A Jacobian matrix , sometimes simply called a Jacobian , is a matrix of first order partial derivatives (in some cases, the term "Jacobian" also refers to the determinant of the Jacobian matrix).
For a function
f
:
R
n
→
R
m
{\displaystyle \mathbf f:\R^n\to\R^m}
, the Jacobian is the following
m
×
n
{\displaystyle m \times n}
matrix:
J
=
∂
(
u
1
,
…
,
u
m
)
∂
(
x
1
,
…
,
x
n
)
=
[
∂
u
1
∂
x
1
⋯
∂
u
1
∂
x
n
⋮
⋱
⋮
∂
u
m
∂
x
1
⋯
∂
u
m
∂
x
n
]
{\displaystyle \mathbf J=\frac{\part(u_1,\ldots,u_m)}{\part(x_1,\ldots,x_n)}=
\begin{bmatrix}
\dfrac{\part u_1}{\part x_1}&\cdots&\dfrac{\part u_1}{\part x_n}\\
\vdots&\ddots &\vdots\\
\dfrac{\part u_m}{\part x_1}&\cdots&\dfrac{\part u_m}{\part x_n}
\end{bmatrix}}
or, in Einstein notation ,
J
i
j
=
∂
j
f
i
{\displaystyle \mathbf J_i^j=\part^j f_i}
Note that in some conventions, the Jacobian is the transpose of the above matrix.
Jacobians where
m
=
n
{\displaystyle m= n}
are square matrices, and are commonly used when changing coordinates , especially when taking multiple integrals and determining whether complex functions are holomorphic . For example, a Jacobian representing a change in variables from
x
{\displaystyle x}
to
x
(
u
,
v
)
{\displaystyle x(u, v)}
and
y
{\displaystyle y}
to
y
(
u
,
v
)
{\displaystyle y(u, v)}
in two dimensions is represented as
J
=
∂
(
u
,
v
)
∂
(
x
,
y
)
=
[
∂
u
∂
x
∂
u
∂
y
∂
v
∂
x
∂
v
∂
y
]
{\displaystyle \mathbf J=\frac{\part(u,v)}{\part(x,y)}=
\begin{bmatrix}
\dfrac{\part u}{\part x}&\dfrac{\part u}{\part y}\\\\
\dfrac{\part v}{ \part x}&\dfrac{\part v}{\part y}
\end{bmatrix}}
A Jacobian matrix is what is usually meant by the derivative of higher-dimensional functions; indeed, differentiability in the components of a Jacobian guarantees differentiability in the function itself. In the case of a multivariable function
f
:
R
N
→
R
{\displaystyle f:\R^N\to\R}
, the Jacobian matrix with respect to the input variables is simply the gradient of the function. The Jacobian is also related to the Hessian matrix by
H
(
f
)
=
J
(
∇
f
)
{\displaystyle \mathbf{H}(f)=\mathbf{J}(\nabla f)}
Applications [ ]
Jacobian matrices are useful in integration when changing coordinate systems. For example, given a two dimensional coordinate transformation, the double integral of
f
(
x
,
y
)
{\displaystyle f(x, y)}
becomes
∬
D
f
(
x
,
y
)
d
x
d
y
=
∬
D
′
f
(
x
(
u
,
v
)
,
y
(
u
,
v
)
)
|
∂
(
x
,
y
)
∂
(
u
,
v
)
|
d
u
d
v
=
∬
D
′
f
(
x
(
u
,
v
)
,
y
(
u
,
v
)
)
|
∂
(
u
,
v
)
∂
(
x
,
y
)
|
−
1
d
u
d
v
{\displaystyle \begin{align}\iint_D f(x,y)\,dx\,dy&=\iint_{D'}f\big(x(u,v),y (u,v)\big)\left|\dfrac{\part(x,y)}{\part(u,v)}\right|\,du\,dv\\
&=\iint_{D'}f\big(x(u,v),y(u,v)\big)\left|\dfrac{\part(u,v)}{\part(x,y)}\right|^{-1}\,du\,dv\end{align}}
When working with one independent variable, this becomes
|
∂
u
∂
x
|
=
|
d
x
d
u
|
−
1
=
d
u
d
x
{\displaystyle \left|\dfrac{\part u}{\part x}\right|=\left|\dfrac{dx}{du}\right|^{-1}=\dfrac{du}{dx}}
which, when used to compute an integral, yields the formula known as integration by substitution .
Examples [ ]
Jacobian matrices are useful in integration when changing coordinate systems. For example, given a two dimensional coordinate transformation, the double integral of
f
(
x
,
y
)
{\displaystyle f(x, y)}
becomes
∬
D
f
(
x
,
y
)
d
x
d
y
=
∬
D
′
f
(
x
(
u
,
v
)
,
y
(
u
,
v
)
)
|
∂
(
x
,
y
)
∂
(
u
,
v
)
|
d
u
d
v
=
∬
D
′
f
(
x
(
u
,
v
)
,
y
(
u
,
v
)
)
|
∂
(
u
,
v
)
∂
(
x
,
y
)
|
−
1
d
u
d
v
{\displaystyle \begin{align}\iint_D f(x,y)\,dx\,dy&=\iint_{D'}f\big(x(u,v),y (u,v)\big)\left|\dfrac{\part(x,y)}{\part(u,v)}\right|\,du\,dv\\
&=\iint_{D'}f\big(x(u,v),y(u,v)\big)\left|\dfrac{\part(u,v)}{\part(x,y)}\right|^{-1}\,du\,dv\end{align}}
When working with one independent variable, this becomes
|
∂
u
∂
x
|
=
|
d
x
d
u
|
−
1
=
d
u
d
x
{\displaystyle \left|\dfrac{\part u}{\part x}\right|=\left|\dfrac{dx}{du}\right|^{-1}=\dfrac{du}{dx}}
which, when used to compute an integral, yields the formula known as integration by substitution .
