在向量分析 中，雅可比矩阵 （也称作Jacobi矩陣 ，英語：Jacobian matrix 函數 的一阶偏导数 以一定方式排列成的矩阵 。
當其為方形矩阵時，其行列式 称为雅可比行列式 （Jacobi determinant）。要注意的是，在英文中雅可比矩陣跟雅可比行列式都可稱作Jacobian 。[ 1]
其重要性在於，如果函數 f : ℝn m x x f x 微分 或者導數 。
在代数几何 中，代数曲线 的雅可比行列式' 表示雅可比簇 代數群 ，曲线可以嵌入其中。
它们全部都以普魯士 数学家 卡爾·雅可比 命名。
假設某函數從 f : ℝn m x ∈ ℝn f (x ) ∈ ℝm m ×n ℝn ℝm
此函數 f J m ×n
J
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∂
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∂
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∂
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=
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∂
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{\displaystyle \mathbf {J} ={\begin{bmatrix}{\dfrac {\partial \mathbf {f} }{\partial x_{1}}}&\cdots &{\dfrac {\partial \mathbf {f} }{\partial x_{n}}}\end{bmatrix}}={\begin{bmatrix}{\dfrac {\partial f_{1}}{\partial x_{1}}}&\cdots &{\dfrac {\partial f_{1}}{\partial x_{n}}}\\\vdots &\ddots &\vdots \\{\dfrac {\partial f_{m}}{\partial x_{1}}}&\cdots &{\dfrac {\partial f_{m}}{\partial x_{n}}}\end{bmatrix}}}
矩陣的分量可表示成：
J
i
j
=
∂
f
i
∂
x
j
{\displaystyle \mathbf {J} _{ij}={\frac {\partial f_{i}}{\partial x_{j}}}}
雅可比矩陣的其他常用符號還有：
D
f
{\displaystyle Df}
D
f
{\displaystyle \mathrm {D} \mathbf {f} }
J
f
(
x
1
,
…
,
x
n
)
{\displaystyle \mathbf {J} _{\mathbf {f} }(x_{1},\ldots ,x_{n})}
∂
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f
1
,
…
,
f
m
)
∂
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x
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,
…
,
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{\displaystyle {\frac {\partial (f_{1},\ldots ,f_{m})}{\partial (x_{1},\ldots ,x_{n})}}}
此矩陣的第
i
{\displaystyle i}
f
i
{\displaystyle f_{i}}
1
≤
i
≤
m
{\displaystyle 1\leq i\leq m}
如果
p
{\displaystyle p}
R
n
{\displaystyle \mathbb {R} ^{n}}
f
{\displaystyle f}
p
{\displaystyle p}
數學分析 ，
J
f
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p
)
{\displaystyle \mathbf {J} _{\mathbf {f} }(p)}
导数 。在此情况下，
J
f
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p
)
{\displaystyle \mathbf {J} _{\mathbf {f} }(p)}
f
{\displaystyle f}
p
{\displaystyle p}
x
{\displaystyle x}
p
{\displaystyle p}
f
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)
≈
f
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+
J
f
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⋅
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x
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p
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{\displaystyle f(x)\approx f(p)+\mathbf {J} _{\mathbf {f} }(p)\cdot (x-p)}
講更詳細點也就是：
f
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+
J
f
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(
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o
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‖
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{\displaystyle \mathbf {f} (\mathbf {x} )=\mathbf {f} (\mathbf {p} )+\mathbf {J} _{\mathbf {f} }(\mathbf {p} )(\mathbf {x} -\mathbf {p} )+o(\|\mathbf {x} -\mathbf {p} \|)}
其中，o 小o符號 ，‖x − p ‖ x p
由球坐标系 到直角坐标系的转化由 F : ℝ+ × [0, π ) × [0, 2π ) → ℝ3
x
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r
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θ
cos
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y
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z
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{\displaystyle {\begin{aligned}x&=r\sin \theta \cos \varphi ;\\y&=r\sin \theta \sin \varphi ;\\z&=r\cos \theta \end{aligned}}}
此坐标变换的雅可比矩阵是
J
F
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=
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sin
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sin
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sin
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r
cos
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sin
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r
sin
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cos
φ
cos
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−
r
sin
θ
0
]
