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행렬 미분소 📂다변수벡터해석

행렬 미분소

정의

n×nn \times n 행렬 X=[xij]a\mathbf{X} = [x_{ij}] a의 미분소 dX\mathrm{d}\mathbf{X}는 다음과 같이 정의된다.

dX=[dx11dx12⋯dx1ndx21dx22⋯dx2n⋮⋮⋱⋮dxn1dxn2⋯dxnn] \mathrm{d} \mathbf{X} = \begin{bmatrix} \mathrm{d} x_{11} & \mathrm{d} x_{12} & \cdots & \mathrm{d} x_{1n} \\ \mathrm{d} x_{21} & \mathrm{d} x_{22} & \cdots & \mathrm{d} x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ \mathrm{d} x_{n1} & \mathrm{d} x_{n2} & \cdots & \mathrm{d} x_{nn} \end{bmatrix}

설명

일변수 함수의 미분소 dx\mathrm{d}x를 벡터로 일반화 한 것이 dx\mathrm{d}\mathbf{x}라면, 이를 행렬로 일반화한 것이 dX\mathrm{d}\mathbf{X}이다.

일변수함수 f:R→Rf: \mathbb{R} \to \mathbb{R}의 전미분:

df=dfdxdx \mathrm{d}f = \dfrac{\mathrm{d}f}{\mathrm{d}x} \mathrm{d}x

다변수함수 f:Rn→Rf: \mathbb{R}^{n} \to \mathbb{R}의 전미분: df=∑i∂f∂xidxi=⟨∇xf,dx⟩=(∇xf)Tdx=[∂f∂x1⋯∂f∂xn][dx1⋮dxn] \mathrm{d}f = \sum_{i} \dfrac{\partial f}{\partial x_{i}} \mathrm{d}x_{i} = \Braket{\nabla_{\mathbf{x}}f, \mathrm{d}\mathbf{x}} = (\nabla_{\mathbf{x}}f)^{\mathsf{T}} \mathrm{d}\mathbf{x} = \begin{bmatrix} \dfrac{\partial f}{\partial x_{1}} & \cdots & \dfrac{\partial f}{\partial x_{n}} \end{bmatrix} \begin{bmatrix} \mathrm{d}x_{1} \\ \vdots \\ \mathrm{d}x_{n} \end{bmatrix}

행렬변수함수function of a matrix f:Rn×n→Rf: \mathbb{R}^{n \times n} \to \mathbb{R}의 전미분: df=∑i,j∂f∂xijdxij=⟨∇Xf,dX⟩=Tr⁡((∇Xf)TdX) \mathrm{d}f = \sum\limits_{i,j} \dfrac{\partial f}{\partial x_{ij}} \mathrm{d}x_{ij} = \Braket{\nabla_{\mathbf{X}}f, \mathrm{d}\mathbf{X}} = \Tr \left( \left( \nabla_{\mathbf{X}}f \right)^{\mathsf{T}} \mathrm{d}\mathbf{X} \right)

스칼라의 미분소 dx\mathrm{d}x에 대한 성질이 그대로 유지된다.

d(ax)=adxd(x+y)=dx+dyd(xy)=ydx+xdy \mathrm{d}(ax) = a \mathrm{d}x \qquad \mathrm{d}(x + y) = \mathrm{d}x + \mathrm{d}y \qquad \mathrm{d}(xy) = y\mathrm{d}x + x\mathrm{d}y

성질

변수인 행렬 X,Y∈Rn×n\mathbf{X}, \mathbf{Y} \in \mathbb{R}^{n \times n}와 스칼라 α∈R\alpha \in \mathbb{R}, 상수 행렬 A∈Rn×n\mathbf{A} \in \mathbb{R}^{n \times n}에 대해서 다음이 성립한다.

  1. d(αX)=αdX\mathrm{d}(\alpha \mathbf{X}) = \alpha \mathrm{d}\mathbf{X}
  2. d(XT)=(dX)T\mathrm{d}(\mathbf{X}^{\mathsf{T}}) = (\mathrm{d}\mathbf{X})^{\mathsf{T}}
  3. d(AX)=AdX\mathrm{d}(\mathbf{A}\mathbf{X}) = \mathbf{A} \mathrm{d}\mathbf{X} 그리고 d(XA)=(dX)A\mathrm{d}(\mathbf{X}\mathbf{A}) = (\mathrm{d}\mathbf{X}) \mathbf{A}
  4. d(X+Y)=dX+dY\mathrm{d}(\mathbf{X} + \mathbf{Y}) = \mathrm{d}\mathbf{X} + \mathrm{d}\mathbf{Y}
  5. d(XY)=(dX)Y+XdY\mathrm{d}(\mathbf{X}\mathbf{Y}) = (\mathrm{d}\mathbf{X})\mathbf{Y} + \mathbf{X} \mathrm{d}\mathbf{Y}

증명

3.

[AX]ij=∑k=1naikxkj[\mathbf{A}\mathbf{X}]_{ij} = \sum_{k=1}^{n} a_{ik} x_{kj}이므로,

[d(AX)]ij=d(∑k=1naikxkj)=∑k=1naik(dxkj) [\mathrm{d}(\mathbf{A}\mathbf{X})]_{ij} = \mathrm{d}\left( \sum_{k=1}^{n} a_{ik} x_{kj} \right) = \sum_{k=1}^{n} a_{ik}(\mathrm{d}x_{kj})

  ⟹  d(AX)=AdX \implies \mathrm{d}(\mathbf{A}\mathbf{X}) = \mathbf{A}\mathrm{d}\mathbf{X}

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5.

[XY]ij=∑k=1nxikykj[\mathbf{X}\mathbf{Y}]_{ij} = \sum_{k=1}^{n} x_{ik} y_{kj}이므로,

[d(XY)]ij=d(∑k=1nxikykj)=∑k=1n(dxik)ykj+∑k=1nxikdykj [\mathrm{d}(\mathbf{X}\mathbf{Y})]_{ij} = \mathrm{d}\left( \sum_{k=1}^{n} x_{ik} y_{kj} \right) = \sum_{k=1}^{n} (\mathrm{d}x_{ik}) y_{kj} + \sum_{k=1}^{n} x_{ik} \mathrm{d}y_{kj}

따라서 다음을 얻는다.

[d(XY)]ij=∑k=1nxikdykj+∑k=1nxikdykj=[(dX)Y]ij+[XdY]ij  ⟹  d(XY)=(dX)Y+XdY [\mathrm{d}(\mathbf{X}\mathbf{Y})]_{ij} = \sum_{k=1}^{n} x_{ik} \mathrm{d}y_{kj} + \sum_{k=1}^{n} x_{ik} \mathrm{d}y_{kj} = [(\mathrm{d}\mathbf{X})\mathbf{Y}]_{ij} + [\mathbf{X}\mathrm{d}\mathbf{Y}]_{ij} \\[1em] \implies \mathrm{d}(\mathbf{X}\mathbf{Y}) = (\mathrm{d}\mathbf{X})\mathbf{Y} + \mathbf{X}\mathrm{d}\mathbf{Y}

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