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The Jacobian determinant and local inverses

The determinant of the Jacobian is the factor by which a smooth map scales small areas near a point. Measure it for polar coordinates and the running warp, and see why a non-zero determinant lets a map be undone near a point, with the inverse matrix as the inverse’s Jacobian.
M05.4.4L2 Intermediate25 min
In review

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Before this lesson: M05.4.3 Composition multiplies Jacobians M02.8 Determinants as volume scaling

What you will be able to do
  • Explain why |det J| is the limit of the area ratio of a small square and its image, and what the sign of det J means.
  • Derive det J = r for polar coordinates and check it against the exact area of a small polar cell.
  • State the inverse function theorem and derive, from the chain rule, that the Jacobian of a local inverse is the inverse matrix.
  • Measure area ratios of small squares numerically and find a local inverse with scipy.optimize.root.

The question

A normalising flow turns simple random points into realistic data by pushing them through a smooth map. To keep the probabilities right it must know, at every point, how much the map stretches or squeezes a tiny region. How do you read that off the map, and when can such a map be run backwards?

From the last lesson

1. What is the Jacobian of \({f \circ g}\) at \(\vect{a}\)?

Show answer

\({\mat{J}_f(g(\vect{a}))\,\mat{J}_g(\vect{a})}\): the outer Jacobian, evaluated at \({g(\vect{a})}\), on the left.

2. What is the Jacobian of tanh applied to each entry of a vector \(\vect{z}\)?

Show answer

\({\mathrm{diag}(1 - \tanh^2 z_i)}\), a diagonal matrix, because each output depends only on its own input.

Intuition

Lay four equal small squares in a row in the \({(r, \theta)}\) plane and send them through the polar map. They come out as curved cells of an annulus, and the further out they sit, the bigger they are.

Two panels. On the left, titled Inputs (r, theta), four equal blue squares sit in a row at the same height on a light grid, centred at r equals 0.5, 1, 1.5 and 2, labelled four equal squares. On the right, titled Outputs (x, y), their images under the polar map: four vermillion curved cells lying along a ray from the origin, each wider than the one before, on a faint grid of circles and rays. They are labelled times 0.5, times 1, times 1.5 and times 2, and a note reads area factor r.
M05.4.4-F1

Four equal squares in \({(r, \theta)}\) and their images. A square centred at radius \(r\) comes out with \(r\) times its area: the polar map’s Jacobian determinant.

Lesson M05.4.1 showed why: the polar Jacobian has a unit column pointing outwards and a column of length \(r\) pointing round the circle. A small square becomes, nearly, a rectangle \(r\) times as wide. The number that measures this stretch at each point is the determinant of the Jacobian.

Build it up

Step 1. A small square becomes a near-parallelogram. Take the square with sides \(s\) along the two axes and a corner at \(\vect{a}\). Its sides are \({s\vect{e}_1}\) and \({s\vect{e}_2}\); by M05.4.1 their images are about \({s\mat{J}\vect{e}_1}\) and \({s\mat{J}\vect{e}_2}\), the two columns of \(\mat{J}\) times \(s\).

Step 2. The parallelogram’s area. The area of the parallelogram spanned by two columns is the absolute value of the determinant of the matrix they form (M02.8). Scaling both columns by \(s\) scales the determinant by \(s^2\), so the area is \({s^2\,\abs{\det\mat{J}}}\).

Step 3. Divide by the square’s area. The square has area \(s^2\), so the ratio of areas is about \({\abs{\det\mat{J}(\vect{a})}}\), and the approximation improves as \(s\) shrinks.

A plot of the area of the image over the area of the square against the side of the square, which shrinks from 1 on the left to 0.001 on the right on a log scale. A solid vermillion line with dots, labelled polar map, square at (2, pi/6), starts at 2.5 and falls onto a black dashed line at 2, labelled det J equals 2. A dashed vermillion line with squares, labelled warp, square with a corner at the origin, starts just above 0.93 and settles onto a black dashed line at 0.91, labelled det J equals 0.91.
M05.4.4-F2

The measured area ratio of a shrinking square, for the polar map at \({(2, \pi/6)}\) and the running warp at the origin. Both settle on the Jacobian determinant.

