frac{\partial\mathbf{f}}{\partial y}(x_i, y_i)\,\delta y,\end{align*}

and, thus, of area

\begin{equation*}\left|\frac{\partial\mathbf{f}}{\partial x}\,\delta x \wedge \frac{\partial\mathbf{f}}{\partial y}\,\delta y\right| = \left|\frac{\partial\mathbf{f}}{\partial x} \wedge \frac{\partial\mathbf{f}}{\partial y}\right|\delta x\,\delta y.\end{equation*}

Now

\mathbf{f} = (u,v)

, so

\mathbf{f}_x = (u_x, v_x)

\mathbf{f}_y = (u_y, v_y)

and thus

\begin{equation*}\left|\frac{\partial\mathbf{f}}{\partial x} \wedge \frac{\partial\mathbf{f}}{\partial y}\right|\delta x\delta y = |(u_x, v_x) \wedge (u_y, v_y)|\delta x\delta y = |(u_x v_y - u_y v_x)\mathbf{k}|\delta x\delta y = \left|\frac{\partial(u,v)}{\partial(x,y)}\right|\delta x\delta y.\end{equation*}

Approximation of the integral

Thus, before taking limits, partitioning

S

by the images

IM_i

, we have an approximation for

\begin{equation*}\iint_{(u,v)\in S} \Psi(u,v)du\,dv\end{equation*}

which is

\begin{equation*}\sum_{i=1}^N \Psi_i\text{Area}(IM_i) = \sum_{i=1}^N \Psi_i\left|\frac{\partial(u,v)}{\partial(x,y)}\right|\delta x\delta y = \sum_{i=1}^N \psi_i\left|\frac{\partial(u,v)}{\partial(x,y)}\right|\delta x\delta y.\end{equation*}

Taking limits gives

\begin{equation*}\iint_{(u,v)\in S} \Psi(u,v)du\,dv = \iint_{(x,y)\in R} \psi(x,y)\left|\frac{\partial(u,v)}{\partial(x,y)}\right|dx\,dy. \quad \Box\end{equation*}

References

[Durrett2019]
Rick Durrett. Probability Theory and Examples. Fifth edition. 2019.