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1.3 Matrices and Graphs

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1.3 Matrices and Graphs Thus Tˆ(bˆj) = ajibˆi, �i and the matrix of Tˆ with respect to Bˆ isA T , the transpose of the matrix ofT with respect toB. Remark: Natural isomorphism betweenV and its double dual Vˆ. We have seen above that everyfinite dimensional vector spaceV has the same dimension as its dual Vˆ, and hence they are isomorphic as vector spaces. Once we choose a basis forV we also define a dual basis for Vˆ and the correspondence between the two bases gives us an explicit isomorphism betweenV and Vˆ. However this isomorphism is not intrinsic to the spaceV, in the sense that it depends upon a choice of basis and cannot be described independently of a choice of basis. ˆ The dual space of Vˆ is denoted Vˆ; it is the space of all linear mappings from Vˆ toF. By all of the above discussion it has the same dimension as Vˆ andV. The reason for mentioning this however is that there is a “natural” isomorphismθ fromV to Vˆ. It is defined in the following way, ˆ forx V andf Vˆ - note thatθ(x) belongs to Vˆ, soθ(x)(f) should be an element ofF. ∈ ∈ θ(x)(f) =f(x). To see thatθ is an isomorphism, suppose thatx 1,...,x k are independent elements ofV. Then ˆ θ(x1),...,θ(x k) are independent elements of Vˆ. To see this leta 1,...,a k be element ofF for which a θ(x ) + +a θ(x ) = 0. This means thatf(a x + +a x ) = 0 for allf Vˆ, which means 1 1 ··· k k 1 1 ··· k k ∈ thata x + +a x = 0, which means that eacha = 0. 1 1 ··· k k i 1.3 Matrices and Graphs A directed graph (or digraph)G consists of a non-empty setV of vertices and a setE of ordered pairs of these vertices, called edges. Each edge is directed from one vertex ofG to another. An undirected graph is similar, except that the edges are not considered to have a direction. A number of square matrices are typcially associated to a graph, the most elementary of which is the adjacency matrix. Definition 1.3.1. LetG be a (directed) graph withn vertices labelledv 1,...,v n. The adjacency matrix A ofG is then n matrix whose entries are given by × 1 there is an edge directed fromv tov inG A = i j ij 0 otherwise � Example 1.3.2. A directed graph and its adjacency matrix. 19 v1 v2 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0 A= 1 0 0 0 0 0 v3 v 4 0 0 1 1 0 1 1 0 0 0 0 0 v6 v5 Ifu andv are vertices in a directed graph, a (directed) walk fromu tov is a sequence of ver- tices that starts atu andfinishes atv, with the property that every pair of consecutive entries is a directed edge. The length of a path is the number of edges in it. In the example above, v5,v 6,v 1,v 2,v 1,v 2,v 3 is a directed walk of length 6 fromv 5 tov 3. The adjacency matrix has the interesting property that its powers count directed walks. Theorem 1.3.3. LetG be a directed graph with adjacency matrixA (with respect to the orderingv 1,v 2,...,v n of the vertices). For every positive integerk, the(i,j) entry ofA k is the number of walks of lengthk from vi tov j inG. Proof. By induction onk. The casek= 1 is just the definition of the adjacency matrix. So assume that the theorem is true fork=m− 1 and considerk=m. Then m m−1 (A )ij = AilAlj . �l Every path of lengthm fromv i tov j must start with an edge fromv i to somev l and follow that with a path of lengthm− 1 fromv l tov j. For eachl, the entryA il is 1 if(v i,v l) is an edge and m−1 0 otherwise. By the induction hypothesisA lj is the number of directed walks of lengthm−1 m−1 fromv l tov j inG. Thus for each vertexl, the integerA ilAlj is the number of walks of length m fromv i tov j that havev l as their second vertex. The sum overl of these is the total number of walks of lengthm fromv i tov j inG. This completes the induction proof. An undirected graph resembles a directed graph except that the edges are unordered pairs of vertices. The adjacency matrix of an undirected graph is symmetric. Note that the adjacency matrix of a (directed or undirected) graphG depends not only on the graph itself but also on the choice of an ordering of the vertices. Suppose thatσ is a permutation of the set{1, . ,n}. LetA be the adjacency matrix ofG with respect to the orderingv 1, dots,v n of the vertices, and letA � be the adjacency matrix with respect to the orderingv σ(1),...,v σ(n). Then A� is obtained fromA by first reordering the columns by replacing Column 1 with Columnσ(1), Column 2 with Col- • umnσ(2), and so on. This means multiplying on the right by the matrixP σ, in which each Columnj (for eachj) has a 1 as itsσ(j)-th entry and is otherwise full of zeros. 