Step 4. Polar cells, exactly. The image of the square \({[r, r + s] \times [\theta, \theta + s]}\) is the part of a sector of radius \({r + s}\) and angle \(s\) outside the sector of radius \(r\). A sector of radius \(R\) and angle \(s\) is the fraction \({s/(2\pi)}\) of a disc of area \({\pi R^2}\), so its area is \({sR^2/2}\), and

\[ \begin{aligned} \text{area} &= \tfrac{s}{2}(r + s)^2 - \tfrac{s}{2}r^2 \\ &= \tfrac{s}{2}(2rs + s^2) = s^2\big(r + \tfrac{s}{2}\big) . \end{aligned} \]

The ratio to the square’s area \(s^2\) is \({r + s/2}\), which tends to \(r\). At \({r = 2}\) it is 2.05 for \({s = 0.1}\) and 2.005 for \({s = 0.01}\).

The formal statement

NoteThe Jacobian determinant as an area factor

Let \(f\) map the plane to the plane with continuous partial derivatives near \(\vect{a}\). For squares \(S\) with a corner at \(\vect{a}\),

\[ \frac{\text{area of } f(S)}{\text{area of } S} \to \abs{\det\mat{J}_f(\vect{a})} \]

as the side of \(S\) shrinks to 0. If \({\det\mat{J}_f(\vect{a}) > 0}\) the map keeps orientation near \(\vect{a}\); if it is negative it turns small shapes over, as a mirror does. The same holds in \(\R^n\) with volumes.

We state this without a full proof. The idea is Steps 1 to 3: the image’s boundary stays within a distance of the parallelogram’s boundary that is small compared with \(s\), so the two areas differ by a small fraction of \(s^2\). Lesson M05.6.4 uses exactly this factor to change variables in an integral.

The polar determinant. From the Jacobian of M05.4.1,

\[ \begin{aligned} \det\mat{J}_P &= \cos\theta \cdot r\cos\theta \\ &\quad - (-r\sin\theta)\sin\theta \\ &= r(\cos^2\theta + \sin^2\theta) = r . \end{aligned} \]

NoteThe inverse function theorem

Let \(f\) map \(\R^n\) to \(\R^n\) with continuous partial derivatives near \(\vect{a}\), and suppose \({\det\mat{J}_f(\vect{a}) \neq 0}\). Then there are open sets \(U\) around \(\vect{a}\) and \(V\) around \({f(\vect{a})}\) such that \(f\) maps \(U\) one-to-one onto \(V\), the inverse \(g\) from \(V\) back to \(U\) has continuous partial derivatives, and

\[ \mat{J}_g(f(\vect{a})) = \mat{J}_f(\vect{a})^{-1} . \]

The existence of \(g\) is the deep part, and we take it from the theorem. The formula for its Jacobian then follows from the chain rule. For every \(\vect{x}\) in \(U\), \({g(f(\vect{x})) = \vect{x}}\). The identity map has Jacobian \(\mat{I}\), and by M05.4.3 the left side has Jacobian \({\mat{J}_g(f(\vect{a}))\,\mat{J}_f(\vect{a})}\). So their product is \(\mat{I}\), and a square matrix whose product with another is \(\mat{I}\) is that matrix’s inverse (M02.7).

Two panels joined by two arrows: a green arrow labelled P pointing right from the first panel to the second, and a sky-blue arrow labelled P inverse pointing back. The first panel, titled U around (2, pi/6), shows a blue disc with an orange centre and, below it, J equals the matrix with rows 0.87, minus 1 and 0.5, 1.73. The second panel, titled V equals P of U around (root 3, 1), shows a vermillion oval leaning to the left with an orange centre and, below it, J inverse equals the matrix with rows 0.87, 0.5 and minus 0.25, 0.43.
M05.4.4-F3

Near \({(2, \pi/6)}\) the polar map carries a small disc one-to-one onto an oval, and a local inverse carries it back, with the inverse matrix as its Jacobian.

