Vector Spaces and Graphs

Given any collection K of vectors on S over R , one can define two ’natural’vector spaces associated with K

S(K) (Span of K) ≡ f : f is a linear combination of vectors in K (to produce a typical f we take some finite set of vectors g1, ...., gn and

linearly combine them using some elements of R )If K is the set of rows of

f1

··fr

S(K) is the set of all vectors which have the form

[

λ1 · · λr

]

f1

··

fr

for an arbitrary choice of λ1, ., ., λr.If f ′, f ′′ ∈ S(K), it is easy to verify that αf ′ + βf ′′ ∈ S(K). So S(K) is a

vector space.We Define the space K⊥ ≡ g : < f , g > = 0, wherever f ∈ K Once again it is easily verified that K⊥ is a vector space.These two ways of generating vector spaces from a given set K of vectors are

very common in network theory.A graph should be visualized as a set of points with pairs of them joined by

means of lines. Formally, to define a particular graph we need to know

1. the set of vertices V

2. the set of edges E

3. for each edge, the ‘end points’.

This could be defined through an incidence function f with, for each e ∈ E,

f(e) ≡ v1, v2

where v1, v2 are the end points of e. We permit an edge to have a single endpoint so that we may have

f(e) ≡ v

Graphs of the above kind are called ’undirected’ since we are not specifyingwhich end point is to be treated as the ‘first’ end point.

The graph in Figure 1 is the triple (V,E, f) where

V = a, b, c, d

1

3

.a

b c

de

e

e

e4

1 2

Figure 1: undirected Graph Gu

E = e1, e2, e3, e4

f(e1) = a, b

f(e2) = a, c

f(e3) = b, c

f(e4) = a

Observe that a vertex can be ‘isolated’ without being the end point of any edgebut every edge has at least one end point.

Graphs can be directed with edge having a first and a second end point. In

3

.a

b c

de

e

e

e4

1 2

Figure 2: Directed Graph Gd

this case the graph can be thought of as a triple (V,E, fd) with

fd(e1) = (a, b), fd(e2) = (a, c), fd(e3) = (b, c), fd(e4) = (a, a).

Graphs can be associated with vector spaces in some nice ways.

2

Incidence matrix of a directed graph

A natural way of associating a graph with vector spaces is to use vectors torepresent incidence relationship at nodes or to indicate which are the ‘positive’and ‘negative’ end points of an edge. The incidence matrix arises out of thisviewpoint. For a directed (or oriented) graph G, the incidence matrix A isdefined as follows. For each node i there is a row i of A and for each edge jthere is a corresponding column j.

A(i, j) =

0 if edge j is not incident at node i or edge j is aself loop (an edge with a single end point)

+1 if edge j is leaving node i−1 if edge j is entering node i

Observe that

1. every column has one +1, one -1 and remaining entries zero or is entirelyzero,

2. if a node is isolated, the corresponding row in the incidence matrix is zero,

3. from the incidence matrix, the graph can be reconstructed except for thenode positions of self loops.

The characteristic feature of an incidence matrix is that every column iseither fully zero or has one +1, one -1 and other entries zero. Given any suchmatrix, one can immediately build a graph corresponding to it. More than onegraph can correspond to a given incidence matrix only if there are zero columns,because the corresponding self loop may be incident at any vertex.

Note: In the above statement, we have used ‘same graph’ in a loose way,regarding two graphs as same if renaming of nodes and edges makes them appearthe same. More formally, if (V1, E1, f1), (V2, E2, f2) are two directed graphs, wecall them ‘isomorphic’ if there exist one-to-one onto functions

η : V1 → V2

ǫ : E1 → E2

s.t.f1(e) = (v1, v2) ⇒ f2(ǫ(e)) = (η(v1), η(v2)).

The word isomorphic is loosely to be interpreted as ‘essentially the same’.

Example

The graphs G1 and G2 of Figure 3 are isomorphic.

η(a) = a′ etc

ǫ(e1) = e′1 etc

3

d’

b’c’

a’

e ’4

e1’e2’

e3’e5’

e4

e1 e2

e3

e5

a

b c

d

f

f ’

Figure 3

f1(e1) = (a, b)

f2(ǫ(e1)) = f2(e′

1) = (a′, b′) = (η(a), η(b)).

The question might arise in the mind of the reader as to the manner in whichself loops are represented, because we seem to be losing information about wherethey are incident. Electrically speaking, the important quantities are currentsand voltages. The position of the self loop is irrelevant as far as currents andvoltages are concerned. So the information lost is not very important. On theother hand, by keeping the column corresponding to the self loop as a zerocolumn, there are advantages which will emerge presently.

e7

e2e1

e3

e5

e6

e4

a

cb

d f

g

Figure 4: Graph G

4

For the graph G of Figure 4, the incidence matrix is

abcdfg

e1 e2 e3 e4 e5 e6 e7

+1 +1 0 0 0 0 0−1 0 +1 +1 0 0 00 −1 −1 0 +1 0 00 0 0 −1 0 +1 00 0 0 0 −1 −1 00 0 0 0 0 0 0

Let us examine the product Ax for graph G of Figure 4.

