- 행렬미분_1
\[f(\mathbf{x}) = w^T\mathbf{x} \]
\[ \nabla{f} = \frac{\partial{w^T\mathbf{x}}}{\partial \mathbf{x}} = \frac{\partial{\mathbf{x}^Tw}}{\partial \mathbf{x}} = w\]
proof)
\[ \frac{\partial{(w^T\mathbf{x})}}{\partial \mathbf{x}} = \begin{bmatrix}
\frac{\partial{(w^T\mathbf{x})}}{\partial x_1}\\
\\
\frac{\partial{(w^T\mathbf{x})}}{\partial x_2}\\
\\
\vdots
\\
\\
\frac{\partial{(w^T\mathbf{x})}}{\partial x_n}\\
\end{bmatrix} = \begin{bmatrix}
\frac{\partial{(w_1x_1 + w_2x_2 + \dots + w_nx_n)}}{\partial x_1}\\
\\
\frac{\partial{(w_1x_1 + w_2x_2 + \dots + w_nx_n)}}{\partial x_2}\\
\\
\vdots
\\
\\
\frac{\partial{(w_1x_1 + w_2x_2 + \dots + w_nx_n)}}{\partial x_n}\\
\end{bmatrix} = \begin{bmatrix}
w_1\\
\\
w_2\\
\\
\vdots
\\
\\
w_n\\
\end{bmatrix} = w \]
- 행렬미분_2
\[ f(\mathbf{x}) = \mathbf{x}^TA\mathbf{x} \]
\[ \nabla{f} = \frac{\partial{\mathbf{x}^TA\mathbf{x}}}{\partial \mathbf{x}} = (A+A^T)\mathbf{x} \]
proof)
\[ \frac{\partial{(\mathbf{x}^TA\mathbf{x})}}{\partial \mathbf{x}} = \begin{bmatrix}
\frac{\partial{(\mathbf{x}^TA\mathbf{x})}}{\partial x_1}\\
\\
\frac{\partial{(\mathbf{x}^TA\mathbf{x})}}{\partial x_2}\\
\\
\vdots
\\
\\
\frac{\partial{(\mathbf{x}^TA\mathbf{x})}}{\partial x_n}\\
\end{bmatrix} =
\begin{bmatrix}
\frac{\partial{(\sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} x_i x_j)}}{\partial x_1}\\
\\
\frac{\partial{(\sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} x_i x_j)}}{\partial x_2}\\
\\
\vdots
\\
\\
\frac{\partial{(\sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} x_i x_j)}}{\partial x_n}\\
\end{bmatrix} \]
\[ = \begin{bmatrix}
\sum_{j=1}^{n} a_{1j}x_j + \sum_{j=1}^{n} a_{i1}x_i\\
\\
\sum_{j=1}^{n} a_{2j}x_j + \sum_{j=1}^{n} a_{i2}x_i\\
\\
\vdots
\\
\\
\sum_{j=1}^{n} a_{nj}x_j + \sum_{j=1}^{n} a_{in}x_i\\
\end{bmatrix} = \begin{bmatrix}
\sum_{j=1}^{n} a_{1j}x_j\\
\\
\sum_{j=1}^{n} a_{2j}x_j\\
\\
\vdots
\\
\sum_{j=1}^{n} a_{nj}x_j\\
\end{bmatrix} +
\begin{bmatrix}
\sum_{j=1}^{n} a_{i1}x_i\\
\\
\sum_{j=1}^{n} a_{i2}x_i\\
\\
\vdots
\\
\\
\sum_{j=1}^{n} a_{in}x_i\\
\end{bmatrix} \]
\[ = A\mathbf{x} + A^T\mathbf{x} = (A+A^T)\mathbf{x} \]
- 벡터를 스칼라 x로 미분하기
\[\mathbf{f}(x) = \begin{bmatrix}
\mathbf{f}_1(x)\\
\\
\mathbf{f}_2(x)\\
\\
\vdots
\\
\\
\mathbf{f}_n(x)\\
\end{bmatrix} \quad \Longrightarrow \quad \frac{\partial \mathbf{f}(x)}{\partial x} = \begin{bmatrix}
\frac{\partial \mathbf{f}_1(x)}{\partial x}\\
\\
\frac{\partial \mathbf{f}_2(x)}{\partial x}\\
\\
\vdots
\\
\\
\frac{\partial \mathbf{f}_(x)}{\partial x}\\
\end{bmatrix} \]
- 벡터를 벡터로 미분하기
\[f = \begin{bmatrix}
f_1\\
\\
f_2\\
\\
\vdots
\\
\\
f_n\\
\end{bmatrix} \quad \Longrightarrow \quad \frac{\partial f}{\partial \mathbf{x}} = \begin{bmatrix}
\frac{\partial f}{\partial x_1}\\
\\
\frac{\partial f}{\partial x_2}\\
\\
\vdots
\\
\\
\frac{\partial f}{\partial x_m}\\
\end{bmatrix} =
\begin{bmatrix}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_2}{\partial x_1} & \dots & \frac{\partial f_n}{\partial x_1}\\
\\
\frac{\partial f_1}{\partial x_2} & \frac{\partial f_2}{\partial x_2} & \dots & \frac{\partial f_n}{\partial x_2}\\
\\
\vdots && \ddots & \vdots
\\
\\
\frac{\partial f_1}{\partial x_m} & \frac{\partial f_2}{\partial x_m} & \dots & \frac{\partial f_n}{\partial x_m}\\
\end{bmatrix}\]
- 행렬 A와 벡터 x의 곱을 벡터 x로 미분하기
\[f(\mathbf{x}) = A \mathbf{x} \]
\[\nabla{f(\mathbf{x})} = \frac{\partial (A \mathbf{x})}{\partial \mathbf{x}} = A^T \]
proof)
Let \[ A \mathbf{x} = c_1x_1 + c_2x_2 + \dots + c_nx_n\]
\[\frac{\partial (A \mathbf{x})}{\partial \mathbf{x}} = \begin{bmatrix}
\frac{\partial (A \mathbf{x})}{\partial x_1}\\
\\
\frac{\partial (A \mathbf{x})}{\partial x_2}\\
\\
\vdots
\\
\\
\frac{\partial (A \mathbf{x})}{\partial x_n}\\
\end{bmatrix} =
\begin{bmatrix}
\frac{\partial (c_1x_1 + c_2x_2 + \dots + c_nx_n)^T}{\partial x_1}\\
\\
\frac{\partial (c_1x_1 + c_2x_2 + \dots + c_nx_n)^T}{\partial x_2}\\
\\
\vdots
\\
\\
\frac{\partial (c_1x_1 + c_2x_2 + \dots + c_nx_n)^T}{\partial x_n}\\
\end{bmatrix} = \begin{bmatrix}
c_1^T\\
\\
c_2^T\\
\\
\vdots
\\
\\
c_n^T\\
\end{bmatrix} = A^T\]
"이 포스팅은 쿠팡 파트너스 활동의 일환으로, 이에 따른 일정액의 수수료를 제공받습니다."