Find the area of a circle of radius
a
{\displaystyle a}
by transforming from Cartesian coordinates to Polar coordinates .
A
=
∬
D
d
A
=
∬
D
d
x
d
y
=
∬
D
′
|
∂
(
x
,
y
)
∂
(
r
,
θ
)
|
d
x
d
y
{\displaystyle A=\iint_D dA=\iint_D dx\,dy=\iint_{D'}\left|\dfrac{\part(x,y)}{\part(r,\theta)}\right|\,dx\,dy}
Since
x
=
r
cos
(
θ
)
{\displaystyle x=r\cos(\theta)}
and
y
=
r
sin
(
θ
)
{\displaystyle y=r\sin(\theta)}
, the Jacobian determinant becomes
|
∂
(
x
,
y
)
∂
(
r
,
θ
)
|
=
|
cos
(
θ
)
−
r
sin
(
θ
)
sin
(
θ
)
r
cos
(
θ
)
|
=
(
cos
(
θ
)
)
(
r
cos
(
θ
)
)
−
(
sin
(
θ
)
)
(
−
r
sin
(
θ
)
)
=
r
(
cos
2
(
θ
)
+
sin
2
(
θ
)
)
=
r
{\displaystyle \begin{align}
\left|\dfrac{\part(x,y)}{\part(r,\theta)}\right|=
\begin{vmatrix}\cos(\theta)&-r\sin(\theta)\\\sin(\theta)&r\cos(\theta)\end{vmatrix}
&=\big(\cos(\theta)\big)\big(r\cos(\theta)\big)-\big(\sin(\theta)\big)\big(-r\sin(\theta)\big)\\
&=r\big(\cos^2(\theta)+\sin^2(\theta)\big)=r
\end{align}}
The integral now becomes
∬
D
′
r
d
r
d
θ
{\displaystyle \iint_{D'}r\,dr\,d\theta}
Since it is now in polar coordinates, we can add the bounds as
∫
0
2
π
∫
0
a
r
d
r
d
θ
{\displaystyle \int\limits_0^{2\pi}\int\limits_0^a r\,dr\,d\theta}
Now we can integrate it.
∫
0
2
π
∫
0
a
r
d
r
d
θ
=
∫
0
2
π
a
2
2
d
θ
=
π
a
2
{\displaystyle \int\limits_0^{2\pi}\int\limits_0^a r\,dr\,d\theta=\int\limits_0^{2\pi}\frac{a^2}{2}d\theta=\pi a^2}
Which is to be expected. We have also found that the differential of
d
A
{\displaystyle dA}
is
r
d
r
d
θ
{\displaystyle r\,dr\,d\theta}
.
This same method can be used to find the volume of a sphere in spherical coordinates . Since
x
=
r
sin
(
ϕ
)
cos
(
θ
)
y
=
r
sin
(
ϕ
)
sin
(
θ
)
z
=
r
cos
(
ϕ
)
{\displaystyle \begin{align}
x&=r\sin(\phi)\cos(\theta)\\y&=r\sin(\phi)\sin(\theta)\\z&=r\cos(\phi)
\end{align}}
the Jacobian determinant evaluates to:
|
∂
(
x
,
y
,
z
)
∂
(
r
,
ϕ
,
θ
)
|
=
|
sin
(
ϕ
)
cos
(
θ
)
r
cos
(
ϕ
)
cos
(
θ
)
−
r
sin
(
ϕ
)
sin
(
θ
)
sin
(
ϕ
)
sin
(
θ
)
r
cos
(
ϕ
)
sin
(
θ
)
r
sin
(
ϕ
)
cos
(
θ
)
cos
(
ϕ
)
−
r
sin
(
ϕ
)
0
|
=
r
2
sin
(
ϕ
)
{\displaystyle \left|\dfrac{\part(x,y,z)}{\part(r,\phi,\theta)}\right|=
\begin{vmatrix}
\sin(\phi)\cos(\theta)&r\cos(\phi)\cos(\theta)&-r\sin(\phi)\sin(\theta)\\
\sin(\phi)\sin(\theta)&r\cos(\phi)\sin(\theta)&r\sin(\phi)\cos(\theta)\\
\cos(\phi)&-r\sin(\phi)&0
\end{vmatrix}=
r^2\sin(\phi)}
This means that for any function in spherical coordinates, the volume element is
d
V
=
r
2
sin
(
ϕ
)
d
r
d
θ
d
ϕ
{\displaystyle dV=r^2\sin(\phi)\,dr\,d\theta\,d\phi}
The integral is therefore
∭
V
d
V
=
∫
0
π
∫
0
2
π
∫
0
a
r
2
sin
(
ϕ
)
d
r
d
θ
d
ϕ
=
∫
0
π
sin
(
ϕ
)
d
ϕ
∫
0
2
π
d
θ
∫
0
a
r
2
d
r
=
(
2
)
(
2
π
)
a
3
3
=
4
3
π
a
3
{\displaystyle \begin{align}\iiint_V dV&=\int\limits_0^\pi\int\limits_0^{2\pi}\int\limits_0^a r^2\sin(\phi)\,dr\,d\theta\,d\phi=\int\limits_0^\pi\sin(\phi)\,d\phi\int\limits_0^{2\pi}d\theta\int\limits_0^a r^2dr\\
&=(2)(2\pi)\frac{a^3}{3}=\frac43\pi a^3\end{align}}