+1 +1 0 0 0 0 0−1 0 +1 +1 0 0 00 −1 −1 0 +1 0 00 0 0 −1 0 +1 00 0 0 0 −1 −1 00 0 0 0 0 0 0

x1

x2

x3

x4

x5

x6

x7

The resulting column vector has one entry per row of A, i.e. per node of G andthe vector is

z =

x1 + x2

−x1 + x3 + x4

−x2 − x3 + x5

−x4 + x6

−x5 − x6

0

.

Clearly, since each row of A has a +1 corresponding to branch leaving the node,-1 corresponding to branch entering the node, 0 when branch does not touch thenode, if we regard the xi as currents in the corresponding branch, the entry ofz corresponding to a node a will be the net current leaving a. Thus, x satisfiesKCL iff Ax = 0. Observe that in the KCE for node ‘a’, the coefficient forcurrent in self loop e7 is zero since the current i7 that enters ‘a’ also leaves it.

Let us next examine the product

[λTA] = yT

We have

[

λa λb λc λd λf λg

]

e1 e2 e3 e4 e5 e6 e7

+1 +1 0 0 0 0 0−1 0 +1 +1 0 0 00 −1 −1 0 +1 0 00 0 0 −1 0 +1 00 0 0 0 −1 −1 00 0 0 0 0 0 0

= [(λa − λb) (λa − λc) (λb − λc) (λb − λd) (λc − λf ) (λd − λf ) 0]

5

Each column of A is either fully zero or has one +1 and one -1. It followstherefore that a typical term in the product vector yT will be either zero orhave the form (λx − λy), where the edge in question (corresponding to thecolumn) leaves node x and enters y. We may interpret this as follows. If weassign potentials λa, λb, λc, λd, λf , λg to the nodes of the graph in Figure 4, thevoltage that will appear in the branches will be, for a typical branch e thatleaves x and enters y, (λx − λy). Hence, λTA gives the voltage vector derived

from the potential vector λT by taking differences of potential at the endpointsof each edge. We claim,

Theorem 1 A vector y on the branches of a graph G satisfies KVL iff

(a) algebraic sum of the values y(e) around each loop in the graph is zero.

equivalently,

(b) the vector y can be ’derived’ from a potential vector λ on the nodes of thegraph. (Here derived means y(e) = λx − λy, when e leaves x and entersy.)

Let us examine the above statement keeping the graph G of Figure 4 in mind.Consider the vector

(y(e1), y(e2), . . . , y(e7)).

If it satisfies condition (a) above, we must have (taking clockwise orientationfor loops)

−y(e1) + y(e2)− y(e3) = 0

y(e7) = 0

−y(e6)− y(e4) + y(e3) + y(e5) = 0

To show condition (a) implies condition (b), we must find a suitable potentialvector (λa λb λc λd λf λg) from which y can be derived. We could proceed’common sensically’. Assign potential λf to node f , travel to d by e6 and setλd = λf + v6, set λb = λd + v4, set λc = v3 + λb, set λa = v2 + λc and setpotential λg to g.

But will the resulting potential vector come out to be the same if we traversethe nodes using a different route? The answer is yes if condition (a) is satisfied.Because if we reach say node a from node f through two different paths andget two different values, say λf + v6 + v4 + v1 and λf + v5 + v2, then v6 + v4 +v1 − v5 − v2 6= 0, i.e. some loop will have algebraic sum not equal to zero. (Ingeneral, two such routes will not give a simple loop but it can always be brokenup into simple loops and edges traversed forward and backward.)

To prove condition (a) from condition (b), consider the LHS of the equation

−y(e1) + y(e2)− y(e3) = 0.

If y is derived from λ, the LHS is −(λb − λc) − (λa − λb) + (λa − λc). Sinceevery term λx occurs once with a + sign and once with a - sign, this reduces to

6

zero.Thus we see that a vector y satisfies condition (a) iff it satisfies condition (b).

The above argument can be made rigorous and general and not based on anexample. However the essential ideas of the proof are as given above. We willtake Theorem 1 as true.

It follows by the discussion preceding the theorem that for a vector y onbranches of a graph to satisfy KVL is equivalent to say

yT = λTA for some λ.

We are now in a position to prove the fundamental theorem of electrical network.

Theorem 2 (Tellegen’s Theorem (weak form)) Let G be a graph. Let x, ybe vectors on E(G). Then, if y satisfies KVL and x satisfies KCL,

yTx = 0.

Proof We have

yTx = (λTA)x (since y satisfies KVL)

= λT (Ax)

= λT (0) (since x satisfies KCL)

Theorem 3 (Tellegen’s Theorem (Strong form)) Let Vv(G),Vi(G) denotethe collection of all branch vectors which satisfy KVL, KCL respectively of thegraph G. Then

(Vv(G))⊥ = Vi(G)

Proof By Tellegen’s theorem(weak form), we have

Vi(G) ⊆ (Vv(G))⊥

Let x ∈ (Vv(G))⊥. We then have yTx = 0 whenever y ∈ Vv(G), i.e., (λ

TA)x =

0 for every λ.Choose λ = [1, 0, . . . , 0], [0, 1, . . . , 0], . . . It follows that x ⊥ each row of A

i.e. Ax = 0. Thus(Vv(G))

⊥ ⊆ Vi(G)

The reverse containment is shown before already. So we have

(Vv(G))⊥ = Vi(G).

