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Inverse Maps and Reciprocal Jacobians

Use the inverse function theorem and the chain rule to calculate and check the Jacobian determinant of an invertible map or its inverse.

Advanced 9 min read

What You'll Learn

  • State when the inverse function theorem guarantees a differentiable local inverse
  • Derive the matrix and determinant relationships for a map and its inverse
  • Keep track of the corresponding points in the original and transformed coordinates
  • Calculate inverse Jacobians for linear and nonlinear coordinate maps
  • Check a reciprocal-Jacobian calculation by composing the maps

Why Inverse Jacobians Are Reciprocal

A coordinate transformation describes how points in one coordinate system correspond to points in another. Its Jacobian matrix records how the map changes coordinates locally, and its determinant summarizes the associated signed local scaling. When a transformation is invertible, its inverse undoes those changes.

This undoing has a precise consequence: at corresponding points, the Jacobian matrices of a differentiable map and its inverse are matrix inverses. Their determinants are therefore reciprocals. The correspondence between the points matters: the determinant of the inverse is evaluated at the image of the original point, not generally at the same coordinate tuple.

Key formula: If \(F\) has a differentiable inverse near \(x\), then \(\det D(F^{-1})(F(x))=1/\det DF(x)\), provided \(\det DF(x)\ne 0\).

The Inverse Function Theorem

Let \(U\) be an open subset of \(\mathbb{R}^n\), and let \(F:U\to\mathbb{R}^n\) be continuously differentiable. The Jacobian matrix \(DF(a)\) is the matrix of first partial derivatives of \(F\) at \(a\). If its determinant is nonzero, the inverse function theorem guarantees that \(F\) has a differentiable inverse on suitable neighborhoods of \(a\) and \(F(a)\).

Theorem: Suppose \(F:U\to\mathbb{R}^n\) is continuously differentiable and \(\det DF(a)\ne 0\). Then there are open neighborhoods \(U_0\) of \(a\) and \(V_0\) of \(F(a)\) such that \(F:U_0\to V_0\) is one-to-one and onto, and \(F^{-1}:V_0\to U_0\) is continuously differentiable. At corresponding points, \[ D(F^{-1})(F(a))=[DF(a)]^{-1}. \]

The nonzero determinant is the condition that makes \(DF(a)\) an invertible matrix. The theorem is local: it guarantees a well-behaved inverse on neighborhoods, not necessarily a single inverse on all of \(U\). If a map is already known to be globally one-to-one, and its Jacobian determinant is nonzero throughout its open domain, the local inverses agree with its global inverse wherever they overlap.

Why the Jacobian Determinants Are Reciprocal

The map followed by its inverse leaves each point unchanged. Differentiating that identity with the multivariable chain rule gives a matrix identity. Taking determinants then produces the reciprocal relationship.

1
Write the identity.
On the neighborhoods where the inverse is defined, \(F^{-1}(F(x))=x\).
2
Apply the chain rule.
Differentiating at \(x\) gives \(D(F^{-1})(F(x))DF(x)=I\), where \(I\) is the \(n\times n\) identity matrix.
3
Take determinants.
The determinant of a product is the product of determinants, and \(\det I=1\).

In full, the last step gives

$$ \det D(F^{-1})(F(x))\,\det DF(x)=1. $$

Because \(\det DF(x)\ne 0\), division by that determinant is valid. Thus

$$ \det D(F^{-1})(F(x)) = \frac{1}{\det DF(x)}. $$

This proves the reciprocal formula. The matrices are inverses, not merely matrices whose determinants happen to be related. The determinant formula follows from that stronger matrix identity.

Worked Example: A Linear Transformation

Worked Example: A Linear Transformation

Consider \(F(x,y)=(2x+y,x+2y)\). Write its output coordinates as \((u,v)\). Find the Jacobian determinants of \(F\) and \(F^{-1}\), and check the inverse relation.

