矩阵向量中两两间欧式距离计算

目标：希望通过的矩阵运算就能得出矩阵向量中两两之间的欧式距离

欧氏距离公式：

• 一般而言，我们常见的欧式距离计算公式如下：

  • a,b 对应的是两组不同的向量
  • $dist(a,b)=\sqrt{(a_1-b_1{)}^2+(a_2-b_2{)}^2+\cdot \cdot \cdot (a_n-b_n{)}^2}$

• 事实上对于上面的公式如果我们通过向量的角度来考虑，就会变成是下列形式：

  • a,b 对应的是两组不同的向量
  • $dist(a,b)=\sqrt{dot(a,a)-2\ast dot(a,b)+dot(b,b)}$

假设有下列矩阵$A$:

• $A=\left[\begin{array}{ccc}a{}_1& a{}_2& a{}_3\\ b{}_1& b{}_2& b{}_3\end{array}\right]$

• 为了凑出上面的公式:

  • 先计算出 $dot(A,A)$ -> $A{A}^T$，

    • $\overline{A}=\left[\begin{array}{ccc}a{}_1& a{}_2& a{}_3\\ b{}_1& b{}_2& b{}_3\end{array}\right]\left[\begin{array}{cc}a{}_1& b{}_1\\ a{}_2& b{}_2\\ a{}_3& b{}_3\end{array}\right]=\left[\begin{array}{cc}(a{}_1{)}^2+(a{}_1{)}^2+(a{}_1{)}^2& (a{}_1)(b{}_1)+(a{}_2)(b{}_2)+(a{}_3)(b{}_3)\\ (a{}_1)(b{}_1)+(a{}_2)(b{}_2)+(a{}_3)(b{}_3)& (b{}_1{)}^2+(b{}_2{)}^2+(b{}_3{)}^2\end{array}\right]$

  • 取$\overline{A}$对角线:

    • $\overline{A}.diag()$ = $\left[\begin{array}{cc}(a{}_1{)}^2+(a{}_2{)}^2+(a{}_3{)}^2& (b{}_1{)}^2+(b{}_2{)}^2+(b{}_3{)}^2\end{array}\right]$

  • 对对角线矩阵进行一些变换

    • $\overline{A}{}_1$ = ${\left[\begin{array}{cc}(a{}_1{)}^2+(a{}_2{)}^2+(a{}_3{)}^2& (b{}_1{)}^2+(b{}_2{)}^2+(b{}_3{)}^2\end{array}\right]}^T$ $\left[\begin{array}{cc}1& 1\end{array}\right]$
    • = $\left[\begin{array}{cc}(a{}_1{)}^2+(a{}_2{)}^2+(a{}_3{)}^2& (a{}_1{)}^2+(a{}_2{)}^2+(a{}_3{)}^2\\ (b{}_1{)}^2+(b{}_2{)}^2+(b{}_3{)}^2& (b{}_1{)}^2+(b{}_2{)}^2+(b{}_3{)}^2\end{array}\right]$
    • $\overline{A}{}_2$ = ${\left[\begin{array}{cc}1& 1\end{array}\right]}^T$ $\left[\begin{array}{cc}(a{}_1{)}^2+(a{}_2{)}^2+(a{}_3{)}^2& (b{}_1{)}^2+(b{}_2{)}^2+(b{}_3{)}^2\end{array}\right]$
    • = $\left[\begin{array}{cc}(a{}_1{)}^2+(a{}_2{)}^2+(a{}_3{)}^2& (b{}_1{)}^2+(b{}_2{)}^2+(b{}_3{)}^2\\ (a{}_1{)}^2+(a{}_2{)}^2+(a{}_3{)}^2& (b{}_1{)}^2+(b{}_2{)}^2+(b{}_3{)}^2\end{array}\right]$

  • 经过了上面的处理,我们就可以得出上述的公式了

    • $dist(A)=\overline{A}{}_1+\overline{A}{}_2-\overline{2A}$