7

v,i

Figure 5

One consequence of Tellegen’s theorem is a result which we often take forgranted: The net power absorbed by all the devices in the network is zero.Recall that if we use the same arrow to represent both voltage and current ofa device, then the power absorbed by the device is v.i. Hence, the net powerabsorbed by all the devices is vT i = Σvjij = 0 since v satisfies KVL and i

satisfies KCL.However, Tellegen’s theorem is not a consequence of conservation of power.

Consider the circuits in Figure 6.

v1, i1

v3,i3 v4,i4

v2,i2

v5

v1’,i1’v2 2’’,i

v4’,i4’v3’,i3’

Figure 6

Tellegen’s Theorem implies not only

vT i = 0 , (v′)T i′ = 0

but alsovT i′ = 0 , (v′)T i = 0

Surely, the latter two equations have nothing to do with power!Next, let us examine the difference between the weak and strong forms of

Tellegen’s theorem. Consider the vectors along x-axis and z-axis in Figure 7.Clearly, all these pairs of vectors are perpendicular to each other. On the otherhand, the set of all vectors perpendicular to every vector along z-axis is morethan the set of all vectors along the x-axis, it is actually all vectors in the x-yplane.

The strong form is like the second statement. If Vv(G) is the collection of allvectors along the z-axis, Vi(G) is the x-y plane.

8

x

y

z

Figure 7

Exercise

Build a 3-element network whose spaces Vi(G), VvG are line through the originand plane through the origin.

Tutorial on Tellegen’s Theorem

1. In a phasor(sinusoidal steady state) circuit, show that

(a) vT i = 0,

(b) (v∗)T i = 0,

(c) (v∗)T i∗ = 0,

(d) vT i∗ = 0.

Recall that only (b) and (c) yield real and reactive power as real andimaginary parts and have ’physical significance’. Nevertheless, the othersare true too.

2. Enclose a subset of nodes of a network in a surface. Show that the netcurrent leaving the surface is zero.

3. Consider a 2-port network made up of resistors.

(a) Let v1, v2, i1, i2 denote the port voltages and currents and v′1, v′

2, i′

1, i′

2;v′′1 , v

′′

2 , i′′

1 , i′′

2 denote distinct port conditions. Show that

v′1i′′

1 + v′2i′′

2 = v′′1 i′

1 + v′′2 i′

2.

(b) Generalize to n-ports.

9

(c) Show that if R or G matrix exists for this network, it must be sym-metric.

(d) What properties must the hybrid ((v1, i2) on one side and (i1, v2) onthe other) and ABCD matrices ((v1, i1) on one side and (v2, i2) onthe other) have?

4. Consider a 2-terminal network made up of positive resistors. Show that ifany of the resistors is increased in value, the overall resistance increases.

5. Let N be a network composed of resistors, voltage sources and currentsources. Let PεJ , Pε, PJ denote the total power absorbed by the resis-tors respectively, when both voltage and current sources are active, onlyvoltage sources are active and only current sources are active. Show that

PεJ = Pε + PJ .

(assume that the network has unique solution for arbitrary values ofsources).

6. (a) In a network with positive resistors and a single current source, showthat no branch current can exceed the current source value.

(b) State and prove the voltage version of the above statement.

(c) What if there were both voltage and current sources?

(d) Will the above results be true for any devices more general thanresistors?

7. Let N be a network with current sources and resistors and which has aunique solution. Consider the solutions [iJ , iR] of KCL agreeing with thegiven currents iJ of the current sources. Among all such solutions, showthat the one that minimizes power loss in the resistors will correspond tothe actual current in the network.

Minty’s Theorem

A result which is as basic to electrical network theory as Tellegen’s theoremis Minty’s theorem. Indeed, it is possible to show that the two theorems areformally equivalent. In this section, we will state and prove this theorem in itsweak form, state the strong form and indicate its proof. We will also considera few applications to electrical networks and graphs. But we need a few pre-liminary definitions of basic notions about graphs. The underlying concepts,however, are important in their own right.

Subgraphs

Unless otherwise stated, let us henceforth use ’graph’ to mean directed graph,i.e., every edge in the graph has an arrow, or equivalently, has a designated pos-itive and negative node.A graph is said to be connected if we can start from any

10

node x and reach any other node y in the graph by traversing a sequence (called‘alternating sequence’) of nodes and edges of the form x, e1, v2, e2, . . . , ek, y,where each node is an endpoint of the succeeding and preceding edges wheneversuch edges exist. Figure 8 shows a disconnected graph.

5

a

b

d

f g

e

e

e

e

ee

e

c

1

6

7

2

3

4

Figure 8: Disconnected graph G

Observe that from the node ’b’, we cannot reach the node ’d’ through analternating sequence.

Let G be a graph on vertices V and edges E. We say a graph G1 on verticesV1 and edges E1 is a subgraph of G iff

(a) V1 ⊆ V

(b) E1 ⊆ E

(c) the endpoints of each edge e ∈ E1 are the same in G and G1.

