THE MATRIX-TREE THEOREM

THE MATRIX-TREE THEOREM 1 The Marix-Tree Theorem. The Marix-Tree Theorem is a formula for he number of spanning rees of a graph in erms of he deerminan of a cerain marix. We begin wih he necessary graph-heoreical background. Le G be a finie graph, allowing muliple edges bu no loops. (Loops could be allowed, bu hey urn ou o be compleely irrelevan.) We say ha G is conneced if here exiss a walk beween any wo verices of G. A cycle is a closed walk wih no repeaed verices or edges, excep for he he firs and las verex. A ree is a conneced graph wih no cycles. In paricular, a ree canno have muliple edges, since a double edge is equivalen o a cycle of lengh wo. The hree nonisomorphic rees wih five verices are given by: A basic heorem of graph heory (whose easy proof we leave as an exercise) is he following. 1.1 Proposiion. Le G be a graph wih p verices. The following condiions are equivalen. (a) G is a ree. (b) G is conneced and has p 1 edges. (c) G is has no cycles and has p 1 edges. (d) There is a unique pah (= walk wih no repeaed verices) beween any wo verices. 1

A spanning subgraph of a graph G is a graph H wih he same verex se as G, and such ha every edge of H is an edge of G. If G has q edges, hen he number of spanning subgraphs of G is equal o 2 q, since we can choose any subse of he edges of G o be he se of edges of H. (Noe ha muliple edges beween he same wo verices are regarded as disinguishable.) A spanning subgraph which is a ree is called a spanning ree. Clearly G has a spanning ree if and only if i is conneced [why?]. An imporan invarian of a graph G is is number of spanning rees, called he complexiy of G and denoed κ(g). 1.2 Example. Le G be he graph illusraed below, wih edges a, b, c, d, e. b a e c d Then G has eigh spanning rees, namely, abc, abd, acd, bcd, abe, ace, bde, and cde (where, e.g., abc denoes he spanning subgraph wih edge se {a, b, c}). 1.3 Example. Le G = K 5, he complee graph on five verices. A simple couning argumen shows ha K 5 has 60 spanning rees isomorphic o he firs ree in he above illusraion of all nonisomorphic rees wih five verices, 60 isomorphic o he second ree, and 5 isomorphic o he hird ree. Hence κ(k 5 ) = 125. I is even easier o verify ha κ(k 1 ) = 1, κ(k 2 ) = 1, κ(k 3 ) = 3, and κ(k 4 ) = 16. Can he reader make a conjecure abou he value of κ(k p ) for any p 1? Our objec is o obain a deerminanal formula for κ(g). For his we need an imporan resul from marix heory which is ofen omied from a beginning linear algebra course. This resul, known as he Bine-Cauchy heorem (or someimes as he Cauchy-Bine heorem), is a generalizaion of he familiar fac ha if A and B are n n marices, hen de(ab) = 2

de(a) de(b) (where de denoes deerminan) 1. We wan o exend his formula o he case where A and B are recangular marices whose produc is a square marix (so ha de(ab) is defined). In oher words, A will be an m n marix and B an n m marix, for some m, n 1. We will use he following noaion involving submarices. Suppose A = (a ij ) is an m n marix, wih 1 i m, 1 j n, and m n. Given an m-elemen subse S of {1, 2,..., n}, le A[S] denoe he m m submarix of A obained by aking he columns indexed by he elemens of S. In oher words, if he elemens of S are given by j 1 < j 2 < < j m, hen A[S] = (a i,jk ), where 1 i m and 1 k m. For insance, if and S = {2, 3, 5}, hen 1 2 3 4 5 A = 6 7 8 9 10 11 12 13 14 15 A[S] = 2 3 5 7 8 10 12 13 15. Similarly, le B = (b ij ) be an n m marix wih 1 i n, 1 j m and m n. Le S be an m-elemen subse of {1, 2,..., n} as above. Then B[S] denoes he m m marix obained by aking he rows of B indexed by S. Noe ha A [S] = A[S], where denoes ranspose. 1.4 Theorem. (he Bine-Cauchy Theorem) Le A = (a ij ) be an m n marix, wih 1 i m and 1 j n. Le B = (b ij ) be an n m marix wih 1 i n and 1 j m. (Thus AB is an m m marix.) If m > n, hen de(ab) = 0. If m n, hen de(ab) = (de A[S])(de B[S]), S 1 In he addiional maerial handou (Theorem 2.4) here is a more general deerminanal formula wihou he use of he Bine-Cauchy heorem. However, he use of he Bine-Cauchy heorem does afford some addiional algebraic insigh. 3

