Exercise 4.7.6

For the 2 by 4 mesh of 8 nodes, I make nodes $x_1-x_4$ on the left half, with $x_1,x_4$ on diagonal. Similarly, $x_5-x_8$ are on the right side, with $x_5,x_8$ on diagonal. With weight $C=I$, we have the matrix $A^T\mathit{CA}=\left[\begin{array}{cccccccc}\hfill 2\hfill & \hfill -1\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill -1\hfill & \hfill 3\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$

From the calculation below, we see that ${\lambda }_2=0.2712864461218309$, the corresponding $z$ has the top half with negative values and the bottom half with positive values, this cuts the nodes in the middle, where $x_1-x_4$ belong to one cluster and $x_5-x_8$ belong to another.

#### Problem IV.7.6 
import numpy as np 
import scipy as sp 
a = np.array([[2, -1, -1, 0, 0, 0, 0, 0], [-1, 3, 0, -1, -1, 0, 0, 0], 
             [-1, 0, 2, -1, 0, 0, 0, 0], [0, -1, -1, 3, 0, 0, -1, 0], 
             [0, -1, 0, 0, 3, -1, -1, 0], [0, 0, 0, 0, -1, 2, 0, -1], 
             [0, 0, 0, -1, -1, 0, 3, -1], [0, 0, 0, 0, 0, -1, -1, 2]]) 
D = np.diag(np.diag(a)) 
v, L = sp.linalg.eig(a, D) 
#np.argmin(v), np.iscomplex(v) 
vi = np.argsort(v) 
print(’lambda␣2:␣’, v[vi[1]]) 
z = L[:, vi[1]] 
print(’second␣eigenvector:␣’, z) 
np.matmul(np.ones(8).reshape(1,8), np.matmul(D, z))

lambda 2:  (0.2712864461218309+0j)
second eigenvector:  [-0.45468835 -0.20798678 -0.45468835 -0.20798678
0.20798678  0.45468835
  0.20798678  0.45468835]
array([1.33226763e-15])