20 then reordering the rows by replacing Row 1 with Rowσ(1), Row 2 with Rowσ(2), etc. This • T −1 means multiplyingA on the left by the matrix(P σ) , which is also equal toP σ . A permutation matrix is a matrix that has exactly one 1 in each row and column and is oth- • erwise full of zeros. Every permutation matrix has the property that its inverse is equal to its transpose. We have shown that adjacency matricesA andA � represent the same graph if and only if T A� =P AP, for some permutation matrixP. This relation is known as permutation equivalence; it is a special case of both similarity and congruence. The adjacency matrix is one of a number of matrices often associated with a graph. We men- tion a few more. Definition 1.3.4. LetG be an undirected graph withn verticesv 1,...,v n andm edgese 1,...,e m. The incidence matrix ofG is then n matrixC that has a1 in position(i,j) if the vertexv is • × i incident with the edgee j, and zeros elsewhere. An oriented incidence matrix ofG is then m matrixB defined byfirst assigning a direction to • × every edge ofG and then setting 1 ifv i is the start vertex ofe j B = −1 ifv is the end vertex ofe ij i j 0 otherwise The oriented incidence matrix depends on a choice of ordering of both the vertices and edges, and on a choice of orientation of the edges. Given matrices that are associated with graphs, a general philosophy is to consider how the properties of the matrix and the graph are related to each other. In the case of an oriented inci- dence matrix, the rank of the matrix tells us about the number of connected components in the graph. Theorem 1.3.5. LetG be a simple graph withn vertices andm edges, and letB be an oriented incidence matrix ofG. Then the rank ofB isn−t, wheret is the number of connected components ofG. Proof. First suppose thatG is connected. This means that for any pair of verticesv i andv j inG, there exists a walk inG fromv i tov j. We consider the left nullspaceN of the matrixB. Note that every column ofB has one entry equal to 1, one equal to−1, and is otherwise full of zeros. This means that the vector(1 1 . 1) belongs to the left nullspace ofV, so this nullspace is at least 1-dimensional. Suppose that(a a ...a ) is a non-zero element ofN, and choosek for whicha = 0, write 1 2 n k � ak =α. Then(a 1 a2 ...a n) must satisfy(a 1 a2 ...a n)v= 0 for every columnv ofB, and in particular for those columns corresponding to edges that are incident with the vertexv k. It follows thata i =a k =α wheneverv i is adjacent tov k. Now by the same reasoning applied to the neighbours ofv k, we must havea j =α wheneverv j is adjacent to a neighbour ofv k. SinceG is connected, repetition of this step reaches all vertices ofG and we conclude thata i =α for alli and thatN is a 1-dimensional space. Thusn=1+ rank(B) and rank(B) =n− 1. Now suppose thatG hast connected components and let their numbers of vertices ben 1,n 2,...,n t, withm 1,m 2,...,m t edges respectively. By ordering the vertices component by component, we can arrange thatB has a rectangular block diagonal structure with an n block in the upper 1 × 1 left, etc. Each block is an oriented incidence matrix of a connected component ofG, and so a block withn i vertices has rankn i − 1. It follows that the total rank is (n −1) + (n −1) + +(n −1) =n−t. 1 2 ··· t 21 Theorem 1.3.6. LetB be an oriented incidence matrix of an undirected simple graphG. Then BBT =D−A, whereD is the diagonal matrix whose diagonal entries are the total degrees of the vertices, andA is the adjacency matrix ofA. Proof. Fori= 1, . ,n, the entry in the(i,i)-position ofBB T is just the ordinary scalar product of Rowi ofB with itself.
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