For the polar map the local inverse is \({r = \sqrt{x^2 + y^2}}\) with \(\theta\) the angle of \({(x, y)}\). Its partial derivatives at \({(\sqrt3, 1)}\), where \({r = 2}\), are \({x/r \approx 0.866}\), \({y/r = 0.5}\), \({-y/r^2 = -0.25}\) and \({x/r^2 \approx 0.433}\): exactly the entries of \({\mat{J}_P(2, \pi/6)^{-1}}\). At \({r = 0}\) the determinant is 0 and the theorem says nothing; indeed every point \({(0, \theta)}\) goes to the origin, so no neighbourhood of \({(0, \theta)}\) is mapped one-to-one.

Frontier note: why normalising flows need log |det J|

A normalising flow writes data as \({\vect{x} = T(\vect{u})}\), with \(\vect{u}\) drawn from a simple density \({p_u}\) and \(T\) invertible and differentiable both ways. Probability is conserved: a small region keeps its probability while its area is multiplied by \({\abs{\det\mat{J}_T}}\), so the density is divided by that factor, \({p_x(\vect{x}) = p_u(\vect{u})\,\abs{\det\mat{J}_T(\vect{u})}^{-1}}\) (Papamakarios et al., Equation 2). In logs, \({\log p_x = \log p_u - \log\abs{\det\mat{J}_T}}\). A flow is a composition of many layers, so by the chain rule its log-determinant is a sum over layers (their Equation 28). Designs such as autoregressive and coupling layers make each Jacobian triangular, so its determinant is the product of the diagonal entries and costs only \(D\) operations in dimension \(D\) (their Equation 32). Change of variables for densities is M05.6.4 and M07.1; flows are DL06.7.

Worked example

The running warp. From \({\mat{J}_W = \begin{bmatrix} 1 & 0.3\cos y \\ 0.3\cos x & 1 \end{bmatrix}}\),

\[ \det\mat{J}_W = 1 - 0.09\cos x\cos y . \]

At \({(0, 0)}\) it is \({1 - 0.09 = 0.91}\): near the origin the warp shrinks areas by 9%. At \({(\pi/2, 0)}\) it is \({1 - 0 = 1}\): the shear there keeps areas. Since the product of the two cosines lies between \(-1\) and 1, the determinant always lies between 0.91 and 1.09. It is never 0, so the inverse function theorem gives a local inverse near every point.

Two panels. On the left, titled Two squares of side 0.4, two blue squares on a light grid with orange lower-left corners at (0, 0) and (pi/2, 0). On the right, titled Their images under W, the warped grid in grey and two vermillion parallelograms, each with a black outline of the exact image almost on top of it. The first, from (0, 0), leans to the right and is labelled det J equals 0.91 below and measured 0.915 above; the second, from (pi/2, 0), is a sheared square labelled det J equals 1.00 and measured 1.017.
M05.4.4-F4

The worked example as a trace. Squares of side 0.4 at the two points, and their images: the measured area ratios, 0.915 and 1.017, are already close to the determinants 0.91 and 1.

Measured. The exact image of a square of side 0.4 with a corner at \({(0, 0)}\) has 0.915 times its area; with side 0.01 the ratio is 0.91000. At \({(\pi/2, 0)}\) the ratios are 1.017 and 1.00045.

A local inverse. Near \({W(\pi/2, 0) = (\pi/2, 0.3)}\), which point is sent to \({(\pi/2 + 0.05,\; 0.28)}\), a step of \({\vect{\delta} = (0.05, -0.02)}\)? The inverse Jacobian answers to first order. Here

\[ \mat{J}_W^{-1} = \begin{bmatrix} 1 & -0.3 \\ 0 & 1 \end{bmatrix} , \]

so the input moves by about \({\mat{J}_W^{-1}\vect{\delta}} = {(0.05 + 0.006,\; -0.02)} = {(0.056, -0.02)}\). Solving \({W(\vect{x}) = (\pi/2 + 0.05,\; 0.28)}\) exactly, as the code does, gives a move of about \({(0.0559, -0.0195)}\).

