## Linear Algebra and Its Applications, Exercise 2.5.2

Exercise 2.5.2. Given the incidence matrix $A$ from exercise 2.5.1 and any vector $b$ in the column space of $A$ show that $b_1 + b_2 - b_3 = 0$. Prove the same result based on the rows of $A$. What is the implication for the potential differences around a loop?

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 $b = (b_1, b_2, b_3)$ is in the column space of $A$ then we have $\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} + c_3 \begin{bmatrix} 0 \\ -1 \\ -1 \end{bmatrix}$

for some set of scalar coefficients $c_1$, $c_2$, and $c_3$ so that $\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} = \begin{bmatrix} c_1 - c_2 \\ c_2 - c_3 \\ c_1 - c_3 \end{bmatrix}$

We then have $b_1 + b_2 - b_3 = (c_1 - c_2) + (c_2 - c_3) - (c_1 - c_3)$ $= (c_1 - c_1) + (c_2 - c_2) + (c_3 - c_3) = 0 + 0 + 0 = 0$

We therefore have $b_1 + b_2 - b_3 = 0$ for all vectors $b$ in the column space of $A$.

Turning to the rows of $A$ if $Ax = b$ we have $\begin{bmatrix} 1&-1&0 \\ 0&1&-1 \\ 1&0&-1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$

which corresponds to the system of equations $\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr} x_1&-&x_2&&&=&b_1 \\ &&x_2&-&x_3&=&b_2 \\ x_1&&&-&x_3&=&b_3 \end{array}$

We then have $b_1 + b_2 - b_3 = (x_1 - x_2) + (x_2 - x_3) - (x_1 - x_3)$ $= (x_1 - x_1) + (x_2 - x_2) + (x_3 - x_3) = 0 + 0 + 0 = 0$

We also have $(x_1 - x_2) + (x_2 - x_3) - (x_1 - x_3)$ $= (x_1 - x_2) + (x_2 - x_3) + (x_3 - x_1)$

so that $(x_1 - x_2) + (x_2 - x_3) + (x_3 - x_1) = 0$

The 3 by 3 incidence matrix $A$ represents a graph with three nodes and three edges and hence one loop. Each node of the graph is represented by a column of $A$ and each edge by a row of $A$. 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.

If the vector $x$ represents potentials at the nodes ( $x_1$ at node 1, $x_2$ at node 2, and $x_3$ at node 3) then $x_1 - x_2$ is the potential difference along edge 1 (from node 2 to node 1), $x_2 - x_3$ is the potential difference along edge 2 (from node 3 to node 2) and $x_3 - x_1$ is the potential difference along edge 3 (from node 3 to node 1). From the equations above we see that the sum of the potential differences around the loop is zero (Kirchoff‘s Voltage 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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