{\displaystyle \mathbf {J} _{\mathbf {F} }(r,\theta ,\varphi )={\begin{bmatrix}{\dfrac {\partial x}{\partial r}}&{\dfrac {\partial x}{\partial \theta }}&{\dfrac {\partial x}{\partial \varphi }}\\[1em]{\dfrac {\partial y}{\partial r}}&{\dfrac {\partial y}{\partial \theta }}&{\dfrac {\partial y}{\partial \varphi }}\\[1em]{\dfrac {\partial z}{\partial r}}&{\dfrac {\partial z}{\partial \theta }}&{\dfrac {\partial z}{\partial \varphi }}\end{bmatrix}}={\begin{bmatrix}\sin \theta \cos \varphi &r\cos \theta \cos \varphi &-r\sin \theta \sin \varphi \\\sin \theta \sin \varphi &r\cos \theta \sin \varphi &r\sin \theta \cos \varphi \\\cos \theta &-r\sin \theta &0\end{bmatrix}}}
其雅可比行列式為 r 2 sin θ dV = dx dy dz dV dV = r 2 sin θ dr dθ dφ
F : ℝ3 → ℝ4
y
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x
1
{\displaystyle y_{1}=x_{1}\,}
y
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5
x
3
{\displaystyle y_{2}=5x_{3}\,}
y
3
=
4
x
2
2
−
2
x
3
{\displaystyle y_{3}=4x_{2}^{2}-2x_{3}\,}
y
4
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x
3
sin
x
1
{\displaystyle y_{4}=x_{3}\sin x_{1}\,}
其雅可比矩阵为:
J
F
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,
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,
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=
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∂
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∂
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∂
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∂
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∂
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=
[
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5
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2
−
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3
cos
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1
0
sin
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{\displaystyle J_{F}(x_{1},x_{2},x_{3})={\begin{bmatrix}{\frac {\partial y_{1}}{\partial x_{1}}}&{\frac {\partial y_{1}}{\partial x_{2}}}&{\frac {\partial y_{1}}{\partial x_{3}}}\\[3pt]{\frac {\partial y_{2}}{\partial x_{1}}}&{\frac {\partial y_{2}}{\partial x_{2}}}&{\frac {\partial y_{2}}{\partial x_{3}}}\\[3pt]{\frac {\partial y_{3}}{\partial x_{1}}}&{\frac {\partial y_{3}}{\partial x_{2}}}&{\frac {\partial y_{3}}{\partial x_{3}}}\\[3pt]{\frac {\partial y_{4}}{\partial x_{1}}}&{\frac {\partial y_{4}}{\partial x_{2}}}&{\frac {\partial y_{4}}{\partial x_{3}}}\\\end{bmatrix}}={\begin{bmatrix}1&0&0\\0&0&5\\0&8x_{2}&-2\\x_{3}\cos x_{1}&0&\sin x_{1}\end{bmatrix}}}
此例子说明雅可比矩阵不一定为方阵。
如果 m = n F ℝn ℝn 方陣 。于是我们可以取它的行列式，称为雅可比行列式 。
在某个给定点的雅可比行列式提供了 F F p F 反函数 。这称为反函数定理 。更进一步，如果 p 正数 ，则 F p F 绝对值 ，就可以知道函数 F p 换元积分法 中的原因。
设有函数 F : ℝ3 → ℝ3
y
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{\displaystyle y_{1}=5x_{2}\,}
y
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−
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sin
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{\displaystyle y_{2}=4x_{1}^{2}-2\sin(x_{2}x_{3})\,}
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{\displaystyle y_{3}=x_{2}x_{3}\,}
则它的Jacobi行列式为：
|
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0
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40
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{\displaystyle {\begin{vmatrix}0&5&0\\8x_{1}&-2x_{3}\cos(x_{2}x_{3})&-2x_{2}\cos(x_{2}x_{3})\\0&x_{3}&x_{2}\end{vmatrix}}=-8x_{1}\cdot {\begin{vmatrix}5&0\\x_{3}&x_{2}\end{vmatrix}}=-40x_{1}x_{2}}
从中我们可以看到，當 x 1 x 2 F x 1 = 0x 2 = 0
这是一个与巴塞尔问题
∑
n
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1
∞
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{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}
级数
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{\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(2n+1)^{2}}}={\frac {\pi ^{2}}{8}}}
二重积分 （在这里 D 1 指 x 与 y 皆为从 0 到 1 的正方形区域）：
∑
n
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∞
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2
n
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{\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(2n+1)^{2}}}=\iint \limits _{D_{1}}\sum _{n=1}^{\infty }(xy)^{2n}\mathrm {d} x\mathrm {d} y=\iint \limits _{D_{1}}{\frac {\mathrm {d} x\mathrm {d} y}{1-x^{2}y^{2}}}}
此时定义映射 F : ℝ2 → ℝ2
{
u
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arctan
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⟺
{