Code

From scratch: the exact image of a square is traced along its boundary and its area found with the shoelace formula, \({\tfrac12\sum (x_k y_{k+1} - x_{k+1}y_k)}\).

import numpy as np
from scipy.optimize import root

def warp(v):
    x, y = v
    return np.array([x + 0.3 * np.sin(y), y + 0.3 * np.sin(x)])

def polar(v):
    r, th = v
    return np.array([r * np.cos(th), r * np.sin(th)])

def square_boundary(corner, s, n=400):
    t = np.linspace(0, s, n, endpoint=False)
    x0, y0 = corner
    xs = np.concatenate([x0 + t, np.full(n, x0 + s), x0 + s - t, np.full(n, x0)])
    ys = np.concatenate([np.full(n, y0), y0 + t, np.full(n, y0 + s), y0 + s - t])
    return np.array([xs, ys])

def area_ratio(f, corner, s):
    x, y = f(square_boundary(corner, s))
    area = 0.5 * np.sum(x * np.roll(y, -1) - np.roll(x, -1) * y)
    return area / s ** 2

det_warp = lambda x, y: 1 - 0.09 * np.cos(x) * np.cos(y)
for corner in [(0.0, 0.0), (np.pi / 2, 0.0)]:
    ratios = [round(area_ratio(warp, corner, s), 5) for s in (0.4, 0.1, 0.01)]
    print("warp at", np.round(corner, 3), "ratios", ratios, " det J", round(det_warp(*corner), 5))
print("polar at (2, pi/6):", [round(area_ratio(polar, (2, np.pi / 6), s), 5) for s in (0.1, 0.01)])
warp at [0. 0.] ratios [np.float64(0.9147), np.float64(0.9103), np.float64(0.91)]  det J 0.91
warp at [1.571 0.   ] ratios [np.float64(1.01729), np.float64(1.00449), np.float64(1.00045)]  det J 1.0
polar at (2, pi/6): [np.float64(2.05), np.float64(2.005)]

Now a local inverse. scipy.optimize.root solves \({W(\vect{x}) = \vect{y}}\) for \(\vect{x}\), starting from a guess; we start at \({(\pi/2, 0)}\). Differencing the solver’s answer in each coordinate of \(\vect{y}\) gives the Jacobian of the inverse, to compare with the inverse matrix.

a = np.array([np.pi / 2, 0.0])
J = np.array([[1.0, 0.3 * np.cos(a[1])], [0.3 * np.cos(a[0]), 1.0]])
delta = np.array([0.05, -0.02])

def inverse(y):
    sol = root(lambda v: warp(v) - y, a, tol=1e-13)
    assert sol.success
    return sol.x

move = inverse(warp(a) + delta) - a
print("exact move", move.round(4), " first-order", np.linalg.solve(J, delta).round(4))

h = 1e-5
J_inv = np.column_stack([(inverse(warp(a) + h * e) - inverse(warp(a) - h * e)) / (2 * h) for e in np.eye(2)])
print("Jacobian of the inverse:\n", J_inv.round(6), "\ninverse of J:\n", np.linalg.inv(J).round(6))
exact move [ 0.0559 -0.0195]  first-order [ 0.056 -0.02 ]
Jacobian of the inverse:
 [[ 1.  -0.3]
 [ 0.   1. ]] 
inverse of J:
 [[ 1.  -0.3]
 [-0.   1. ]]

Explore

This module’s interactive, the grid warp, lives in lesson M05.4.1. Choose the polar map and drag the point outwards: the measured area ratio follows \(r\). Then press Shrink the square and watch the ratio settle on \({\det\mat{J}}\), from \({r + s/2}\) down to \(r\).

Pitfalls

Two square panels, each with faint axes. The left panel, marked with a cross and titled Strip of 1.25 turns, shows a vermillion annulus between radii 0.6 and 1.4, with its upper-right quarter, covered by the first quarter turn and again by the last, shaded reddish purple and a black dot in it, and the words covered twice in the middle; the caption reads theta from 0 to 2.5 pi. The right panel, marked with a tick and titled Strip of 0.75 turn, shows three quarters of the same annulus, a gap in the lower right quarter, captioned theta from 0 to 1.5 pi. A line below reads: both strips have 0.6 at most r at most 1.4, so det J equals r, greater than 0 at every point of both.
M05.4.4-F5

Wrong beside right. The polar map has \({\det\mat{J} = r > 0}\) on both strips, but the longer strip wraps round and lands on the same points twice.

