Jacobian matrix: Difference between revisions

Latest revision as of 02:32, 13 May 2012

This article describes an analogue for functions of multiple variables of the following term/fact/notion for functions of one variable: derivative

Importance

The Jacobian matrix is the appropriate notion of derivative for a function that has multiple inputs (or equivalently, vector-valued inputs) and multiple outputs (or equivalently, vector-valued outputs).

Definition at a point

Direct epsilon-delta definition

Definition at a point in terms of gradient vectors as row vectors

Suppose $f$ is a vector-valued function with $n$ -dimensional inputs and $m$ -dimensional outputs. Explicitly, suppose $f$ is a function with inputs $x_{1},x_{2},\dots ,x_{n}$ and outputs $f_{1}(x_{1},x_{2},\dots ,x_{n}),f_{2}(x_{1},x_{2},\dots ,x_{n}),\dots ,f_{m}(x_{1},x_{2},\dots ,x_{n})$ . Suppose $(a_{1},a_{2},\dots ,a_{n})$ is a point in the domain of $f$ such that $f_{i}$ is differentiable at $(a_{1},a_{2},\dots ,a_{n})$ for $i\in \{1,2,\dots ,m\}$ . Then, the Jacobian matrix of $f$ at $(a_{1},a_{2},\dots ,a_{n})$ is a $m\times n$ matrix of numbers whose $i^{th}$ row is given by the gradient vector of $f_{i}$ at $(a_{1},a_{2},\dots ,a_{n})$ .

Explicitly, in terms of rows, it looks like:

${\begin{pmatrix}\nabla (f_{1})(a_{1},a_{2},\dots ,a_{n})\\\nabla (f_{2})(a_{1},a_{2},\dots ,a_{n})\\\cdot \\\cdot \\\cdot \\\nabla (f_{m})(a_{1},a_{2},\dots ,a_{n})\\\end{pmatrix}}$

Definition at a point in terms of partial derivatives

Suppose $f$ is a vector-valued function with $n$ -dimensional inputs and $m$ -dimensional outputs. Explicitly, suppose $f$ is a function with inputs $x_{1},x_{2},\dots ,x_{n}$ and outputs $f_{1}(x_{1},x_{2},\dots ,x_{n}),f_{2}(x_{1},x_{2},\dots ,x_{n}),\dots ,f_{m}(x_{1},x_{2},\dots ,x_{n})$ . Suppose $(a_{1},a_{2},\dots ,a_{n})$ is a point in the domain of $f$ such that $f_{i}$ is differentiable at $(a_{1},a_{2},\dots ,a_{n})$ for $i\in \{1,2,\dots ,m\}$ . Then, the Jacobian matrix of $f$ at $(a_{1},a_{2},\dots ,a_{n})$ is a $m\times n$ matrix of numbers whose $(ij)^{th}$ entry is given by:

${\frac {\partial f_{i}}{\partial x_{j}}}(x_{1},x_{2},\dots ,x_{n})|_{(x_{1},x_{2},\dots ,x_{n})=(a_{1},a_{2},\dots ,a_{n})}$

Here's how the matrix looks:

${\begin{pmatrix}{\frac {\partial f_{1}}{\partial x_{1}}}(x_{1},x_{2},\dots ,x_{n})|_{(x_{1},x_{2},\dots ,x_{n})=(a_{1},a_{2},\dots ,a_{n})}&{\frac {\partial f_{1}}{\partial x_{2}}}(x_{1},x_{2},\dots ,x_{n})|_{(x_{1},x_{2},\dots ,x_{n})=(a_{1},a_{2},\dots ,a_{n})}&\dots &{\frac {\partial f_{1}}{\partial x_{n}}}(x_{1},x_{2},\dots ,x_{n})|_{(x_{1},x_{2},\dots ,x_{n})=(a_{1},a_{2},\dots ,a_{n})}\\{\frac {\partial f_{2}}{\partial x_{1}}}(x_{1},x_{2},\dots ,x_{n})|_{(x_{1},x_{2},\dots ,x_{n})=(a_{1},a_{2},\dots ,a_{n})}&{\frac {\partial f_{2}}{\partial x_{2}}}(x_{1},x_{2},\dots ,x_{n})|_{(x_{1},x_{2},\dots ,x_{n})=(a_{1},a_{2},\dots ,a_{n})}&\dots &{\frac {\partial f_{2}}{\partial x_{n}}}(x_{1},x_{2},\dots ,x_{n})|_{(x_{1},x_{2},\dots ,x_{n})=(a_{1},a_{2},\dots ,a_{n})}\\\cdot &\cdot &\cdot &\cdot \\{\frac {\partial f_{m}}{\partial x_{1}}}(x_{1},x_{2},\dots ,x_{n})|_{(x_{1},x_{2},\dots ,x_{n})=(a_{1},a_{2},\dots ,a_{n})}&{\frac {\partial f_{m}}{\partial x_{2}}}(x_{1},x_{2},\dots ,x_{n})|_{(x_{1},x_{2},\dots ,x_{n})=(a_{1},a_{2},\dots ,a_{n})}&\dots &{\frac {\partial f_{m}}{\partial x_{n}}}(x_{1},x_{2},\dots ,x_{n})|_{(x_{1},x_{2},\dots ,x_{n})=(a_{1},a_{2},\dots ,a_{n})}\\\end{pmatrix}}$