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{\displaystyle {\begin{cases}u=\arctan \left(x{\sqrt {\dfrac {1-y^{2}}{1-x^{2}}}}\right)\\v=\arctan \left(y{\sqrt {\dfrac {1-x^{2}}{1-y^{2}}}}\right)\end{cases}}\iff {\begin{cases}x={\dfrac {\sin u}{\cos v}}\\y={\dfrac {\sin v}{\cos u}}\end{cases}}}
于是有相应的雅可比行列式：
|
∂
x
∂
u
∂
x
∂
v
∂
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∂
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∂
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∂
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=
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cos
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cos
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sin
u
sin
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cos
2
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sin
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sin
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cos
2
u
cos
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cos
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−
sin
2
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sin
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v
cos
2
u
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2
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1
−
x
2
y
2
{\displaystyle {\begin{vmatrix}{\dfrac {\partial x}{\partial u}}&{\dfrac {\partial x}{\partial v}}\\{\dfrac {\partial y}{\partial u}}&{\dfrac {\partial y}{\partial v}}\end{vmatrix}}={\begin{vmatrix}{\dfrac {\cos u}{\cos v}}&{\dfrac {\sin u\sin v}{\cos ^{2}v}}\\{\dfrac {\sin u\sin v}{\cos ^{2}u}}&{\dfrac {\cos v}{\cos u}}\end{vmatrix}}=1-{\frac {\sin ^{2}u\sin ^{2}v}{\cos ^{2}u\cos ^{2}v}}=1-x^{2}y^{2}}
因此
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{\displaystyle \mathrm {d} x\mathrm {d} y=(1-x^{2}y^{2})\mathrm {d} u\mathrm {d} v}
D 1 映射成 u >0、v >0、u +v <π/2 的等腰直角三角形，记为 D 2 ，得到：
∬
D
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d
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−
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∬
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∫
0
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∫
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π
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−
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d
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π
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8
{\displaystyle \iint \limits _{D_{1}}{\frac {\mathrm {d} x\mathrm {d} y}{1-x^{2}y^{2}}}=\iint \limits _{D_{2}}\mathrm {d} u\mathrm {d} v=\int _{0}^{\frac {\pi }{2}}\left(\int _{0}^{{\frac {\pi }{2}}-v}\mathrm {d} u\right)\mathrm {d} v={\frac {\pi ^{2}}{8}}}
^ W., Weisstein, Eric. Jacobian . mathworld.wolfram.com. [2 May 2018] . （原始内容存档 于3 November 2017）.
![{\displaystyle \mathbf {J} _{\mathbf {F} }(r,\theta ,\varphi )={\begin{bmatrix}{\dfrac {\partial x}{\partial r}}&{\dfrac {\partial x}{\partial \theta }}&{\dfrac {\partial x}{\partial \varphi }}\\[1em]{\dfrac {\partial y}{\partial r}}&{\dfrac {\partial y}{\partial \theta }}&{\dfrac {\partial y}{\partial \varphi }}\\[1em]{\dfrac {\partial z}{\partial r}}&{\dfrac {\partial z}{\partial \theta }}&{\dfrac {\partial z}{\partial \varphi }}\end{bmatrix}}={\begin{bmatrix}\sin \theta \cos \varphi &r\cos \theta \cos \varphi &-r\sin \theta \sin \varphi \\\sin \theta \sin \varphi &r\cos \theta \sin \varphi &r\sin \theta \cos \varphi \\\cos \theta &-r\sin \theta &0\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db70522712b56ee256247603ef074d20313e5ba1)
![{\displaystyle J_{F}(x_{1},x_{2},x_{3})={\begin{bmatrix}{\frac {\partial y_{1}}{\partial x_{1}}}&{\frac {\partial y_{1}}{\partial x_{2}}}&{\frac {\partial y_{1}}{\partial x_{3}}}\\[3pt]{\frac {\partial y_{2}}{\partial x_{1}}}&{\frac {\partial y_{2}}{\partial x_{2}}}&{\frac {\partial y_{2}}{\partial x_{3}}}\\[3pt]{\frac {\partial y_{3}}{\partial x_{1}}}&{\frac {\partial y_{3}}{\partial x_{2}}}&{\frac {\partial y_{3}}{\partial x_{3}}}\\[3pt]{\frac {\partial y_{4}}{\partial x_{1}}}&{\frac {\partial y_{4}}{\partial x_{2}}}&{\frac {\partial y_{4}}{\partial x_{3}}}\\\end{bmatrix}}={\begin{bmatrix}1&0&0\\0&0&5\\0&8x_{2}&-2\\x_{3}\cos x_{1}&0&\sin x_{1}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebf91f57907e1656c70decff4f02a0acf8fbdacf)