“The determinant is never zero, so the map is one-to-one everywhere.”

The inverse function theorem is local. The polar map on the strip \({0.6 \le r \le 1.4}\), \({0 \le \theta \le 2.5\pi}\) has determinant \({r > 0}\) at every point, yet \({(1, \pi/4)}\) and \({(1, \pi/4 + 2\pi)}\) land on the same point. Near any one point it can be undone; over the whole strip it cannot.

“Where the determinant is zero, there is no inverse at all.”

A zero determinant only removes the guarantee. The map \({x \mapsto x^3}\) has derivative 0 at 0, yet it is one-to-one, with inverse the cube root. What fails is the inverse’s derivative: the cube root is not differentiable at 0. The polar map at \({r = 0}\), by contrast, really cannot be undone.

“The area factor is \({\det\mat{J}}\), so it can be negative.”

Area is never negative. The factor is \({\abs{\det\mat{J}}}\); the sign tells you about orientation. The swap \({(x, y) \mapsto (y, x)}\) has determinant \(-1\): it keeps every area and turns every shape over. Forgetting the absolute value gives negative probabilities in a change of variables.

Check yourself

Recall

1. What does \({\abs{\det\mat{J}_f(\vect{a})}}\) measure?

Show answer

The factor by which \(f\) scales small areas near \(\vect{a}\): the limit of the area of the image of a small square over the area of the square.

2. What is the Jacobian determinant of the polar map, and where is it zero?

Show answer

\(r\). It is zero at \({r = 0}\), where every angle is sent to the origin.

3. State the inverse function theorem in one sentence.

Show answer

If \(f\) has continuous partial derivatives near \(\vect{a}\) and \({\det\mat{J}_f(\vect{a}) \neq 0}\), then \(f\) can be undone near \(\vect{a}\) by a smooth local inverse, whose Jacobian at \({f(\vect{a})}\) is \({\mat{J}_f(\vect{a})^{-1}}\).

4. How does the chain rule give the Jacobian of the inverse?

Show answer

From \({g(f(\vect{x})) = \vect{x}}\): \({\mat{J}_g(f(\vect{a}))\,\mat{J}_f(\vect{a}) = \mat{I}}\), so \({\mat{J}_g(f(\vect{a})) = \mat{J}_f(\vect{a})^{-1}}\).

5. What does a negative Jacobian determinant mean?

Show answer

The map turns small shapes over near the point, as a mirror does; it still scales area by the absolute value.

Exercises

1. ★ Find the Jacobian determinants of \({(u, v) \mapsto (2u, 3v)}\) and of \({(u, v) \mapsto (u + v^2, v)}\).

Show solution

The first has Jacobian \({\mathrm{diag}(2, 3)}\), determinant 6: every area is multiplied by 6. The second has Jacobian \({\begin{bmatrix} 1 & 2v \\ 0 & 1 \end{bmatrix}}\), determinant 1 at every point: it bends the grid but keeps every area, like a shear.

2. ★★ Let \({f(x, y) = (x^2 - y^2,\; 2xy)}\), the squaring of complex numbers (M01.9). Find \({\det\mat{J}_f}\). Where does the inverse function theorem fail, and is \(f\) one-to-one away from that point?

Show solution

\[ \mat{J}_f = \begin{bmatrix} 2x & -2y \\ 2y & 2x \end{bmatrix} , \]

so \({\det\mat{J}_f = 4x^2 + 4y^2}\), which is zero only at the origin. Everywhere else \(f\) has a local inverse. But it is not one-to-one: \({f(1, 0) = f(-1, 0) = (1, 0)}\), since \(z\) and \(-z\) have the same square. Again the theorem is only local.

