## 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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