where S ranges over all m-elemen subses of {1, 2,..., n}. Before proceeding o he proof, le us give an example. We wrie a ij for he deerminan of he marix (a ij ). Suppose [ ] c 1 d a a a 1 1 2 3 A =, B = c b1 b 2 b 2 d 2. 3 c 3 d 3 Then a a 1 2 c1 d de( AB) = 1 a 1 a 3 c 1 d 1 a 2 a 3 c2 d 2 + + b 1 b 2. c 2 d 2 b 1 b 3 c 3 d 3 b 2 b 3 c 3 d 3 Proof of Theorem 1.4 (skech). Firs suppose m > n. Since from linear algebra we know ha rank(ab) rank(a) and ha he rank of an m n marix canno exceed n (or m), we have ha rank(ab) n < m. Bu AB is an m m marix, so de(ab) = 0, as claimed. Now assume m n. We use noaion such as M rs o denoe an r s marix M. I is an immediae consequence of he definiion of marix muliplicaion (which he reader should check) ha [ ][ ] [ ] Rmm S mn Vmn W mm RV + SX RW + SY =. (1) T nm U nn X nn Y nm TV + UX TW + UY In oher words, we can muliply block marices of suiable dimensions as if heir enries were numbers. Noe ha he enries of he righ-hand side of (1) all have well-defined dimensions (sizes), e.g., RV + SX is an m n marix since boh RV and SX are m n marices. Now in equaion (1) le R = I m (he m m ideniy marix), S = A, T = O nm (he n m marix of 0 s), U = I n, V = A, W = O mm, X = I n, and Y = B. We ge [ Im A ][ ] [ ] A O mm Omn AB =. (2) O nm I n In B I n B Take he deerminan of boh sides of (2). The firs marix on he lef-hand side is an upper riangular marix wih 1 s on he main diagonal. Hence is 4

deerminan is one. Since he deerminan of a produc of square marices is he produc of he deerminans of he facors, we ge A Omm O = mn AB ( 3 I n B. ) I n B I is easy o see [why?] ha he deerminan on he righ-hand side of (3) is equal o ± de(ab). So consider he lef-hand side. A nonzero erm in he expansion of he deerminan on he lef-hand side is obained by aking he produc (wih a cerain sign) of m + n nonzero enries, no wo in he same row and column (so one in each row and each column). In paricular, we mus choose m enries from he las m columns. These enries belong o m of he boom n rows [why?], say rows m + s 1, m + s 2,...,m + s m. Le S = {s 1, s 2,..., s m } {1, 2,..., n}. We mus choose n m furher enries from he las n rows, and we have no choice bu o choose he 1 s in hose rows m+i for which i S. Thus every erm in he expansion of he lef-hand side of (3) uses exacly n m of he 1 s in he boom lef block I n. Wha is he conribuion o he expansion of he lef-hand side of (3) from hose erms which use exacly he 1 s from rows m + i where i S? We obain his conribuion by deleing all rows and columns o which hese 1 s belong (in oher words, delee row m + i and column i whenever i {1, 2,..., n} S), aking he deerminan of he 2m 2m marix M S ha remains, and muliplying by an appropriae sign [why?]. Bu he marix M S is in block-diagonal form, wih he firs block jus he marix A[S] and he second block jus B[S]. Hence de M S = (de A[S])(de B[S]) [why?]. Taking all possible subses S gives de AB = ±(de A[S])(de B[S]). S {1,2,...,n} S =m I is sraighforward bu somewha edious o verify ha all he signs are +; we omi he deails. This complees he proof. Le G be a graph wih verices v 1,...,v p. The adjacency marix of G is he p p marix A = A(G), over he field of complex numbers, whose (i, j)-enry a ij is equal o he number of edges inciden o v i and v j. Thus 5

