## Linear Algebra and Its Applications, Exercise 2.5.3

Exercise 2.5.3. Given the incidence matrix $A$ from exercise 2.5.1 and any vector $f$ in the row space of $A$ show that $f_1 + f_2 + f_3 = 0$. Prove the same result based on the linear system $A^Ty = f$. What is the implication if $f_1$, $f_2$, and $f_3$ are currents into each node?

Answer: From exercise 2.5.1 we have the incidence matrix

$A = \begin{bmatrix} 1&-1&0 \\ 0&1&-1 \\ 1&0&-1 \end{bmatrix}$

If $f = (f_1, f_2, f_3)$ is in the row space of $A$ then we have

$\begin{bmatrix} f_1 \\ f_2 \\ f_3 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$

for some set of scalar coefficients $c_1$, $c_2$, and $c_3$ so that

$\begin{bmatrix} f_1 \\ f_2 \\ f_3 \end{bmatrix} = \begin{bmatrix} c_1 + c_3 \\ -c_1 + c_2 \\ -c_2 - c_3 \end{bmatrix}$

We then have

$f_1 + f_2 + f_3 = (c_1 + c_3) + (-c_1 + c_2) + (-c_2 - c_3)$

$= (c_1 - c_1) + (c_2 - c_2) + (c_3 - c_3) = 0 + 0 + 0 = 0$

We therefore have $f_1 + f_2 + f_3 = 0$ for all vectors $f$ in the row space of $A$.

Turning to the system $A^Ty = f$ we have

$\begin{bmatrix} 1&0&1 \\ -1&1&0 \\ 0&-1&-1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} f_1 \\ f_2 \\ f_3 \end{bmatrix}$

which corresponds to the system of equations

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr} y_1&&&+&y_3&=&f_1 \\ -y_1&+&y_2&&&=&f_2 \\ &&-y_2&-&y_3&=&f_3 \end{array}$

We then have

$f_1 + f_2 + f_3 = (y_1 + y_3) + (-y_1 + y_2) + (-y_2 - y_3)$

$= (y_1 - y_1) + (y_2 - y_2) + (y_3 - y_3) = 0$

The 3 by 3 incidence matrix $A$ represents a graph with three nodes and three edges. The first row represents edge 1 from node 2 to node 1 (i.e., leaving node 2 and entering node 1). The second row represents edge 2 from node 3 to node 2. The third row represents edge 3 from node 3 to node 1.

Each node of the graph is represented by a column of $A$ and thus by a row of $A^T$. If the vector $f$ represents current sources at each node ($f_1$ at node 1, $f_2$ at node 2, and $f_3$ at node 3) then the fact that $f_1 + f_2 + f_3 = 0$ means that the net current into each node is zero (Kirchoff’s Current Law).

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.

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