Chapter 10.1, Problem 1PS

To determine

The $3$ by $3$ incidence matrix $A$ for the triangle graph. What vectors $\left(x_1, x_2, x_3\right)$ are in the null space? How do you know that $\left(1,0,0\right)$ is not in its row space?

Answer to Problem 1PS

   $A=\left(\begin{array}{ccc}-1& 1& 0\\ -1& 0& 1\\ 0& -1& 1\end{array}\right)$

Null space contains $\left(\begin{array}{c}c\\ c\\ c\end{array}\right)$

   $\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$ is not in the row space

Explanation of Solution

Given information:

  

The first row has $-1$ in column $1$ and $+1$ in column $2$.

Calculation:

To find the incidence matrix, we put a $"1"$ whenever a vertex is incident to an edge and we put a $"0"$ if that vertex is not incident to that edge.

For the directed graphs, $"1"$ s refer to arcs incident from a vertex while $"-1"$ s refer to arcs incident to a vertex.

  

The first row has $-1$ in column $1$ and $+1$ in column $2$ (given).

Therefore, incidence matrix $A=\left(\begin{array}{ccc}-1& 1& 0\\ -1& 0& 1\\ 0& -1& 1\end{array}\right)$

To find the null-space we solve the equation $Ax=0$

   $\left(\begin{array}{ccc}-1& 1& 0\\ -1& 0& 1\\ 0& -1& 1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$

   $-x_1+x_2=0\to x_1=x_2$and $-x_1+x_3=0\to x_1=x_3$

Therefore, $x_1=x_2=x_3$

   $N\left(A\right)=\begin{array}{c}c\\ c\\ c\end{array}=c\langle \begin{array}{c}1\\ 1\\ 1\end{array}\rangle$

   $\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$is not in the row space since it is not orthogonal to that null-space.