What is the Jacobian matrix?

What are its applications?

What is its physical and geometrical meaning?

Can someone please explain with examples?

Rodrigo de Azevedo
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Pratik Deoghare
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    It's the derivative of a vector-valued function. Anyway, what in [here](http://mathworld.wolfram.com/Jacobian.html) or [here](http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant) don't you understand? – J. M. ain't a mathematician Dec 20 '10 at 13:29
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    I have read both the articles but its still unclear to me. I would like to see Jacobian in action. Why was it invented in the first place? – Pratik Deoghare Dec 20 '10 at 13:32
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    One instance of its use is in generalizing Newton-Raphson to $n$ equations in $n$ unknowns. Letting $\mathbf J(\mathbf x)$ be the Jacobian of the vector-valued function $\mathbf f(\mathbf x)$, Newton-Raphson for $\mathbf f(\mathbf x)=\mathbf 0$ goes like $\mathbf x_{i+1}=\mathbf x_i-\mathbf J(\mathbf x_i)^{-1}\mathbf f(\mathbf x_i)$. (Yes, that's a matrix inverse.) – J. M. ain't a mathematician Dec 20 '10 at 13:38
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    look here: http://www.youtube.com/watch?v=mR-WDILxx24 –  Mar 15 '12 at 19:15
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    Here is the link for Jacobian Matrix explanation: http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant – Spider Oct 30 '13 at 10:08

9 Answers9


The Jacobian $df_p$ of a differentiable function $f : \mathbb{R}^n \to \mathbb{R}^m$ at a point $p$ is its best linear approximation at $p$, in the sense that $f(p + h) = f(p) + df_p(h) + o(|h|)$ for small $h$. This is the "correct" generalization of the derivative of a function $f : \mathbb{R} \to \mathbb{R}$, and everything we can do with derivatives we can also do with Jacobians.

In particular, when $n = m$, the determinant of the Jacobian at a point $p$ is the factor by which $f$ locally dilates volumes around $p$ (since $f$ acts locally like the linear transformation $df_p$, which dilates volumes by $\det df_p$). This is the reason that the Jacobian appears in the change of variables formula for multivariate integrals, which is perhaps the basic reason to care about the Jacobian. For example this is how one changes an integral in rectangular coordinates to cylindrical or spherical coordinates.

The Jacobian specializes to the most important constructions in multivariable calculus. It immediately specializes to the gradient, for example. When $n = m$ its trace is the divergence. And a more complicated construction gives the curl. The rank of the Jacobian is also an important local invariant of $f$; it roughly measures how "degenerate" or "singular" $f$ is at $p$. This is the reason the Jacobian appears in the statement of the implicit function theorem, which is a fundamental result with applications everywhere.

Qiaochu Yuan
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    This is the explanation I was looking for, general and abstract and to the point! What is the interpretation of a singular $f$ in terms of the Jacobian? What happens to the volume change? – user578 Oct 28 '17 at 18:26
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    @Ahmad: if $df_p$ is singular this means the volume is being multiplied by zero, so a higher-dimensional thing is being squashed into a lower-dimensional thing. – Qiaochu Yuan Oct 28 '17 at 18:57
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    in some textbook I find that the *jacobian* is the determinant of the *jacobian matrix*, so the use of the term jacobian alone must cause confusion. – Masacroso Dec 02 '17 at 01:24
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    Yeah, it's pretty bad, and "Jacobian" has at least one other meaning in mathematics as well. Nowadays I would just call $df_p$ the derivative. – Qiaochu Yuan Dec 02 '17 at 01:30

Here is an example. Suppose you have two implicit differentiable functions

$$F(x,y,z,u,v)=0,\qquad G(x,y,z,u,v)=0$$

and the functions, also differentiable, $u=f(x,y,z)$ and $v=g(x,y,z)$ such that

$$F(x,y,z,f(x,y,z),g(x,y,z))=0,\qquad G(x,y,z,f(x,y,z),g(x,y,z))=0.$$

If you differentiate $F$ and $G$, you get

\begin{eqnarray*} \frac{\partial F}{\partial x}+\frac{\partial F}{\partial u}\frac{\partial u}{ \partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial x} &=&0\qquad \\ \frac{\partial G}{\partial x}+\frac{\partial G}{\partial u}\frac{\partial u}{ \partial x}+\frac{\partial G}{\partial v}\frac{\partial v}{\partial x} &=&0. \end{eqnarray*}