A is a real symmeric marix (and hence has real eigenvalues) whose race is he number of loops in G. We now define wo marices relaed o A(G). Assume for simpliciy ha G has no loops. (This assumpion is harmless since loops have no effec on κ(g).) 1.5 Definiion. Le G be as above. Give G an orienaion o, i.e, for every edge e wih verices u, v, choose one of he ordered pairs (u, v) or (v, u). (If we choose (u, v), say, hen we hink of puing an arrow on e poining from u o v; and we say ha e is direced from u o v, ha u is he iniial verex and v he final verex of e, ec.) (a) The incidence marix M(G) of G (wih respec o he orienaion o) is he p q marix whose (i, j)-enry M ij is given by 1, if he edge ej has iniial verex v i M ij = 1, if he edge e j has final verex v i 0, oherwise. (b) The laplacian marix L(G) of G is he p p marix whose (i, j)-enry L ij is given by { mij, if i j and here are m ij edges beween v i and vj L ij = deg(v i ), if i = j, where deg(v i ) is he number of edges inciden o vi. (Thus L(G) is symmeric and does no depend on he orienaion o.) Noe ha every column of M(G) conains one 1, one 1, and q 2 0 s; and hence he sum of he enries in each column is 0. Thus all he rows sum o he 0 vecor, a linear dependence relaion which shows ha rank(m(g)) < p. Two furher properies of M(G) and L(G) are given by he following lemma. 1.6 Lemma. (a) We have MM = L. (b) If G is regular of degree d (i.e., every verex of G has degree d), hen L(G) = di A(G), where A(G) denoes he adjacency marix of G. 6

Hence if G (or A(G)) has eigenvalues λ 1,...,λ p, hen L(G) has eigenvalues d λ 1,..., d λ p. Proof. (a) This is immediae from he definiion of marix muliplicaion. Specifically, for v i, v j V (G) we have ( MM ) ij = M ikm jk. ek E(G) If i = j, hen in order for M ik M jk 0, we mus have ha he edge e k connecs he verices v i and v j. If his is he case, hen one of M ik and M jk will be 1 and he oher 1 [why?], so heir produc is always 1. Hence ( MM ) ij = m ij, as claimed. There remains he case i = j. Then M ik M ik will be 1 if e k is an edge wih v i as one of is verices and will be 0 oherwise [why?]. So now we ge ( MM ) ii = deg(v i ), as claimed. This proves (a). (b) Clear by (a), since he diagonal elemens of MM are all equal o d. Now assume ha G is conneced, and le M 0 (G) be M(G) wih is las row removed. Thus M 0 (G) has p 1 rows and q columns. Noe ha he number of rows is equal o he number of edges in a spanning ree of G. We call M 0 (G) he reduced incidence marix of G. The nex resul ells us he deerminans (up o sign) of all (p 1) (p 1) submarices N of M 0. Such submarices are obained by choosing a se S of p 1 edges of G, and aking all columns of M 0 indexed by he edges in S. Thus his submarix is jus M 0 [S]. 1.7 Lemma. Le S be a se of p 1 edges of G. If S does no form he se of edges of a spanning ree, hen de M 0 [S] = 0. If, on he oher hand, S is he se of edges of a spanning ree of G, hen de M 0 [S] = ±1. Proof. If S is no he se of edges of a spanning ree, hen some subse R of S forms he edges of a cycle C in G. Consider he submarix M 0 [R] of M 0 [S] obained by aking he columns indexed by edges in R. Suppose ha he cycle C defined by R has edges f 1,..., f s in ha order. Muliply he column of M 0 [R] indexed by f j by 1 if in going around C we raverse 7

