样本矩阵
X=x11x12⋯x1dx21x22⋯x2d⋮⋮⋮xN1xN2⋯xNd∈RN×d{\bm X} = \begin{bmatrix} x_{11} & x_{12} & \cdots & x_{1d} \\ x_{21} & x_{22} & \cdots & x_{2d} \\ \vdots & \vdots & & \vdots \\ x_{N1} & x_{N2} & \cdots & x_{Nd} \\ \end{bmatrix} \in \mathbb{R}^{N \times d} X= x11x21⋮xN1x12x22⋮xN2⋯⋯⋯x1dx2d⋮xNd ∈RN×d
增广样本矩阵
X‾=1x11x12⋯x1d1x21x22⋯x2d⋮⋮⋮⋮1xN1xN2⋯xNd∈RN×(d+1){\overline {\bm X}} = \begin{bmatrix} 1 & x_{11} & x_{12} & \cdots & x_{1d} \\ 1& x_{21} & x_{22} & \cdots & x_{2d} \\ \vdots & \vdots& \vdots & & \vdots \\ 1 & x_{N1} & x_{N2} & \cdots & x_{Nd} \\ \end{bmatrix} \in \mathbb{R}^{N \times (d+1)} X= 11⋮1x11x21⋮xN1x12x22⋮xN2⋯⋯⋯x1dx2d⋮xNd ∈RN×(d+1)
规范化增广样本矩阵
Z=1x11(1)⋯x1,d(1)1x21(1)⋯x2,d(1)⋮⋮⋮1xN1,1(1)⋯xN1,d(1)−1−x11(2)⋯−x1,d(2)−1−x21(2)⋯−x2,d(2)⋮⋮⋮−1−xN2,1(2)⋯−xN2,d(2)∈RN×(d+1) {\bm Z} = \begin{bmatrix} \begin{array}{c|ccc} 1&x_{11} ^{(1)}& \cdots & x_{1,d} ^{(1)} \\ 1&x_{21} ^{(1)}& \cdots & x_{2,d}^{(1)} \\ \vdots&\vdots & & \vdots \\ 1&x_{N_1,1}^{(1)} & \cdots & x_{N_1,d}^{(1)} \\ \hline -1 & -x_{11}^{(2)} & \cdots & -x_{1,d}^{(2)} \\ -1 & -x_{21}^{(2)} & \cdots & -x_{2,d}^{(2)} \\ \vdots & \vdots && \vdots \\ -1 & -x_{N_2,1}^{(2)} & \cdots & -x_{N_2,d}^{(2)} \end{array} \end{bmatrix} \in \mathbb{R}^{N \times (d+1)} Z= 11⋮1−1−1⋮−1x11(1)x21(1)⋮xN1,1(1)−x11(2)−x21(2)⋮−xN2,1(2)⋯⋯⋯⋯⋯⋯x1,d(1)x2,d(1)⋮xN1,d(1)−x1,d(2)−x2,d(2)⋮−xN2,d(2) ∈RN×(d+1)
N=N1+N2N=N_1+N_2N=N1+N2
这样矩阵Z\boldsymbol{Z}Z可以写成分块矩阵
Z=11X1−12−X2 \boldsymbol{Z}= \begin{bmatrix} \boldsymbol{1}_1 & \boldsymbol{X}_1 \\ -\boldsymbol{1}_2 & -\boldsymbol{X}_2 \end{bmatrix} Z=11−12X1−X2
其中，1i\boldsymbol{1}_i1i是NiN_iNi个 1 的列向量，Xi\boldsymbol{X}_iXi是一个Ni×dN_i \times dNi×d矩阵，它的行是标记为CiC_iCi的样本。