002-矩阵求导

2021-01-20  本文已影响0人  不懂球的2大业

对于\frac {df(x)} {dx},分以下三种情况讨论:
1.f(x)为标量,x为向量;
2.f(x)为向量,x为标量;
3.f(x)为向量,x为标量。

1.f(x)为标量,x为向量

其中,x = (x_{1},x_{2},...,x_{n})^{T}f(x) = f(x_{1},x_{2},...,x_{n})
则:\frac {df(x)} {dx} = \begin{bmatrix} \frac {\partial {f(x)}} {x_{1}} \\ \frac {\partial {f(x)}} {x_{2}} \\...\\ \frac {\partial {f(x)}} {x_{n}} \end{bmatrix}

2.f(x)为矢量,x为标量

其中,f(x) = (f_{1}(x),f_{2}(x),...,f_{n}(x))
则:\frac {df(x)} {dx} = \begin{bmatrix} \frac {df_{1}(x)} {dx} , \frac {df_{2}(x)} {dx} , ..., \frac {df_{n}(x)} {dx}\\ \end{bmatrix}

3.f(x)为矢量,x为矢量

其中,x = (x_{1},x_{2},...,x_{n})^{T}f(x) = [f_{1}(x_{1},x_{2},...,x_{n}),f_{2}(x_{1},x_{2},...,x_{n}),...,f_{m}(x_{1},x_{2},...,x_{n})]
则:\frac {df(x)} {dx} = \begin{pmatrix} \frac {\partial {f_{1}(x)}}{\partial {x_{1}}}&\frac {\partial {f_{2}(x)}}{\partial {x_{1}}}&\frac {\partial {f_{3}(x)}}{\partial {x_{1}}}&\cdots&\frac {\partial {f_{m}(x)}}{\partial {x_{1}}}\\ \frac {\partial {f_{1}(x)}}{\partial {x_{2}}}& \frac {\partial {f_{2}(x)}}{\partial {x_{2}}} &\frac {\partial {f_{3}(x)}}{\partial {x_{2}}}&\cdots&\frac {\partial {f_{m}(x)}}{\partial {x_{2}}}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \frac {\partial {f_{1}(x)}}{\partial {x_{n}}}&\frac {\partial {f_{2}(x)}}{\partial {x_{n}}}&\frac {\partial {f_{3}(x)}}{\partial {x_{n}}}&\cdots&\frac {\partial {f_{m}(x)}}{\partial {x_{n}}}\\ \end{pmatrix}

4.公式

(1)\frac {d A^{T}X} {dX} = \frac {d X^{T}A} {dX} = A
(2)\frac {dX^{T}AX} {dX} = (A+A^{T})X

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