3

e2

b c

d f

e

ee

e

e

1 2

3

4 5e

6

a

cb

2e1e

a

3ec3e

b

a

1e

.d

g g g1 2

Figure 9: Subgraphs

We often use terms such as subgraph on an edge set E1 or subgraph on avertex set V1. By a subgraph on E1, we mean a subgraph of the original graphwith edges in E1 and vertices, those that are endpoints of edges in E1. Similarly,by a subgraph on V1, we mean a subgraph with vertices V1 and edges those thathave both end points in V1. When we deal with directed graphs, we will alsomaintain the same arrow direction for an edge e ∈ E1 in both G and G1. Wesay a subgraph is proper if it is not the same as the original graph.

11

Maximal and Minimal

We often need to use the words ’maximal’ and ’minimal’ as adjectives qualifyingsubsets with respect to a specified property. Consider the ’family’ of subsetsshown within curly brackets below :-

1, 2, 3, 4, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

In this family, the subset 1 is maximal in the sense that no other subset inthe family properly contains this subset. The subset 2, 3, 4 is not maximalsince 2, 3, 4, 5 ⊇ 2, 3, 4 and 2, 3, 4, 5 also belongs to the family. The subset6, 7, 8, 9, 10, 11, 12 is, ofcourse, maximal in this family. On the other hand, 1is minimal in the sense that no other subset in the family is a proper subset ofthis subset. (The only such subset is the null subset which is not in this family).The subset 2, 3, 4, 5 is not minimal since 2, 3, 4 ⊆ 2, 3, 4, 5. Both subsets2, 3, 4 and 6, 7, 8, 9, 10, 11, 12 are minimal. Observe that the words minimaland maximal do not imply minimum size and maximum size respectively, since1 is maximal with respect to the property of being in the above mentionedfamily and 6, 7, 8, 9, 10, 11, 12 is minimal.We can extend the use of these adjectives to objects other than subsets byextending the idea of ’contained in’. If G1 is a subgraph of G, it is, in anintuitive sense, ’contained in’ G.

For instance, consider the graph G in Figure 8. Let us call the subgraph onthe left as Gleft. Clearly, Gleft is a connected subgraph of G with the property thatit is not a proper subgraph of any other connected subgraph of G. Therefore,one can say Gleft is a ’maximal connected subgraph of G’. A maximal connectedsubgraph of a given graph is called a connected component of the graph.

In general, ’maximal’ and ’minimal’ are with respect to a ’partial ordering’of objects. A partial order ’≥’ on a set S has the following properties. For everya, b ∈ S, we have a ≥ b or b ≥ a. If both a ≥ b and b ≥ a, then a = b. Further,if a, b, c ∈ S such that a ≥ b, b ≥ c, then a ≥ c. With respect to the partialorder ’≥’, an element x is maximal iff no y ∈ S is such that x ≤ y and x 6= y.Minimal is defined similarly.

Circuits, Crossing edge sets and Cutsets

A circuit graph is a connected graph with every vertex having exactly two edgesincident on it. In a given graph, a set of edges forms a circuit iff the subgraphon it is a circuit graph. For the graph G in Figure 9, the subgraph G2 in thesame figure is a circuit subgraph. The set of edges e1, e2, e3 is a circuit.

We use the term ’directed circuit’ as follows. First in the given circuit graphdefine an orientation as shown in Figure 10. If all arrows support or all arrowsoppose this orientation, then we say that the circuit graph is oriented and thecorresponding circuit is directed.

The circuit graph of Figure 10 is not oriented, while the one in Figure 11is. [If one wishes to be formal, ’orientation’ can be defined to be an ’alternating

12

3

b

f

ee

e

e e

c

d

a

1

2

5

4

Figure 10: Orientation for a circuit graph

sequence’ of edges and vertices starting and ending at the same vertex with eachsucceeding edge being incident at the preceding vertex.]

3

b

f

ee

e

e e

c

d

a

1

2

5

4

Figure 11: Oriented circuit graph

A crossing edge set of a graph on vertex set V is defined as the set of alledges with one end point in V1 and another in V2, where V1, V2 is a partitionof V (i.e. V1 ∪ V2 = V and V1 ∩ V2 = φ). (In Figure 12, e7, e8, e9 is a crossingedge set.)

1

e e

e

e

e

e

ee

e

v

v

1 2

3

4

56

7

89

2

Figure 12: Crossing edge set

We say the crossing edge set is directed if all the arrows are directed from

13

V1 to V2 or from V2 to V1. Figure 13 shows a directed crossing edge set.

1

e

ee

v

v

7

89

2

Figure 13: Directed crossing edge set

A cutset is a minimal crossing edge set of the graph, i.e., it is a crossing edgeset which does not properly contain another. Figure 14(a) shows a crossing edgeset that is not minimal since it contains the ones in Figure 14(b).

(b)

1

e2

e2

..

.

a

c

b

a.

.

.

b

c. .

.

a b

c

e e1 2

(a)

e

Figure 14: Non minimal crossing edge set

Given a crossing edge set that is not minimal, how does one extract a cutsetout of it?

