Exercise 4.6.7

• 1.

  From problem IV.6.2, we have $A^{T}A=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right]$

• 1.

  From loops we find the $6-4+1=3$ solutions to Kirchhoff’s Law $A^{T}w=0$:

We assume we have a matrix $A=\left[\begin{array}{cccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$, from this matrix, we can tell the flow of current (from which node to which) and use them to generate the solutions to Kirchhoff’s law.

From the four loops of $(x_1,x_3,x_4)$, $(x_3,x_4,x_2)$, $(x_1,x_4,x_2)$ and $(x_1,x_3,x_2)$ we find 4 solutions:

$y_1=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill -1\hfill \end{array}\right]$, $y_2=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right]$, $y_3=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right]$, and $y_4=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right]$

It’s easy to see that $y_1+y_3=y_2+y_4$, so there are only 3 independent solutions. Pick any of three should be fine.