f i in he direcion of is arrow; oherwise muliply he column by 1. These column muliplicaions will muliply he deerminan of M 0 [R] by ±1. I is easy o see (check a few small examples o convince yourself) ha every row of his modified M 0 [R] has he sum of is elemens equal o 0. Hence he sum of all he columns is 0. Thus in M 0 [S] we have a se of columns for which a linear combinaion wih coefficiens ±1 is 0 (he column vecor of all 0 s). Hence he columns of M 0 [S] are linearly dependen, so de M 0 [S] = 0, as claimed. Now suppose ha S is he se of edges of a spanning ree T. Le e be an edge of T which is conneced o v p (he verex which indexed he boom row of M, i.e., he row removed o ge M 0 ). The column of M 0 [S] indexed by e conains exacly one nonzero enry [why?], which is ±1. Remove from M 0 [S] he row and column conaining he nonzero enry of column e, obaining a (p 2) (p 2) marix M 0. Noe ha de(m 0[S]) = ± de(m 0 ) [why?]. Le T be he ree obained from T by conracing he edge e o a single verex (so ha v p and he remaining verex of e are merged ino a single verex u). Then M 0 is jus he marix obained from he incidence marix M(T ) by removing he row indexed by u [why?]. Hence by inducion on he number p of verices (he case p = 1 being rivial), we have de(m 0) = ±1. Thus de(m 0 [S]) = ±1, and he proof follows. We have now assembled all he ingrediens for he main resul of his secion (due originally o Borchard). Recall ha κ(g) denoes he number of spanning rees of G. 1.8 Theorem. (he Marix-Tree Theorem) Le G be a finie conneced graph wihou loops, wih laplacian marix L = L(G). Le L 0 denoe L wih he las row and column removed (or wih he ih row and column removed for any i). Then de(l 0 ) = κ(g). Proof. Since L = M M (Lemma 1.6(a)), i follows immediaely ha L = M M. Hence by he Bine-Cauchy heorem (Theorem 1.4), we have 0 0 0 de(l 0 ) = (de M 0 [S])(de M 0 [S]), (4) S where S ranges over all (p 1)-elemen subses of {1, 2..., q} (or equivalenly, over all (p 1)-elemen subses of he se of edges of G). Since in general 8

A [S] = A[S], equaion (4) becomes de( L ) = 0 (de M0 [S]) 2. (5) S According o Lemma 1.7, de(m 0 [S]) is ±1 if S forms he se of edges of a spanning ree of G, and is 0 oherwise. Therefore he erm indexed by S in he sum on he righ-hand side of (5) is 1 if S forms he se of edges of a spanning ree of G, and is 0 oherwise. Hence he sum is equal o κ(g), as desired. The operaion of removing a row and column from L(G) may seem somewha conrived. We would prefer a descripion of κ(g) direcly in erms of L(G). Such a descripion will follow from he nex lemma. 1.9 Lemma. Le M be a p p marix wih real enries such ha he sum of he enries in every row and column is 0. Le M 0 be he marix obained from M by removing he las row and las column (or more generally, any row and any column). Then he coefficien of x in he characerisic polynomial de(m xi) of M is equal o p de(m 0 ). (Moreover, he consan erm of de(m xi) is 0.) Proof. The consan erm of de(m xi) is de(m), which is 0 since he rows of M sum o 0. For simpliciy we prove he res of he lemma only for removing he las row and column, hough he proof works jus as well for any row and column. Add all he rows of M xi excep he las row o he las row. This doesn effec he deerminan, and will change he enries of he las row all o x (since he rows of M sum o 0). Facor ou x from he las row, yielding a marix N(x) saisfying de(m xi) = x de(n(x)). Hence he coefficien of x in de(m xi) is given by de(n(0)). Now add all he columns of N(0) excep he las column o he las column. This does no effec de(n(0)). Because he columns of M sum o 0, he las column of N(0) becomes he column vecor [0, 0,..., 0, p]. Expanding he deerminan by he las column shows ha de(n(0)) = p de(m 0 ), and he proof follows. 1.10 Corollary. (a) Le G be a conneced (loopless) graph wih p verices. Suppose ha he eigenvalues of L(G) are µ 1,...,µ p 1, µ p, wih µ p = 9