We first obtain a neat way of visualizing cutsets. Let us begin with a con-nected graph G. Let C be a crossing edge set corresponding to a partition(N1, N2) (i.e. C is the set of all edges with one end point in N1 and anotherin N2). Then C is a crossing edge set iff the subgraphs of G on N1, N2 areconnected.

Suppose the subgraphs on N1, N2 are connected. We claim that C is aminimal crossing edge set. To see this, note that removal of a crossing edgeset (say corresponding to N1, N2) disconnects the graph since there no longerwould be any edges between N1 and N2. Now let C be a proper subset of C.Observe that removal of C cannot disconnect the graph since the subgraphs on

14

2

Connected

Connected

e ee

N

N

1 23

1

Figure 15: Cutset

N1 and N2 are connected and C − C has atleast one edge which now connectsthese two connected graphs. For instance, in Figure 16, to go from a in N1 tod in N2, go from a to b, which is possible since subgraph on N1 is connected,travel from b to c through edge e and travel from c to d in N2, which is possiblesince subgraph on N2 is connected. This proves that C is a minimal crossingedge set, i.e. it is a cutset.

d

N

N

1

2.

.

..

e

ab

c

Figure 16

Suppose the subgraph on N1 is not connected and N1 can be partitioned intonode sets N11,N12, where the subgraph on N11 is connected. Now, consider the

15

node partition (N11,N2 ∪ N12). The corresponding crossing edge set is clearlycontained in the original crossing edge set. (Note that there are edges betweenN11 and the rest of the nodes since the graph is connected. All these edges arebetween N11 and N2, since there are no edges between N11 and N12). Now wehave a crossing edge set C2 between N11 and N2 ∪N12 where the subgraph onN11 is connected but that on N2 ∪N12 may not be. If the latter case happensto be true, N2 ∪ N12 can be partitioned as N21, N22, . . . , N2k such that thesubgraphs on each subset of these nodes are connected.

(a)

N11 N12N11

N12N3

N4 N3

N4

C C1 2

N U NN

2 12

(b)

Figure 17

Since the original graph G is connected, there must be edges between N11

and each of the sets N21, N22, . . . Consider the partition of vertices N−N21, N21

The set of edges C3 between these node sets is clearly contained in the previousset C2. But we know that the subgraph on N − N21 is connected (since thesubset of vertices N11 is linked to all the vertices in N − N21 and further thesubgraph on N11 is connected). This sequence of steps is illustrated in Figures17 and 18.

We are now in a position to state and prove Minty’s theorem weak form.

Theorem 4 Minty’s Theorem (weak form) Let G be a directed graph. Lete be an edge in G. Then one and only one of the following is true.

(a) e belongs to a directed cutset.

(b) e belongs to a directed circuit.

Proof We first note the (a) and (b) cannot both be true. Figure 19 should makeit clear. If e belongs to both a directed cutset and a directed circuit, movingfrom a to l along e and back again to a is not possible without encountering anarrow in the opposite direction.

Next, let us show that one of the two eventualities must necessarily occur.Start from b and move away from it always choosing outward arrows. Let N2

be the set of all vertices reached by this process. (Whenever you reach a vertex,

16

C

(c)

NN

N N

114

3 12

3

Figure 18: Extracting a cutset

4ee e e

.b

.a

12 3

Figure 19: Directed circuit and cutset

record, next to it, the last vertex from which it was reached. This way thedirected path by which a vertex was reached is also remembered.) One of twothings must happen. Either a ∈ N2 or a /∈ N2. If a ∈ N2, we can start froma, move to b through the directed edge and then get back to a by a directedpath, which will give us a directed circuit. If a /∈ N2, consider the partitionof the vertex V (G) into (V (G)−N2,N2). Clearly, a ∈ V (G)−N2. All arrowsmust be directed (including ofcourse e) from V (G)−N2 to N2. If e′ is directed

17

from N2 to V (G)−N2, the head of the arrow of e′, which is outside N2, isclearly reachable by a directed path. But this contradicts the definition of N2.We conclude, therefore, that the partition ((V (G)−N2),N2) corresponds to adirected crossing edge set.

Minty’s Theorem (Strong form)

The strong form of Minty’s theorem is ‘colourful’. It is usually called colouredgraph theorem.

Theorem 5 Minty’s Coloured Graph Theorem Let G be a directed graph.Let edges of G be coloured red, blue or green. Let e be a green edge. Then oneand only one of the following must occur.

1. e belongs to a directed cutset with only green and red edges, in which allthe green edges are directed the same way, with respect to an orientationof the cutset.

2. e belongs to a circuit with only green and blue edges, in which all thegreen edges are directed the same way with respect to an orientation of thecircuit.

Proof It is easily seen that both of these situations cannot occur simultane-ously. This part of the proof is essentially the same as for the ‘weak form’case.

To see that one of the two must occur we proceed as follows:Open circuit all the red edges (i.e., remove the edges, leaving the end points

in place) and short circuit all the blue edges (remove the edge, but fuse the endpoints).

The resulting graph, Greduced, satisfies Minty’s Theorem (weak form). So ebelongs to a directed circuit or a directed cutset in Greduced.