Solving this system you obtain

$$\frac{\partial u}{\partial x}=-\frac{\det \begin{pmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial v} \\ \frac{\partial G}{\partial x} & \frac{\partial G}{\partial v} \end{pmatrix}}{\det \begin{pmatrix} \frac{\partial F}{\partial u} & \frac{\partial F}{\partial v} \\ \frac{\partial G}{\partial u} & \frac{\partial G}{\partial v} \end{pmatrix}}$$

and similar for $\dfrac{\partial u}{\partial y}$, $\dfrac{\partial u}{\partial z}$, $\dfrac{\partial v}{\partial x}$, $\dfrac{\partial v}{\partial y}$, $% \dfrac{\partial v}{\partial z}$. The compact notation for the denominator is

$$\frac{\partial (F,G)}{\partial (u,v)}=\det \begin{pmatrix} \frac{\partial F}{\partial u} & \frac{\partial F}{\partial v} \\ \frac{\partial G}{\partial u} & \frac{\partial G}{\partial v} \end{pmatrix}$$

and similar for the numerator. Then

$$\dfrac{\partial u}{\partial x}=-\dfrac{\dfrac{\partial (F,G)}{\partial (x,v)}}{% \dfrac{\partial (F,G)}{\partial (u,v)}}$$

where $\dfrac{\partial (F,G)}{\partial (x,y)},\dfrac{\partial (F,G)}{\partial (u,v)}$ are Jacobians (after the 19th century German mathematician Carl Jacobi).

The absolute value of the Jacobian of a coordinate system transformation is also used to convert a multiple integral from one system into another. In $\mathbb{R}^2$ it measures how much the unit area is distorted by the given transformation, and in $\mathbb{R}^3$ this factor measures the unit volume distortion, etc.

Another example: the following coordinate transformation (due to Beukers, Calabi and Kolk)

$$x=\frac{\sin u}{\cos v}$$

$$y=\frac{\sin v}{\cos u}$$

transforms (see this question of mine) the square domain $0\lt x\lt 1$ and $0\lt y\lt 1$ into the triangle domain $u,v>0,u+v<\pi /2$ (in Proofs from the BOOK by M. Aigner and G. Ziegler).

For this transformation you get (see Proof 2 in this collection of proofs by Robin Chapman)

$$\dfrac{\partial (x,y)}{\partial (u,v)}=1-x^2y^{2}.$$

Jacobian sign and orientation of closed curves. Assume you have two small closed curves, one around $(x_0,y_0)$ and another around $u_0,v_0$, this one being the image of the first under the mapping $u=f(x,y),v=g(x,y)$. If the sign of $\dfrac{\partial (x,y)}{\partial (u,v)}$ is positive, then both curves will be travelled in the same sense. If the sign is negative, they will have opposite senses. (See Oriented Regions and their Orientation.)

Américo Tavares
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    This answer is a lengthy application of Jacobian (derivative) matrix's concept, instead, the Qiaochu Yuan's answer aside offers less confusion. – janmarqz Jan 17 '14 at 23:09
  • this question is mixing the concepts of *jacobian* and *jacobian matrix*, where the first is the determinant of the second. – Masacroso Dec 02 '17 at 01:22
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    Sir, I have a question. When we see that the determinant of a Jacobian matrix is 1 then the area is preserved by the map (not necessarily linear). But when we cannot compute the determinant of the Jacobian matrix (non-square) what are the ways of determining if the area is not preserved? – MathCosmo Jul 23 '18 at 14:11

In single variable calculus, if $f:\mathbb R \to \mathbb R$, then \begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}. \end{equation} A very useful way to think about $f'(x)$ is this: \begin{equation} \tag{$\spadesuit$} f(x + \Delta x) \approx f(x) + f'(x) \Delta x. \end{equation}