Note that for this definition to be correct, it is still necessary that the gradient vectors exist. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.

Definition as a function

Definition in terms of gradient vectors as row vectors

Suppose $f$ is a vector-valued function with $n$ -dimensional inputs and $m$ -dimensional outputs. Explicitly, suppose $f$ is a function with inputs $x_{1},x_{2},\dots ,x_{n}$ and outputs $f_{1}(x_{1},x_{2},\dots ,x_{n}),f_{2}(x_{1},x_{2},\dots ,x_{n}),\dots ,f_{m}(x_{1},x_{2},\dots ,x_{n})$ . Then, the Jacobian matrix of $f$ is a $m\times n$ matrix of functions whose $i^{th}$ row is given by the gradient vector of $f_{i}$ . Explicitly, it looks like this:

${\begin{pmatrix}\nabla (f_{1})\\\nabla (f_{2})\\\cdot \\\cdot \\\cdot \\\nabla (f_{m})\\\end{pmatrix}}$

Note that the domain of this function is the set of points at which all the $f_{i}$ s individually are differentiable.

Definition in terms of partial derivatives

Suppose $f$ is a vector-valued function with $n$ -dimensional inputs and $m$ -dimensional outputs. Explicitly, suppose $f$ is a function with inputs $x_{1},x_{2},\dots ,x_{n}$ and outputs $f_{1}(x_{1},x_{2},\dots ,x_{n}),f_{2}(x_{1},x_{2},\dots ,x_{n}),\dots ,f_{m}(x_{1},x_{2},\dots ,x_{n})$ . Then, the Jacobian matrix of $f$ is a $m\times n$ matrix of functions whose $(ij)^{th}$ entry is given by:

${\frac {\partial f_{i}}{\partial x_{j}}}(x_{1},x_{2},\dots ,x_{n})$

wherever all the $f_{i}$ s individually are differentiable in the sense of the gradient vectors existing. Here's how the matrix looks:

${\begin{pmatrix}{\frac {\partial f_{1}}{\partial x_{1}}}&{\frac {\partial f_{1}}{\partial x_{2}}}&\dots &{\frac {\partial f_{1}}{\partial x_{n}}}\\{\frac {\partial f_{2}}{\partial x_{1}}}&{\frac {\partial f_{2}}{\partial x_{2}}}&\dots &{\frac {\partial f_{2}}{\partial x_{n}}}\\\cdot &\cdot &\cdot &\cdot \\{\frac {\partial f_{m}}{\partial x_{1}}}&{\frac {\partial f_{m}}{\partial x_{2}}}&\dots &{\frac {\partial f_{m}}{\partial x_{n}}}\\\end{pmatrix}}$

If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.