The Jacobian matrix and determinant of \(F\) are

$$ DF(x,y)= \begin{pmatrix} 2&1\\ 1&2 \end{pmatrix}, \qquad \det DF(x,y)=2\cdot2-1\cdot1=3. $$

Since the determinant is nonzero, solve for \(x\) and \(y\) in terms of \(u\) and \(v\). The equations are \(u=2x+y\) and \(v=x+2y\). Subtracting twice the second equation from the first gives \(u-2v=-3y\), so \(y=(-u+2v)/3\). Subtracting the first equation from twice the second gives \(2v-u=3y\); to find \(x\), use \(2u-v=3x\), so \(x=(2u-v)/3\). Therefore

$$ F^{-1}(u,v)= \left(\frac{2u-v}{3},\frac{-u+2v}{3}\right). $$

Differentiating the inverse gives

$$ D(F^{-1})(u,v)= \begin{pmatrix} 2/3&-1/3\\ -1/3&2/3 \end{pmatrix}, \qquad \det D(F^{-1})(u,v)=\frac{4}{9}-\frac{1}{9}=\frac{1}{3}. $$

The reciprocal formula predicts \(1/3\), since the forward determinant is \(3\). To check the inverse itself, compose the maps:

$$ \begin{aligned} F(F^{-1}(u,v)) &=\left( 2\frac{2u-v}{3}+\frac{-u+2v}{3}, \frac{2u-v}{3}+2\frac{-u+2v}{3} \right)\\ &=(u,v). \end{aligned} $$

The derivative matrices also multiply to the identity:

$$ \begin{pmatrix} 2&1\\1&2 \end{pmatrix} \begin{pmatrix} 2/3&-1/3\\-1/3&2/3 \end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}. $$

Worked Example: A Nonlinear Transformation

Worked Example: A Nonlinear Transformation

Let \(F:\mathbb{R}^2\to (0,\infty)\times\mathbb{R}\) be \(F(x,y)=(e^x,y)\). Find both Jacobian determinants and verify the reciprocal relationship.

The Jacobian matrix of \(F\) is

$$ DF(x,y)= \begin{pmatrix} e^x&0\\ 0&1 \end{pmatrix}, \qquad \det DF(x,y)=e^x. $$

To invert the map, set \(u=e^x\) and \(v=y\). Since \(u>0\), these equations give \(x=\ln u\) and \(y=v\). Hence

$$ F^{-1}(u,v)=(\ln u,v). $$

Its Jacobian matrix and determinant are

$$ D(F^{-1})(u,v)= \begin{pmatrix} 1/u&0\\ 0&1 \end{pmatrix}, \qquad \det D(F^{-1})(u,v)=\frac{1}{u}. $$

At the corresponding point \(F(x,y)=(e^x,y)\), the inverse determinant is \(1/e^x\), exactly \(1/\det DF(x,y)\). As a composition check,

$$ F^{-1}(F(x,y))=F^{-1}(e^x,y)=(\ln(e^x),y)=(x,y). $$

In derivative form, the matrices multiply to the identity:

$$ \begin{pmatrix}1/e^x&0\\0&1\end{pmatrix} \begin{pmatrix}e^x&0\\0&1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}. $$

Worked Example: A Nonlinear Map with Two Mixed Coordinates

Worked Example: A Nonlinear Map with Two Mixed Coordinates

On the domain \(U=\{(x,y)\in\mathbb{R}^2:x>y\}\), define \(F(x,y)=(x+y,xy)=(u,v)\). Find the inverse and its Jacobian determinant, and compare it with the forward determinant.

