📂다변수벡터해석

벡터와 행렬의 도함수 표

스칼라 함수의 그래디언트

스칼라 함수 $f : \mathbb{R}^{n} \to \mathbb{R}$의 그래디언트는 다음과 같다. $$\frac{ \partial f(\mathbf{x})}{ \partial \mathbf{x} } := \nabla f(\mathbf{x}) = \begin{bmatrix} \dfrac{ \partial f(\mathbf{x})}{ \partial x_{1} } & \dfrac{ \partial f(\mathbf{x})}{ \partial x_{2} } & \cdots & \dfrac{ \partial f(\mathbf{x})}{ \partial x_{n} } \end{bmatrix}^{T}$$

여기서 $\dfrac{ \partial f(\mathbf{x})}{ \partial x_{i} }$는 $f$의 $x_{i}$에 대한 편도함수이다.

내적

고정된 $\mathbf{w} \in \mathbb{R}^{n}$에 대해서, $f(\mathbf{x}) = \left\langle \mathbf{w}, \mathbf{x} \right\rangle = \mathbf{w}^{T}\mathbf{x}$라고 하면,

$$\frac{ \partial f(\mathbf{x})}{ \partial \mathbf{x} } = \frac{ \partial (\mathbf{w}^{T}\mathbf{x})}{ \partial \mathbf{x} } = \frac{ \partial (\mathbf{x}^{T}\mathbf{w})}{ \partial \mathbf{x} } = \mathbf{w}$$

놈

$f(\mathbf{x}) = \left\| \mathbf{x} \right\|^{2} = \mathbf{x}^{T} \mathbf{x}$이라 하면,

$$\nabla f(\mathbf{x}) = \dfrac{\partial \left\| \mathbf{x} \right\|^{2}}{\partial \mathbf{x}} = 2\mathbf{x}$$

이차형식

$n \times n$ 행렬 $\mathbf{R}$에 대해서, $f(\mathbf{x}) = \mathbf{x}^{T}\mathbf{R}\mathbf{x}$라 하면,

$$\dfrac{\partial f(\mathbf{x})}{\partial \mathbf{x}} = \dfrac{\partial (\mathbf{x}\mathbf{R}\mathbf{x})}{\partial \mathbf{x}} = (\mathbf{R} + \mathbf{R}^{T})\mathbf{x}$$

증명

내적

$$\begin{align*} \frac{ \partial f(\mathbf{x})}{ \partial \mathbf{x} } =\frac{ \partial (\mathbf{w}^{T}\mathbf{x})}{ \partial \mathbf{x} } &=\begin{bmatrix} \dfrac{ \partial \left( \sum _{i=1} ^{n} w_{i}x_{i}\right)}{ \partial x_{1} } & \dfrac{ \partial \left( \sum _{i=1} ^{n} w_{i}x_{i}\right)}{ \partial x_{2} } & \cdots & \dfrac{ \partial \left( \sum _{i=1} ^{n} w_{i}x_{i}\right)}{ \partial x_{n} } \end{bmatrix}^{T} \\ &= \begin{bmatrix} w_{1} & w_{2} & \cdots & w_{n} \end{bmatrix}^{T} \\ &= \mathbf{w} \end{align*}$$

또한 $\mathbf{w}^{T}\mathbf{x} = \mathbf{x}^{T}\mathbf{w}$이므로,

$$\frac{ \partial \mathbf{x}^{T}\mathbf{w}}{ \partial \mathbf{x} } = \mathbf{w}$$

■

놈

내적에 대한 증명과 마찬가지로,

$$\begin{align*} \frac{ \partial f(\mathbf{x})}{ \partial \mathbf{x} } = \frac{ \partial (\mathbf{x}^{T}\mathbf{x})}{ \partial \mathbf{x} } &=\begin{bmatrix} \dfrac{ \partial \left( \sum _{i=1} ^{n} x_{i}x_{i}\right)}{ \partial x_{1} } & \dfrac{ \partial \left( \sum _{i=1} ^{n} x_{i}x_{i}\right)}{ \partial x_{2} } & \cdots & \dfrac{ \partial \left( \sum _{i=1} ^{n} x_{i}x_{i}\right)}{ \partial x_{n} } \end{bmatrix}^{T} \\ &=\begin{bmatrix} \dfrac{ \partial \left( \sum _{i=1} ^{n} x_{i}^{2}\right)}{ \partial x_{1} } & \dfrac{ \partial \left( \sum _{i=1} ^{n} x_{i}^{2}\right)}{ \partial x_{2} } & \cdots & \dfrac{ \partial \left( \sum _{i=1} ^{n} x_{i}^{2}\right)}{ \partial x_{n} } \end{bmatrix}^{T} \\ &= \begin{bmatrix} 2x_{1} & 2x_{2} & \cdots & 2x_{n} \end{bmatrix}^{T} \\ &= 2\mathbf{x} \end{align*}$$

■

이차형식

미분 계산이 쉽도록 먼저 계산을 다음과 같이 해준다. 임의의 $k\in\left\{1,\dots,n\right\}$에 대해서 다음의 식을 얻는다.