One of the advantages of equation $(\spadesuit)$ is that it still makes perfect sense in the case where $f:\mathbb R^n \to \mathbb R^m$:

\begin{equation} f(\underbrace{x}_{n \times 1} + \underbrace{\Delta x}_{n\times 1}) \approx \underbrace{f(x)}_{m \times 1} + \underbrace{f'(x)}_{?} \underbrace{\Delta x}_{n \times 1}. \end{equation} You see, if $f'(x)$ is now an $m \times n$ matrix, then this equation makes perfect sense. So, with this idea, we can extend the idea of the derivative to the case where $f:\mathbb R^n \to \mathbb R^m$. This is the first step towards developing calculus in a multivariable setting. The matrix $f'(x)$ is called the "Jacobian" of $f$ at $x$, but maybe it's more clear to simply call $f'(x)$ the derivative of $f$ at $x$.

The matrix $f'(x)$ allows us to approximate $f$ locally by a linear function (or, technically, an "affine" function). Linear functions are simple enough that we can understand them well (using linear algebra), and often understanding the local linear approximation to $f$ at $x$ allows us to draw conclusions about $f$ itself.

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    I found your explanation very clear! Thank you :) – valentin Apr 24 '17 at 12:39
  • I don't understand, In this interpretation, why do we take then the absolute value of the Jacobian (i.e. of f'(x))? a derivative can be negative. – blz Apr 05 '18 at 13:48
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    @blz The multivariable change of variables formula for integration involves the quantity $| \det f'(x) |$. But there are many situations throughout calculus where the matrix $f'(x)$ is useful by itself, without considering its determinant. One potential source of confusion is that the term "Jacobian" sometimes refers to the matrix $f'(x)$, and sometimes refers to the determinant $\det f'(x)$. Hopefully the meaning is always clear from context. Note that $f'(x)$ is often a rectangular matrix, in which case it does not have a determinant. – littleO Apr 05 '18 at 21:28
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    @blz By the way, I wrote an explanation of the multivariable change of variables formula for integration here: https://math.stackexchange.com/a/464972/40119 – littleO Apr 05 '18 at 21:30
  • From the above, it seems that we can take the Jacobian also for a matrix with respect another matrix. Whereas in most places it is written that Jacobian is only valid for Vector over matrix. Can you please clarify – Alex Punnen Mar 05 '22 at 01:58
  • @AlexPunnen I'm not sure what you mean by "take the Jacobian also for a matrix with respect to another matrix." In my answer here, the function $f$ takes a vector as input and returns a vector as output. Do you mean that if we have a function $g$ which takes a matrix as input and returns a matrix as output then we could compute the 'Jacobian' of $g$? My equation ($\spadesuit$) does suggest a way to define the "derivative" of such a function $g$. Namely, $g'(x)$ is a linear transformation $L:\mathbb R^{m \times n} \to \mathbb R^{p \times q}$ such that $g(x+\Delta x) \approx g(x) + L(\Delta x)$. – littleO Mar 05 '22 at 06:37
  • @littleO I have framed it as a new question here with proper context https://math.stackexchange.com/questions/4397390/jacobian-matrix-of-an-element-wise-operation-on-a-matrix – Alex Punnen Mar 06 '22 at 17:52

(I know this is slightly late, but I think the OP may appreciate this)

As an application, in the field of control engineering the use of Jacobian matrices allows the local (approximate) linearisation of non-linear systems around a given equilibrium point and so allows the use of linear systems techniques, such as the calculation of eigenvalues (and thus allows an indication of the type of the equilibrium point).

Jacobians are also used in the estimation of the internal states of non-linear systems in the construction of the extended Kalman filter, and also if the extended Kalman filter is to be used to provide joint state and parameter estimates for a linear system (since this is a non-linear system analysis due to the products of what are then effectively inputs and outputs of the system).

