## Linear Algebra and Its Applications, Exercise 2.5.16

Exercise 2.5.16. Suppose we have a directed graph with four nodes and five edges as follows:

• edge 1 from node 1 to node 2
• edge 2 from node 2 to node 3
• edge 3 from node 1 to node 3
• edge 4 from node 1 to node 4
• edge 5 from node 3 to node 4

Also assume that the graph is grounded at node 4. Do the following:

1. Describe the current laws $A^Ty = 0$ at the three ungrounded nodes 1, 2, and 3.
2. Describe how the current law at the grounded node 4 follows from the current laws at the other three nodes.
3. Specify the rank of $A^T$.
4. Relate the solutions to $A^Ty = 0$ to the loops in the graph.

Answer: We start by constructing the incidence matrix for the network; it contains 5 rows corresponding to the edges and 4 columns corresponding to the nodes: $A = \begin{bmatrix} -1&1&0&0 \\ 0&-1&1&0 \\ -1&0&1&0 \\ -1&0&0&1 \\ 0&0&-1&1\end{bmatrix}$

The system $A^Ty = 0$ can be expressed in matrix form as $\begin{bmatrix} -1&0&-1&-1&0 \\ 1&-1&0&0&0 \\ 0&1&1&0&-1 \\ 0&0&0&1&1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \\ y_5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

a) The equations corresponding to the current laws for the first three (ungrounded) nodes are $\begin{array}{rcrcrcrcrcrcr} -y_1&&&-&y_3&-&y_4&&&=&0 \\ y_1&-&y_2&&&&&&&=&0 \\ &&y_2&+&y_3&&&-&y_5&=&0 \end{array}$

b) The equation for the current law at the grounded node is $y_4+y_5=0$. It can be derived from the three equations above by adding the first equation to the second equation to produce the equation $-y_2-y_3-y_4 = 0$ and then adding this equation to the third equation to produce $-y_4-y_5=0$ or $y_4+y_5=0$.

c) From inspection $A^T$ appears to have three pivots (in columns 1, 2, and 4) and therefore has rank 3. We can confirm this via elimination: $\begin{bmatrix} -1&0&-1&-1&0 \\ 1&-1&0&0&0 \\ 0&1&1&0&-1 \\ 0&0&0&1&1 \end{bmatrix} \Rightarrow \begin{bmatrix} -1&0&-1&-1&0 \\ 0&-1&-1&-1&0 \\ 0&1&1&0&-1 \\ 0&0&0&1&1 \end{bmatrix}$ $\Rightarrow \begin{bmatrix} -1&0&-1&-1&0 \\ 0&-1&-1&-1&0 \\ 0&0&0&-1&-1 \\ 0&0&0&1&1 \end{bmatrix} \Rightarrow \begin{bmatrix} -1&0&-1&-1&0 \\ 0&-1&-1&-1&0 \\ 0&0&0&-1&-1 \\ 0&0&0&0&0 \end{bmatrix}$

As noted above, there are three pivots in columns 1, 2, and 4, and thus the rank of $A^T$ is 3.

d) From the elimination above we see that $y_1$, $y_2$, and $y_4$ are basic variables and $y_3$ and $y_5$ are free variables. The resulting echelon matrix corresponds to the following system of equations: $\begin{array}{rcrcrcrcrcrcr} -y_1&&&-&y_3&-&y_4&&&=&0 \\ &&-y_2&-&y_3&-&y_4&&&=&0 \\ &&&&&&-y_4&-&y_5&=&0 \end{array}$

If we set $y_3 =1$ and $y_5 = 0$ then from the third equation we have $y_4 = 0$. Substituting into the second equation we have $y_2 = -1$. Substituting into the first equation we have $y_1 = -1$.

If we set $y_3 =0$ and $y_5 = 1$ then from the third equation we have $y_4 = -1$. Substituting into the second equation we have $y_2 = 1$. Substituting into the first equation we have $y_1 = 1$.

The general solution to $A^Ty = 0$ is thus given by $y_3 \begin{bmatrix} -1 \\ -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + y_5 \begin{bmatrix} 1 \\ 1 \\ 0 \\ -1 \\ 1 \end{bmatrix}$

The first term can be interpreted as a set of currents around the upper loop in the graph: In the bottom edge we have a current of $y_3$ from node 1 to node 3. We then have a current of $-y_3$ from node 2 to node 3, equivalent to a current of $y_3$ from node 3 to node 2. Finally we have a current of $-y_3$ from node 1 to node 2, equivalent to a current of $y_3$ from node 2 to node 1. The net current around the loop is zero.

The second term can be interpreted as a set of currents around the outer loop in the graph: In the top left edge we have a current of $y_5$ from node 1 to node 2. We then have a current of $y_5$ from node 2 to node 3 along the upper right edge, and a current of $y_5$ from node 3 to node 4 along the bottom right edge. Finally we have a current of $-y_5$ from node 1 to node 4, equivalent to a current of $y_5$ from node 4 to node 1. Again the net current around the loop is zero.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

This entry was posted in linear algebra. Bookmark the permalink.