本文探讨了矩阵D的相关矩阵C及其对角化矩阵P的关系，通过推导证明了公式(2tr[D^T(P-C)D]=sum_{i,j}
1. 问题
对于一个矩阵D=[d1;d2;… ;dn]∈Rn×lD = [d_1;d_2;\dots;d_n] \in R^{n \times l}D=[d1;d2;…;dn]∈Rn×l, 其 相关矩阵为 C∈Rn×nC \in R^{n \times n}C∈Rn×n, 是否有以下表达式成立(经第2部分推导，基本成立，最正确的见式3\textcolor{red}{基本成立，最正确的见式3}基本成立，最正确的见式3):
tr[D(P−C)DT]=∑i,jcij∥di−dj∥2(1) tr[D(P-C)D^T] =\sum_{i,j} c_{ij} \|d_i-d_j\|^2 \tag 1 tr[D(P−C)DT]=i,j∑cij∥di−dj∥2(1)
tr[D(P−C)DT]=∑i=1n{ 12∑p,qlCpq[dicp−dicq]2}(2) tr[D(P-C)D^T] = \sum_{i=1}^n \{ \frac{1}{2}\sum_{p,q}^lC_{pq}[d_ic_p-d_ic_q]^2 \} \tag 2 tr[D(P−C)DT]=i=1∑n{ 21p,q∑lCpq[dicp−dicq]2}(2)
其中C是对称矩阵，P=diag(C×1n×1)∈Rn×nP=diag(C\times 1^{n \times 1}) \in R^{n \times n}P=diag(C×1n×1)∈Rn×n表示C按行累加形成的列向量，构成的对角矩阵。
2. 试着推导公式1
令
D=[d11d12⋯d1ld21d22⋯d2l⋮⋮⋱⋮dn1dn2⋯dnl],C=[1c12⋯c1nc211⋯c2n⋮⋮⋱⋮cn1cn2⋯1] D=\left[ \begin{matrix} d_{11} & d_{12} & \cdots & d_{1l} \\ d_{21} & d_{22} & \cdots & d_{2l} \\ \vdots & \vdots & \ddots & \vdots \\ d_{n1} & d_{n2} & \cdots & d_{nl} \\ \end{matrix} \right], C=\left[ \begin{matrix} 1 & c_{12} & \cdots & c_{1n} \\ c_{21} & 1 & \cdots & c_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n1} & c_{n2} & \cdots & 1 \\ \end{matrix} \right] D=
d11d21⋮dn1d12d22⋮dn2⋯⋯⋱⋯d1ld2l⋮dnl
,C=
1c21⋮cn1c121⋮cn2⋯⋯⋱⋯c1nc2n⋮1
则
DT=[d11d21⋯dn1d12d22⋯dn2⋮⋮⋱⋮d1ld2l⋯dnl]，P=[∑j=1nc1j0⋯00∑j=1nc2j⋯0⋮⋮⋱⋮00⋯∑j=1ncnj], D^T=\left[ \begin{matrix} d_{11} & d_{21} & \cdots & d_{n1} \\ d_{12} & d_{22} & \cdots & d_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ d_{1l} & d_{2l} & \cdots & d_{nl} \\ \end{matrix} \right] ， \\ P=\left[ \begin{matrix} \sum_{j=1}^n c_{1j} & 0 & \cdots & 0 \\ 0 & \sum_{j=1}^n c_{2j} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \sum_{j=1}^n c_{nj} \\ \end{matrix} \right],\\ DT=
d11d12⋮d1ld21d22⋮d2l⋯⋯⋱⋯dn1dn2⋮dnl
，P=
∑j=1nc1j0⋮00∑j=1nc2j⋮0⋯⋯⋱⋯00⋮∑j=1ncnj
,
从右侧开始：
∑i,jcij∥di−dj∥2=∑i,jcij∥di−dj∥22=∑i,jcij∑k=1l(dik−djk)2=∑i=1n∑j=1ncij∑k=1l(dik−djk)2 \sum_{i,j} c_{ij} \|d_i-d_j\|^2=\sum_{i,j} c_{ij} \|d_i-d_j\|_2^2 \\ =\sum_{i,j} c_{ij} \sum_{k=1}^l(d_{ik}-d_{jk})^2\\ =\sum_{i=1}^n \sum_{j=1}^n c_{ij} \sum_{k=1}^l(d_{ik}-d_{jk})^2\\ i,j∑cij∥di−dj∥2=i,j∑cij∥di−dj∥22=i,j∑cijk=1∑l(dik−djk)2=i=1∑nj=1∑ncijk=1∑l(dik−djk)2
左侧：
tr[D(P−C)DT]无法直接运算，试着更正为：tr[DT(P−C)D] tr[D(P-C)D^T] 无法直接运算，试着更正为：tr[D^T(P-C)D] tr[D(P−C)DT]无法直接运算，试着更正为：tr[DT(P−C)D]
则
tr[DT(P−C)D]=tr[DTPD−DTCD]=tr(DTPD)−tr(DTCD) tr[D^T(P-C)D] \\ =tr[D^TPD-D^TCD] \\ =tr(D^TPD)-tr(D^TCD)\\ tr[DT(P−C)D]=tr[DTPD−DTCD]=tr(DTPD)−tr(DTCD)
DTP=[d11d21⋯dn1d12d22⋯dn2⋮⋮⋱⋮d1ld2l⋯dnl][∑j=1nc1j0⋯00∑j=1nc2j⋯0⋮⋮⋱⋮00⋯∑j=1ncnj]=[d11∑j=1nc1jd21∑j=1nc2j⋯dn1∑j=1ncnjd12∑j=1nc1jd22∑j=1nc2j⋯dn2∑j=1ncnj⋮⋮⋱⋮d1l∑j=1nc1jd2l∑j=1nc2j⋯dnl∑j=1ncnj] D^TP = \left[ \begin{matrix} d_{11} & d_{21} & \cdots & d_{n1} \\ d_{12} & d_{22} & \cdots & d_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ d_{1l} & d_{2l} & \cdots & d_{nl} \\ \end{matrix} \right] \left[ \begin{matrix} \sum_{j=1}^n c_{1j} & 0 & \cdots & 0 \\ 0 & \sum_{j=1}^n c_{2j} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \sum_{j=1}^n c_{nj} \\ \end{matrix} \right] \\ =\left[ \begin{matrix} d_{11}\sum_{j=1}^n c_{1j} & d_{21}\sum_{j=1}^n c_{2j} & \cdots & d_{n1}\sum_{j=1}^n c_{nj} \\ d_{12}\sum_{j=1}^n c_{1j} & d_{22}\sum_{j=1}^n c_{2j} & \cdots & d_{n2}\sum_{j=1}^n c_{nj} \\ \vdots & \vdots & \ddots & \vdots \\ d_{1l}\sum_{j=1}^n c_{1j} & d_{2l}\sum_{j=1}^n c_{2j} & \cdots & d_{nl}\sum_{j=1}^n c_{nj} \\ \end{matrix} \right] DTP= d
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