$$\begin{align*} f(\mathbf{x}) &= \mathbf{x}^{T}\mathbf{R}\mathbf{x} =\sum \limits _{j=1} ^{n} x_{j} \sum \limits _{i=1} ^{n}r_{ji}x_{i} \\ & = x_{k} \sum \limits _{i=1} ^{n}r_{ki}x_{i} +\sum \limits _{j\ne k}x_{j} \sum \limits _{i=1} ^{n}r_{ji}x_{i} \\ &= x_{k}\left(r_{kk}x_{k} + \sum \limits _{i\ne k} r_{ki}x_{i}\right) +\sum \limits _{j\ne k}x_{j} \left(r_{jk}x_{k}+ \sum \limits _{i \ne k} r_{ji}x_{i} \right) \\ &= x_{k}^{2}r_{kk} + x_{k}\sum \limits _{i\ne k} r_{ki}x_{i} + \sum \limits _{j\ne k}x_{j}r_{jk}x_{k}+ \sum \limits _{j\ne k}\sum \limits _{i \ne k}x_{j} r_{ji}x_{i} \end{align*}$$

$\dfrac{ \partial f(\mathbf{x})}{ \partial x_{k}}$를 계산하면 다음과 같다.

$$\begin{align*} \frac{ \partial f(\mathbf{x})}{ \partial x_{k} }&=\frac{ \partial }{ \partial x_{k} } \left( x_{k}^{2}r_{kk} + x_{k}\sum \limits _{i\ne k} r_{ki}x_{i} + \sum \limits _{j\ne k}x_{j}r_{jk}x_{k}+ \sum \limits _{j\ne k}\sum \limits _{i \ne k}x_{j} r_{ji}x_{i} \right) \\ &=2w_{k}r_{kk} + \sum \limits _{i\ne k} r_{ki}x_{i} + \sum \limits _{j\ne k}x_{j}r_{jk} \\ &=\sum \limits _{i=1}^{n} r_{ki}x_{i} + \sum \limits _{j=1}^{n}r_{jk}x_{j} \end{align*}$$

따라서 $f$의 그래디언트를 계산하면 다음과 같다.

$$\begin{align*} \frac{ \partial f(\mathbf{x})}{ \partial \mathbf{x} } = \dfrac{\partial \left( \mathbf{x}^{T}\mathbf{R}\mathbf{x} \right)}{\partial \mathbf{x}} &= \begin{bmatrix} \dfrac{ \partial f (\mathbf{x})}{ \partial x_{1} } & \dfrac{ \partial f (\mathbf{x})}{ \partial x_{2} } & \dots & \dfrac{ \partial f (\mathbf{x})}{ \partial x_{n} } \end{bmatrix}^{T} \\[1em] &= \begin{bmatrix} \sum _{i=1} ^{n} r_{1i}x_{i} + \sum _{j=1}^{n} r_{j1}x_{j} \\[0.5em] \sum _{i=1} ^{n} r_{2i}x_{i} + \sum _{j=1}^{n} r_{j2}x_{j} \\[0.5em] \vdots \\[0.5em] \sum _{i=1} ^{n} r_{ni}x_{i} + \sum _{j=1}^{n} r_{jn}x_{j} \end{bmatrix} \\[3.5em] &= \begin{bmatrix} \sum _{i=1} ^{n} r_{1i}x_{i} \\[0.5em] \sum _{i=1} ^{n} r_{2i}x_{i} \\[0.5em] \vdots \\[0.5em] \sum _{i=1} ^{n} r_{ni}x_{i} \end{bmatrix} + \begin{bmatrix} \sum _{j=1}^{n} r_{j1}x_{j} \\[0.5em] \sum _{j=1}^{n} r_{j2}x_{j} \\[0.5em] \vdots \\[0.5em] \sum _{j=1}^{n} r_{jn}x_{j} \end{bmatrix} \\[3.5em] &=\mathbf{R} \mathbf{x} + \mathbf{R}^{T}\mathbf{x}=\left( \mathbf{R}+\mathbf{R}^{T} \right)\mathbf{x} \end{align*}$$

행렬-벡터 곱

$\mathbf{X} \in M_{n\times n}$, $\mathbf{y} \in M_{n \times 1}$에 대해서,

$$\mathbf{X}\mathbf{y} =\begin{bmatrix} \sum _{i=1} ^{n}x_{1i}y_{i} \\ \sum _{i=1} ^{n}x_{2i}y_{i} \\ \vdots \\ \sum _{i=1} ^{n}x_{ni}y_{i} \end{bmatrix},\qquad \mathbf{X}^{T}\mathbf{y} =\begin{bmatrix} \sum _{i=1} ^{n}x_{i1}y_{i} \\ \sum _{i=1} ^{n}x_{i2}y_{i} \\ \vdots \\ \sum _{i=1} ^{n}x_{in}y_{i} \end{bmatrix}$$

만약 $\mathbf{R}$이 대칭행렬이면,

$$\frac{ \partial }{ \partial \mathbf{x} }\left( \mathbf{x}^{T}\mathbf{R}\mathbf{x} \right)=2\mathbf{R}\mathbf{x}$$