Suppose e belongs to a directed circuit, Lreduced in Greduced. Observe thateach node of Greduced is a group of nodes connected through blue edges in G.So Lreduced can be expanded into a circuit in G which has only green and blueedges and where all the edges of Lreduced (green) are directed the same waywith respect to an orientation of the circuit.

Suppose e belongs to a directed cutset in Greduced. There will be a partitionof its vertex set corresponding to this cutset. Each vertex of Greduced is a groupof vertices of G fused together. Also each vertex in G belongs to one suchcluster which is a vertex of Greduced. Therefore, when vertices of Greduced arepartitioned, this partition can be made into a partition of the vertices of Gby making the vertices of Greduced into the corresponding cluster of G. Nowreintroduce the blue edges. Clearly these must have both endpoints within oneof the two blocks of the partition. (Otherwise when the blue edge is shorted itsend points will lie in both blocks in Greduced which contradicts the fact that theblocks form a partition of vertices of Greduced).However some red edges may now

18

lie with one end point in one block and another in the other block. Thereforethe original green cutset now becomes a green red cutset. (Observe that insidethe two blocks of vertices we will continue to have connected subgraphs evenafter the fused vertices of Greduced are enlarged). Since originally the greenedges were directed the same way they will continue to be directed the sameway even in G with all the green edges directed the same way with respect tothe corresponding partition of vertices.

Planar Networks and Duality

A graph is said to be planar if it can be drawn on a plane without edges crossingeach other. The graph in Figure 20a is planar because although it is drawn sothat edges are crossing each other, it can be redrawn as in Figure 20 so thatthey do not.

(b)(a)

e e

7

.

.

..

m

m

m

0

1m

2

3

e e

ee

ee

e

e

1

2

3

4 5

6

8

Figure 20: Planar Graph

When a planar graph is drawn on a plane without edges crossing, it divides theplane into regions (or windows or meshes) including the outermost regions.

Consider the graph in Figure 21.Choose a clockwise orientation for all the interior meshes and an anticlockwiseorientation for the outermost mesh. Construct the mesh matrix where each rowcorresponds to a mesh with the entry corresponding to edge e equal to +1 if theorientation of e agrees with the orientation of the mesh, equal to -1 if it opposes

19

8

m

m

mm

0

1 2

3

e

eee

e

e

e

e

1

2

4

3

5

6

7

Figure 21: Planar Graph with Oriented Meshes

and zero if the edge e is not in the mesh. For the graph of Figure 21, the meshmatrix is as follows

M =

−1 1 1 −1 0 0 0 00 0 −1 0 −1 1 1 00 0 0 1 1 0 0 −11 −1 0 0 0 −1 −1 1

Observe that each edge is in exactly two meshes (including the outermost mesh)and its orientation agrees with that of one of the meshes and opposes that ofthe other. It follows that each column has exactly one +1 and one -1 in it.This makes the mesh matrix of the original graph into the incidence matrix ofanother graph.

Theorem: Let a planar graph G be connected. Then the number of meshesof G (including the outermost) is e− v + 2, where e is the number of edges andv the number of vertices.

Proof: Observe that a tree graph is a planar graph and has v−1 edges and a sin-gle (outermost) region. Thus for a tree, the number of regions is e− v+2(= 1).For any graph G, let m(G) ≡ e− v + 2. Now, let the theorem be true for graphfor which m(G) ≤ k, where k ≥ 1.

Let the given graph G satisfy m(G) = k + 1. Since G is not a tree, it mustcontain atleast one circuit and therefore, the graph when drawn on the planewill divide it into atleast two regions. Let e lie in one of these meshes (i.e. re-gions). Open circuit e. In the resulting graph G, the two regions in which e lieswill merge into a single region. Hence, the number of rows in the mesh matrixwill reduce by 1. However, G satisfies m(G) =k (since the number of edges hasreduced by 1 and the number of vertices remains the same). Hence, the meshmatrix of G has k rows. It follows that the mesh matrix of G has k + 1 rows.

Q.E.D.

The mesh matrix of G is the incidence matrix of another graph Gd, which we

20

will call the dual of G. This graph has m(G) nodes and the same number ofedges as G. Gd can be built by inspection as follows:-

1. Spread G on the plane

2. Place one node nj in each mesh mj of G

3. Give each mesh of G a clockwise orientation and the outermost mesh ananticlockwise orientation.

4. Whenever edge e in G lies at the boundary of meshes mi and mj , the edgee′ in Gd joins nodes ni and nj . If direction of e agrees with orientationmi, it will oppose the orientation of mj and e′ will then be directed fromni to nj .

Observe that when G is spread on the plane, we can move from any region toany other region by repeatedly moving from a region to another which shares aboundary edge with it. It follows that Gd is connected (this is true even whenG is not). Hence dropping one row of the incidence matrix of Gd will result ina set of independant rows. Hence the space spanned by the rows of the meshmatrix of G has dimension e − v + 1. But each row of this matrix is a currentvector of G. Hence the rows of the mesh matrix of G span Vi(G) (since Vi(G)has dimension e− v + 1). We know that this matrix is the incidence matrix ofG. It follows that

Vi(G) = Vv(Gd)

and therefore by Tellegen’s Theorem (strong form),

Vv(G) = (Vi(G))⊥ = (Vv(G

d))⊥ = Vi(Gd)

The above fact can be used to build ‘dual’ networks, where currents of onenetwork behave like voltages of the other.