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I found the most beautiful usage of jacobian matrices in studying differential geometry, when one abandons the idea that analysis can be done "only on balls of $\mathbb{R}^n$". The definition of tangent space in a point $p$ of a manifold $M$ can be given via the kernel of the jacobian of a suitable submersion, or via the image of the differential of a suitable immersion from an open set $U\subseteq\mathbb{R}^{\dim M}$. Quite a simple example, but when I was an undergrad four years ago it gave me the "right" idea of what a linear transformation does in a differential (analytical) framework.

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This is not a rigorous explanation, but here is the best intuitive explanation/motivation for the Jacobian Matrix. Start with an interval $[x_1,x_2] \subset \mathbb{R}$. What is a common measurement of space for this interval ? It is length. To find the length of $[x_1,x_2]$, take $x_2-x_1$. Now suppose I define an invertible linear transformation $T:\mathbb{R} \rightarrow \mathbb{R}$, where $$T(x)=\begin{bmatrix}a\end{bmatrix}x,$$ where $\begin{bmatrix}a\end{bmatrix}$ is a $1\times 1$ matrix with a nonzero entry $a$. The image of $[x_1,x_2]$ under $T$ is the interval $[ax_1,ax_2]$, and the length of this new interval is $ax_2-ax_1=a(x_2-x_1)$. Now we ask ourselves this question. How does the length of the new interval relate to the length of the old interval ? The length of the $[ax_1,ax_2]$ is $|a|$ times the length of $[x_1,x_2]$. But notice that: $$|a|=\left |\det\begin{bmatrix}a\end{bmatrix}\right |.$$ Now suppose you are doing u substitution to evaluate an integral in the form $$\int_{S} f(x) dx.$$ We define $x=x(u)$ and the differential $dx$ becomes $\frac{dx}{du}du$. If you view $dx$ and $du$ as vectors in $\mathbb{R}$, you get $$dx=\begin{bmatrix}\frac{dx}{du}\end{bmatrix}du.$$ The determinant of $\begin{bmatrix}\frac{dx}{du}\end{bmatrix}$ plays the same role as $a$ in that it is a scaling factor between different "infinitesimal" interval lengths.

The higher dimensional analogue of the interval in $\mathbb{R}$ is a parallelepiped in $\mathbb{R}^n$. Measurement of space in $\mathbb{R}^n$ is the $n$-dimensional volume. If you define an invertible linear transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$, and if you write $T(x)=Ax$, where $A$ is an $n \times n$ matrix, the absolute value of $\det A$, scales the volume of a parallelepiped. Similarly, if you are dealing with the multidimensional integral: $$\int_{S}f(x_1,...,x_n)dx_1...dx_n$$ and wish to use change of variables: $$x_i=x_i(u_1,...,u_n),1 \leq i \leq n$$ you can regard $dx=(dx_1,...,dx_n),du=(du_1,...,du_n)$ as vectors in $\mathbb{R}^n$ and relate them by $$dx=\begin{bmatrix}\frac{\partial x_i}{\partial u_j}\end{bmatrix}_{ij}du.$$ The Jacobian Matrix here is: $$\begin{bmatrix}\frac{\partial x_i}{\partial u_j}\end{bmatrix}_{ij},$$ and the notation means the $i$th row and $j$th column entry is $\frac{\partial x_i}{\partial u_j}$. The absolute value of the determinant of the Jacobian Matrix is a scaling factor between different "infinitesimal" parallelepiped volumes. Again, this explanation is merely intuitive. It is not rigorous as one would present it in a real analysis course.

Vivek Kaushik
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A very short contribution for the applicability question: it is a matrix of partial derivatives. One of the applications is to find local solutions of a system of nonlinear equations. When you have a system of nonlinear equation, the x`s that are solutions of the system are not easy to find, because it is difficult to invert a the matrix of nonlinear coefficients of the system. However, you can take the partial derivative of the equations, find the local linear approximation near some value, and then solve the system. Because the system becomes locally linear, you can solve it using linear algebra.

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I don't know much about this, but I know is used in programming robotics for transforming between two frame of references. The equations become very simple. So moving from one frame to another to another is just the matrix product of Jacobian matrix.

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The simplest answer I can give is - Jacobian Matrix is used when there is a change of variable requirement in the greater than one dimensional space.

One of the explanations above explains it simplistically in the single variable concept.

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