Particular cases

Case	What happens in that case?
$m=n=1$	$f$ is a real-valued function of one variable. The Jacobian matrix is a $1\times 1$ matrix whose entry is the ordinary derivative.
$n=1$ , $m>1$	$f$ is a vector-valued function of one variable. We can think of it as a parametric curve in $\mathbb {R} ^{m}$ . The Jacobian matrix is a $m\times 1$ matrix which, read as a column vector, is the parametric derivative of the vector-valued function.
$m=1$ , $n>1$	$f$ is a real-valued function of multiple variables. The Jacobian matrix is a $1\times n$ matrix which, read as a row vector, is the gradient vector function.
$f$ is a linear or affine map.	The Jacobian matrix is the same as the matrix describing $f$ (or, if $f$ is affine, the matrix describing the linear part of $f$ ).
$m=n$ , and we are identifying the spaces of inputs and outputs of $f$ .	The Jacobian matrix can then be thought of as a linear self-map from the $n$ -dimensional space to itself. In this context, we can consider the Jacobian determinant.

@@ Line 11: / Line 11: @@
 ===Definition at a point in terms of gradient vectors as row vectors===
-Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix'''of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math> at <math>(a_1,a_2,\dots,a_n)</math>.
+Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix''' of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math> at <math>(a_1,a_2,\dots,a_n)</math>.
+Explicitly, in terms of rows, it looks like:
+<math>\begin{pmatrix} \nabla(f_1)(a_1,a_2,\dots,a_n) \\ \nabla(f_2)(a_1,a_2,\dots,a_n) \\ \cdot \\ \cdot \\ \cdot \\ \nabla(f_m)(a_1,a_2,\dots,a_n) \\\end{pmatrix}</math>
+<center>{{#widget:YouTube|id=O8isoxng_9g}}</center>
 ===Definition at a point in terms of partial derivatives===
-Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix'''of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>(ij)^{th}</math> entry is given by:
+Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix''' of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>(ij)^{th}</math> entry is given by:
 <math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)}</math>
+Here's how the matrix looks:
+<math>\begin{pmatrix} \frac{\partial f_1}{\partial x_1}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \frac{\partial f_1}{\partial x_2}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \dots & \frac{\partial f_1}{\partial x_n}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} \\
+\frac{\partial f_2}{\partial x_1}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \frac{\partial f_2}{\partial x_2}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \dots & \frac{\partial f_2}{\partial x_n}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} \\
+\cdot & \cdot & \cdot & \cdot \\
+\frac{\partial f_m}{\partial x_1}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \frac{\partial f_m}{\partial x_2}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \dots & \frac{\partial f_m}{\partial x_n}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} \\\end{pmatrix}</math>
 Note that for this definition to be correct, it is still necessary that the gradient vectors exist. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.
+<center>{{#widget:YouTube|id=VCM4RVM09_I}}</center>
 ==Definition as a function==
@@ Line 25: / Line 40: @@
 ===Definition in terms of gradient vectors as row vectors===
-Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix'''of <math>f</math>is a <math>m \times n</math> matrix of ''functions'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math>.
+Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix''' of <math>f</math>is a <math>m \times n</math> matrix of ''functions'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math>. Explicitly, it looks like this:
+<math>\begin{pmatrix} \nabla(f_1) \\ \nabla(f_2)\\ \cdot \\ \cdot \\ \cdot \\ \nabla(f_m) \\\end{pmatrix}</math>
 Note that the domain of this function is the set of points at which all the <math>f_i</math>s individually are differentiable.
@@ Line 31: / Line 49: @@
 ===Definition in terms of partial derivatives===
-Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix'''of <math>f</math> is a <math>m \times n</math> matrix of ''functions'' whose <math>(ij)^{th}</math> entry is given by:
+Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix''' of <math>f</math> is a <math>m \times n</math> matrix of ''functions'' whose <math>(ij)^{th}</math> entry is given by:
 <math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)</math>
-wherever all the <math>f_i</math>s individually are differentiable in the sense of the gradient vectors existing. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.
+wherever all the <math>f_i</math>s individually are differentiable in the sense of the gradient vectors existing. Here's how the matrix looks:
+<math>\begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2}& \dots & \frac{\partial f_1}{\partial x_n}\\
+\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \dots & \frac{\partial f_2}{\partial x_n}\\
+\cdot & \cdot & \cdot & \cdot \\
+\frac{\partial f_m}{\partial x_1} & \frac{\partial f_m}{\partial x_2} & \dots & \frac{\partial f_m}{\partial x_n}\\\end{pmatrix}</math>
+If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.
+<center>{{#widget:YouTube|id=jTmwUMnuUec}}</center>
 ==Particular cases==