Let N = (G,D) be a network defined on a planar graph. (Here the devicecharacteristic is a collection of pairs of vectors (v, i), which could be taken asfunctions of time). Define Dd ≡ (x,y) : (y,x) ∈ D.

The dual network N d is defined to be (G,Dd). Thus the constraints of thenetwork N and N d are

N : v ∈ Vv(G)i ∈ Vi(G)(v, i) ∈ D

N d : v ∈ Vv(Gd) = Vi(G)

i ∈ Vi(Gd) = Vv(G)

(v, i) ∈ Dd i.e., (i, v) ∈ D

21

To build the dual of a planar network, we first build the dual graph. Thenwe dualize each device or group of coupled devices. Table gives dual pairs ofdevices.

N N d

1. (resistor) (conductance)

v = Ri i = Rv

2. (capacitor) (inductor)

i = C dvdt

v = C didt

3. (inductor) (capacitor)

v = L didt

i = Ldvdt

4. (forward biased diode) (reverse biased diode)

v ≤ 0 i ≤ 0i ≥ 0 v ≥ 0

v.i = 0 v.i = 0

5. (vcvs) (cccs)i1 = 0 v1 = 0

v2 − αv1 = 0 i2 − αi1

6. (ideal transformer) (ideal transformer, turns ratio inverted,one of the windings’ polarity reversed)

v1 − nv2 = 0 i1 − ni2 = 0ni1 + i2 = 0 nv1 + v2 = 0

7. v = ε volts i = ε Amps8. i = J Amps v = J volts

While it is true that dual networks can be constructed only for planar net-works, it is worth noticing that duality ideas are more generally applicable thanjust for planar networks.

Let us define a generalized network M ≡ (V,D), where V is a vector spaceon a set S over R and D (the device characteristics) is a collection of orderedpairs of vectors (x,y), x,y being on S. To solve this network means to find(x,y) s.t.

x ∈ V

y ∈ V⊥

(x,y) ∈ D

We could define the dual network Md ≡ (V⊥,Dd), where Dd ≡ (x,y) :(y,x) ∈ D. It is clear that (x,y) is a solution of M iff (y,x) is a solution of

22

Md. Suppose the device characteristic D is defined by

xε = ε

xR −RyR = 0

We could define generalized loop analysis for M as follows:Let B be a matrix whose rows form a basis for V⊥. We then have

(

Bε BR

)

(

xε

xR

)

= 0

i.e.,BRxR = −Bεxε

using xR = RyR,BRRyR = −Bεxε

using

[

BTε

BTR

]

σ =

[

yε

yR

]

,

BRRBTRσ = −Bεxε (1)

The equations (1) may be called generalized loop analysis for M. We coulddefine generalized cutset analysis for a network with device characteristic

yJ = J

yG −GxG = 0

as follows:Let Q be a matrix whose rows form a basis for V. We then have

(

QJ QG

)

(

yJ

yG

)

= 0

i.e.,QGyG = −QJyJ

using yG = GxG, we have,

QGGxG = −QJyJ

using

[

QTG

QTJ

]

λ =

[

xG

xJ

]

,

QGGQTGλ = −QJyJ (2)

Equations (2) may be called generalized cutset analysis equations. It is clearthat whenever we can perform generalized loop analysis for M, we are essen-tially performing generalized cutset analysis for Md. Thus the two are ‘dual’procedures.

23

Problems

Set 1

1. Build the incidence matrix for the graphs in Figure 22.

h

f gk m

j

d

a

b c

e13

e11 e12

e7

e1 e2

e8

e3

(a)(b)

a b

d c

f

e3

e5

e2 e4e1

e6 e6

e4 e5

e9 e10

Figure 22

2. Let λ be a column vector with entries λa,λb,.... Premultiply the incidencematrix A by λT for the graphs in Figure 22. Assign λa, λb, ... as potentialsto the nodes and recompute λTA directly as a voltage vector correspondingto these potentials.

3. Let x be a column vector with entries x1,x2,.... Post multiply the incidencematrix A by x for the graphs in Figure 22. Assign x1, x2, ... as currentsin the edges and recompute Ax directly as current injected at the nodes.

4. Show that the collection of vectors of the form vT=λTA (λ arbitrary) andx such that Ax=0 form vector spaces. Show that these two vector spacesare complementary orthogonal.

5. Show that summing the rows of an incidence matrix results in a newincidence matrix - similarly for deletion of columns. How are the corre-sponding graphs related to the original graph?

6. How does the incidence matrix of a disconnected graph look? Characterizethe solution space of λTA=0.(Hint: What interpretation can you give to λTA?)