0. Then 1 κ(g) = µ 1 µ 2 µ p p (b) Suppose ha G is also regular of degree d, and ha he eigenvalues of A(G) are λ 1,...,λ p 1, λ p, wih λ p = d. Then Proof. (a) We have 1. 1 κ(g) = (d λ 1 )(d λ 2 ) (d λ p 1 ). p de(l xi) = (µ 1 x) (µ p 1 x)(µ p x) = (µ 1 x)(µ 2 x) (µ p 1 x)x. Hence he coefficien of x is µ 1 µ 2 µ p 1. By Lemma 1.9, we ge µ 1 µ 2 µ p 1 = p de(l 0 ). By Theorem 1.8 we have de(l 0 ) = κ(g), and he proof follows. (b) Immediae from (a) and Lemma 1.6(b). Le us look a a couple of examples of he use of he Marix-Tree Theorem. 1.11 Example. Le G = K p, he complee graph on p verices. Now K p is regular of degree d = p 1, and A(K p ) + I = J, he p p marix of all 1 s. Noe ha rank(j) = 1 [why?], so p 1 eigenvalues of J are equal o 0. Since race(j) = p and he sum of he eigenvalues equals he race, he remaining eigenvalue of J is p. Thus he eigenvalues of A(K p ) = J I are 1 (p 1 imes) and p 1 (once).hence from Corollary 1.10 here follows 1 κ(k p ) = ((p 1) ( 1)) p 1 = p p 2. p This surprising resul is ofen aribued o Cayley, who saed i wihou proof in 1889 (and even cied Borchard explicily). However, i was in fac saed by Sylveser in 1857, while a proof was published by Borchard in 1860. I is clear ha Cayley and Sylveser could have produced a proof 10

if asked o do so. There are many oher proofs known, including elegan combinaorial argumens due o Prüfer, Joyal, Piman, and ohers. 1.12 Example. The n-cube C n is he graph wih verex se Z n 2 (he se of all n-uples of 0 s and 1 s), and wo verices u and v are conneced by an edge if hey differ in exacly one componen. Now C n is regular of degree ( ) n, and i can be shown ha is eigenvalues are n 2i wih mulipliciy n for i 0 i n. (See he soluion o Exercise 10.18(d) of he ex.) Hence from Corollary 1.10(b) here follows he amazing resul κ(c n ) = n 1 ) ( n (2i i) 2 n i=1 n = 2 2n n n 1 i ( i). To my knowledge a direc combinaorial proof is no known. i=1 11

18.314 (Fall 2011) Problems on he Marix-Tree Theorem 1. The complee biparie graph K rs has verex se A B, where #A = r, #B = s, and A B = Ø. There is an edge beween every verex of A and every verex of B, so rs edges in all. Le L = L(K rs ) be he laplacian marix of K rs. (a) Find a simple upper bound on rank(l ri). Deduce a lower bound on he number of eigenvalues of L equal o r. (b) Assume r s, and do he same as (a) for s insead of r. (c) Find he remaining eigenvalues of L. (Hin. Use he fac ha he rows of L sum o 0, and compue he race of L.) (d) Use (a) (c) o compue κ(k rs ), he number of spanning rees of K rs. (e) (opional) Give a combinaorial proof of he formula for κ(k rs ). 2. Le V be he subse of Z Z on or inside some simple closed polygonal curve whose verices belong o Z Z, such ha every line segmen ha makes up he curve is parallel o eiher he x-axis or y-axis. Draw an edge e beween any wo poins of V a disance one apar, provided e lies on or inside he boundary curve. We obain a planar graph G, an example being Le G be he dual graph G wih he ouside verex deleed. (The verices of G are he regions of G. For each edge e of G, say wih regions R and R on he wo sides of e, here is an edge of G beween R and R.) For he above example, G is given by 1

Le λ 1,...,λ p denoe he eigenvalues of G (i.e., of he adjacency marix A(G )). Show ha p κ(g) = (4 λ i ). i=1 2

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