Exercise 2.5.2. Given the incidence matrix from exercise 2.5.1 and any vector in the column space of show that . Prove the same result based on the rows of . What is the implication for the potential differences around a loop?

Answer: From exercise 2.5.1 we have the incidence matrix

If is in the column space of then we have

for some set of scalar coefficients , , and so that

We then have

We therefore have for all vectors in the column space of .

Turning to the rows of if we have

which corresponds to the system of equations

We then have

We also have

so that

The 3 by 3 incidence matrix represents a graph with three nodes and three edges and hence one loop. Each node of the graph is represented by a column of and each edge by a row of . 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 represents potentials at the nodes ( at node 1, at node 2, and at node 3) then is the potential difference along edge 1 (from node 2 to node 1), is the potential difference along edge 2 (from node 3 to node 2) and 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.