7. Suppose Ax=0 =⇒ x=0 where A is the incidence matrix of graph G.Describe G.

Set 2

1. Prove the rules for working with partitioned matrices. Assume that thepartitions are compatible.

24

(a) A (B1|B2) = (AB1|AB2)

(b)

(

A1

A2

)

B =

(

A1BA2B

)

(c) (A1A2)

(

B1

B2

)

= A1B1 +A2B2

2. Let matrices A, B be row equivalent. Show that a set of columns of Aare linearly independent iff the corresponding columns of B are linearlyindependent.

3. Let E be an elementary matrix. Show that EA has independent rows iffA has independent rows.

4. Show that an elementary matrix E always has an inverse i.e there existsa matrix E′ such that EE′=E′E=I.Hence show that the product of elementary matrices is invertible.Show that a square matrix with linearly independent rows is the productof a sequence of elementary matrices and is therefore invertible.

5. Show that the row rank and column rank of a matrix are the same.

6. Show that all bases (maximally independent subsets of vectors) of a vectorspace always have the same size.

7. Let V be a vector space on a set S over ℜ. Show that

dim(V) + dim(V⊥) = |S|

Set 3

1. For the graphs in Figure 23:

(a) Build all trees using Feussner’s Method.

(b) Construct the incidence matrices.

(c) Find the RRE of the incidence matrices.

2. Solve the circuits in Figure 24.

3. For the graphs in Figure 23:

(a) What are the dimensions of current space and voltage space?

(b) Consider only the collection of all vectors of the form (i1, i2, i3, i4)and of the form (v1, v2, v3, v4).What are the dimensions of the corresponding vector spaces?Can you associate a graph with each of these spaces?

(c) Can you build a basis for the current space which has the form of anincidence matrix?

25

cb

d f

a

e1 e2

e3

(a)

e4 e5

e8

h

ge6 e7

a

h

f

c

ge4

e6

e3

d

b

e1

e7

e5

(b)

e2

a

b

e4

e3e1

e2

(c)

Figure 23

Ω

Ω

3

2

5

1A

4A

2A 3AΩ

Ω1

Ω4

(a)

+

+ +

+−

−

−

−

−+

1

6V

4V

2V 4V

2V

Ω Ω

Ω

Ω

4 2

6

(b)

Figure 24

4. For the vector spaces which are the rowspaces of the matrices given, findthe bases for the complementary orthogonal spaces.

(a)

1 1 2 0 0 02 0 4 1 0 03 0 −1 0 1 04 0 1 0 0 1

(b)

1 0 0 4 5 6−1 1 0 1 −1 00 0 1 0 0 −10 0 0 0 0 0

5. Consider the following collections of matrices.

(a) [0]nxn

(b) All matrices of the formr n− r

26

rn− r

[

0 00 K

]

Do these collections have identity matrices?With respect to the collection which matrices have inverses?

6. Given(

A1 A2

)

(

x1

x2

)

= b, show how you would build an equivalent

system A1x1 = b.

7. Find the inverse of:

(a)

1 2 0 01 0 0 00 0 1 10 0 1 −1

(b)

1 α 00 1 00 0 1

(c)

1 0 0 00 0 1 00 1 0 00 0 0 1

(d)

0 0 1 20 0 1 01 1 0 01 −1 0 0

8. Let A =

[

A11 A12

A21 A22

]

with A11,A invertible. Show that:

A−1 can be expressed in terms of A−1

11explicitly and some other sub-

matrices.

Set 4

1. For the graphs in Figure 25, build the fundamental cutset and fundamentalcircuit matrices with respect to the tree e5, e6, e7, e8, e9.

In each case build the RRE of the matrices.

2. Given two trees t0 and tf of a connected graph, show how to build asequence of trees

t0, t1, t2, ... tf

where each tree differs from the preceding element in a single edge. Illus-trate your scheme by taking t0 = e1, e2, e3, e4, e5 and tf = e5, e6, e7, e8, e9for the graphs in Figure 25.

27

e4

e1 e7e2

e9

e5e6

e8

e3

(b)

e2

e5

e1

e8 e9

e3

e7

e4

(a)

e6

Figure 25

3. For the graphs in Figure 25, build atleast 10 elementary current and volt-age vectors indicating the edges in the support of these vectors.

4. Show that elementary current vector support is a circuit of the graph andthat for elementary voltage vector is a cutset.

5. Show that circuits of a graph satisfy the following properties:

(a) If C1, C2 are distinct circuits of G, neither is properly contained inanother.

(b) If e1 ∈ (C1 − C2) and ec ∈ (C1 ∩ C2)Show that there exists a circuit C3 ⊆ (C1 ∪ C2) s.t e1∈ C3 andec /∈ C3.

Prove that cutsets of G also satisfy the above properties.

6. Show that a circuit of graph G can intersect a cutset of G only in aneven number of edges. Define directed circuits and directed cutsets in a’natural’ manner. Show that no edge in G can belong to a directed circuitas well as a directed cutset.

7. Show the following for a graph G:

tree ≡ minimal set of edges of G for which corresponding induced subgraph isconnected and has all vertices of G.

≡ minimal set of edges intersecting every cutset≡ maximal set of edges containing no circuit

cutset ≡ minimal set of edges intersecting every treecircuit ≡ minimal set of edges intersecting every cotree

28