Math 2270-2 Spring 2012Computer Lab 2This lab was written by Professor Chris Cashen in his Postdoc years at Utah. He is currently in 2012 at the Frenchresearch institute Laboratoire de Mathmatiques Nicolas Oresme, Universit de Caen Basse-Normandie. I am indebted tohim for the valuable sources, and I must record here Thank you Chris!I hope that you will agree with me, that this project is a Linear Algebra Disneyland, an adventure which connects theabstract Four Subspaces of Strang with the physics of Kirchhoff’s voltage and current laws.Some minor display and font size changes were made. It remains exactly the same as Chris Cashen wrote, except fortypo corrections.Submit your project on the due date in class as a worksheet print. Submit study group efforts as one worksheetprint with multiple names.Read Section 8.2 of Strang’s textbook for background. Engineers should read Michael Hanke’s explanation, a copy foundat http://math.utah.edu/~gustafso/MichaelHanke2006.pdf (beware of two critical typos). Therein are physicalexamples plus an explanation of how to remove superfluous variables. In the DC circuit example, you can imagine thecurrent source to be additional information, namely an ammeter reading. Such physical examples have no capacitorsnor inductors, only resistors, wires and batteries, realized as a Radio Shack project.Chris Cashen’s Electrical Network Maple LabThe main idea of this section is that in applications the four fundamental subspaces have a physical interpretation.Consider a graph consisting of nodes and directed edges. To such a graph associate an incidence matrix A with one rowfor each edge and one column for each node. If edge i goes from node j to node k then set the i, j–entry of A to −1,the i, k–entry of A to 1, and the other entries of row i to 0.For example, if the graph is a triangle with nodes at the vertices labeled 1, 2, 3 and edges from 1 to 2, from 1 to 3, and1from 2 to 3, then the incidence matrix would be:A =−1 1 0−1 0 10 −1 1We could multiply A by a vector ~v of dimension equal to the number of columns = number of nodes, and this will resultin a vector of dimension equal to the number of rows of A= number of edges in the graph. Imagine that the graph isan electrical circuit and the entries of the vector ~v are the voltage at the corresponding node. ThenA¯v =v2− v1v3− v1v3− v2So, A~v is the vector whose components are the voltage differences across the various edges.The Column Space of A is the collection of all linear combinations of the columns of A, which is the same as thecollection of vectors A~v for all possible input vectors ~v.Conversely, Kirchhoff’s Voltage Law says that the sum of voltages differences around a loop in the graph must be zero.For any vector~d of voltage differences satisfying Kirchhoff’s Law there is a choice of voltage vector ~v such that A~v =~d.To see this, choose the voltage of one node arbitrarily. The voltage differences then tell you the voltage at adjacentnodes, and so on.You should think about why Kirchhoff’s Law implies that this procedure works.So, the column space of A has a physical interpretation as the set of voltage differences satisfying Kirchhoff’s VoltageLaw.The Nullspace of A is the set of all vectors that give zero voltage difference across each edge. This is true if and onlyif all the voltages are the same, so the nullspace is the one dimensional space consisting of vectors with all componentsequal.Ohm’s Law says that Voltage = Current × Resistance. If we assume that each edge has resistance 1Ω (1 Ohm) thenthe current through the edge in Amps is equal to the voltage difference across the edge in Volts.Kirchhoff’s Current Law says that the current flowing in to node equals the current flowing out. If ~c is a vector whosecomponents are the currents through the various edges, thenAT~c =−1 −1 01 0 −10 1 1c1c2c3=−c1− c2c1− c3c2+ c3Kirchhoff’s Current Law is satisfied if and only if AT~c =~0, so the Left Nullspace of A has a physical interpretation,it is the set of vectors whose components are currents satisfying this Law.Now suppose that we introduce an external current s flowing into node 1 and ground node 3. The grounded node hasvoltage zero, and has net current s flowing out. For the other nodes, the current flowing in matches the current flowingout, so the circuit satisfies the equation:ATA~v =s0−sWe can find the voltage at each node by solving this equation. Recall that we know the nullspace is one dimensional,so there is exactly one free column. We have grounded a node, which means we have set the free variable v3= 0, andwe then solve for a particular solution.Finally, we could have different resistances across the edges. We account for this using a square matrix C whose i, ientry is1resistance of edge i.Suppose edge 3 has resistance 2Ω. Then:C =1 0 00 1 00 012To find the voltages solve the equation:ATCAv1v20=s0−sThen, once the voltages are known, find the currents through each edge by computing CA~v.Problem: For the following network diagram, find the voltage at each numbered node and the current flow acrosseach numbered edge, given the following assumptions:Figure 2. The resistance across each edge is 1Ω exceptfor edges e2 and e9, which have resistance 5Ω. There isa current of 12A that enters node 1 from the top, and acurrent of 12A the exits node 7 from the bottom.End of Chris Cashen’s maple lab.Physical Interpretation of the Four Fundamental SubspacesImage(A) = Column Space of A All vectors ~v with component vi= voltage difference acrossedge i. This is Kirchhoff’s Voltage Law (KVL).Kernel(A) = Nullspace of A Set of node potential vectors ~x such that all edge voltagedifferences are zero.1Kernel(AT) = Left Nullspace of A Kirchhoff’s Current Law (KCL).Image(AT) = Row Space of A No useful physical interpretation (yet).11Literature will write voltage at node i, which means voltage difference between node i and the reference node. The reference (mass)node could be a diagram node, but in engineering literature it is likely removed from the diagram, in order to decrease the variable count. Apossible way to remove a node with assigned variable xiis to set it to zero, a physically unmotivated but effective algebraic trick. The logic?If the plan is to find the voltage differences, then the value at the reference node will be