The forward Jacobian matrix is

$$ DF(x,y)= \begin{pmatrix} 1&1\\ y&x \end{pmatrix}, \qquad \det DF(x,y)=x-y. $$

Because \(x>y\), this determinant is positive and in particular nonzero. From \(u=x+y\) and \(v=xy\), the numbers \(x\) and \(y\) are the roots of \(t^2-ut+v=0\). Define \(s=\sqrt{u^2-4v}\). On the image of \(U\), \(s>0\), and the condition \(x>y\) selects

$$ F^{-1}(u,v)= \left(\frac{u+s}{2},\frac{u-s}{2}\right). $$

Since \(s_u=u/s\) and \(s_v=-2/s\), its partial derivatives are

$$ x_u=\frac{1+u/s}{2},\quad x_v=-\frac{1}{s}, \qquad y_u=\frac{1-u/s}{2},\quad y_v=\frac{1}{s}. $$

Thus the inverse determinant is

$$ \begin{aligned} \det D(F^{-1})(u,v) &=x_u y_v-x_v y_u\\ &=\frac{1+u/s}{2s} +\frac{1-u/s}{2s}\\ &=\frac{1}{s}. \end{aligned} $$

At the corresponding point, \(s=x-y\), because \(u^2-4v=(x+y)^2-4xy=(x-y)^2\) and \(x-y>0\). Therefore \(\det D(F^{-1})(F(x,y))=1/(x-y)=1/\det DF(x,y)\). Finally, substituting the inverse formulas gives \(x+y=u\) and \(xy=(u^2-s^2)/4=v\), since \(s^2=u^2-4v\). The inverse and determinant checks both agree.

Keep Track of the Evaluation Point

The reciprocal statement compares determinants at corresponding points:

$$ x\in U,\qquad (u,v,\ldots)=F(x),\qquad \det D(F^{-1})(u,v,\ldots)=\frac{1}{\det DF(x)}. $$

For a nonlinear transformation, the determinant may vary from point to point. The inverse determinant is a function of the transformed coordinates, while the forward determinant is a function of the original coordinates. They are related by evaluating them at matching points, not by substituting the same tuple into both formulas.

Common mistake: Treating the two coordinate tuples as if they were interchangeable can give the wrong value. For \(F(x,y)=(e^x,y)\), take \((x,y)=(\ln 2,0)\). The image is \((u,v)=(2,0)\), so \(\det D(F^{-1})(2,0)=1/2\). Evaluating the inverse determinant formula \(1/u\) at the original tuple \((\ln 2,0)\) instead gives \(1/\ln 2\), which is not \(1/2\). The correction is to evaluate the inverse determinant at \(F(x,y)\), the image point.

When the Reciprocal Formula Applies

The reciprocal formula requires an inverse that is differentiable at the corresponding point. The inverse function theorem supplies such a local inverse when \(F\) is continuously differentiable near the point and its Jacobian determinant there is nonzero. If the determinant is zero, the theorem does not guarantee a differentiable local inverse, and the reciprocal formula cannot be justified by this argument.

A nonzero determinant at one point guarantees a local inverse near that point, not global one-to-one behavior on the entire domain. For a globally defined inverse, global injectivity must be established separately. Conversely, if a differentiable inverse exists and the chain rule applies, differentiating the composition forces the derivative matrices to be inverses and hence forces the determinants to be nonzero.

These relationships are useful whenever a coordinate map is easier to differentiate in one direction than the other. One may calculate the determinant for the simpler map, then obtain the inverse determinant by taking its reciprocal at the corresponding point. The chain-rule identity supplies a direct check that the calculation is consistent.

Check Your Understanding

Use the inverse function theorem and the chain-rule relationship to answer these questions.

  1. What condition on \(DF(a)\) ensures that the inverse function theorem gives a differentiable local inverse near \(a\)?
  2. If \(\det DF(x)=-4\), what is \(\det D(F^{-1})(F(x))\), assuming the inverse exists and is differentiable there?
  3. For \(F(x,y)=(e^x,y)\), what is the inverse determinant at the image of \((0,3)\)?
  4. Why must the inverse determinant be evaluated at \(F(x)\) in the reciprocal formula?
  5. Does a nonzero Jacobian determinant at one point establish that \(F\) is one-to-one on its entire domain? Explain briefly.

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