3. ★★★ Find the Jacobian of the local inverse of the running warp at \({W(0, 0) = (0, 0)}\). Then show that the warp is one-to-one on the whole plane. (Use \({\abs{\sin p - \sin q} \le \abs{p - q}}\).)

Show solution

\({\mat{J}_W(0, 0) = \begin{bmatrix} 1 & 0.3 \\ 0.3 & 1 \end{bmatrix}}\) with determinant 0.91, so the inverse’s Jacobian is

\[ \begin{aligned} &\frac{1}{0.91}\begin{bmatrix} 1 & -0.3 \\ -0.3 & 1 \end{bmatrix} \\ &\quad \approx \begin{bmatrix} 1.099 & -0.330 \\ -0.330 & 1.099 \end{bmatrix} . \end{aligned} \]

Suppose \({W(\vect{p}) = W(\vect{q})}\). Subtracting the two outputs gives

\[ \begin{aligned} \vect{p} - \vect{q} = -0.3\,(&\sin p_2 - \sin q_2, \\ &\sin p_1 - \sin q_1) . \end{aligned} \]

Each entry of the bracket is at most the matching entry of \({(p_2 - q_2,\; p_1 - q_1)}\) in size, a vector of length \({\norm{\vect{p} - \vect{q}}}\). So \({\norm{\vect{p} - \vect{q}} \le 0.3\,\norm{\vect{p} - \vect{q}}}\), which forces \({\vect{p} = \vect{q}}\).

Summary card

The Jacobian determinant is the local area factor; where it is not zero, the map can be undone nearby.

The anchor figure again: four equal squares in the r, theta plane and their images under the polar map, curved cells that grow with r, labelled times 0.5, times 1, times 1.5 and times 2.
M05.4.4-F1

Area factor \(r\) for polar cells.

  • Area of \({f(S)}\) over area of \(S\) tends to \({\abs{\det\mat{J}_f(\vect{a})}}\); the sign is orientation.
  • Polar: \({\det\mat{J} = r}\). Warp: \({1 - 0.09\cos x\cos y}\), from 0.91 to 1.09.
  • \({\det\mat{J}_f(\vect{a}) \neq 0}\): a local inverse \(g\), with \({\mat{J}_g(f(\vect{a})) = \mat{J}_f(\vect{a})^{-1}}\).
  • Local, not global: the polar map wraps round.

Sources and next steps

Sources. Each was opened and checked against this lesson on 15 November 2026.

  • Gilbert Strang and Edwin Herman, Calculus Volume 3 (OpenStax), Section 5.7: the book’s “Jacobian” is the determinant of this matrix; Example 5.67 finds \({J(r, \theta) = r}\) for polar coordinates, and Theorem 5.14 asks for a one-to-one map with a non-zero Jacobian.
  • Marc Peter Deisenroth, A. Aldo Faisal and Cheng Soon Ong, Mathematics for Machine Learning, Section 5.3: the absolute value of the Jacobian determinant is the factor by which areas or volumes are scaled, exact for a linear map and a local approximation for a nonlinear one.
  • Rasul Shafikov, Real Analysis lecture notes, Section 2 (University of Western Ontario): Theorem 2.4, the inverse function theorem, and the example \({x^3}\), whose inverse exists but is not differentiable at 0.
  • George Papamakarios, Eric Nalisnick, Danilo Jimenez Rezende, Shakir Mohamed and Balaji Lakshminarayanan, Normalizing Flows for Probabilistic Modeling and Inference, JMLR 22 (2021), Section 2.1 (Equations 2 to 6: the density under a map, the determinant as a volume ratio, composition), Equation 28 and Section 3.1 (Equation 32, triangular Jacobians).
  • SciPy 1.18 reference, scipy.optimize.root: arguments fun and x0, default method hybr, and the result’s x and success.

Next. Lesson M05.5.1 differentiates the gradient itself: the Jacobian of the gradient is the Hessian. Lesson M05.6.4 uses the area factor to change variables in an integral.

This lesson unlocks

M05.5 Hessians, curvature and second-order Taylor M05.6 Multiple integrals and change of variables

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