Problem set 6 The Perron Frobenius theorem.

Transcription

1 Probem set 6 The Perron Frobenius theorem. Math 22a4 Oct 2 204, Due Oct.28 In a future probem set I want to discuss some criteria which aow us to concude that that the ground state of a sef-adjoint operator is nondegenerate. In the physics iterature this is known as the non-degeneracy of the vacuum.these criteria are extensions to the infinite dimensiona setting of theorems about finite matrices which go under the name of Perron-Frobenius. Unfortunatey, as far as I can te, these important theorems are not taught in any course in the math department here at Harvard. So the purpose of this probem set is to famiiarize you with the the Perron Frobenius theorem. This probem set is taken (with some modifications) from a probem set in my Lie agebras course. What I do here is give my version of a proof of the cassification of the extended Dynkin diagrams. From this cassification it is immediate to obtain the cassification of the ordinary Dynkin diagrams and then with considerabe Lie agebra work the cassification of a the simpe finite dimensiona Lie agebras, due to Kiing. John Coeman cas Kiing s paper The Greatest Mathematica Paper of A Time. Contents Graphs. 2 Perron-Frobenius. 3 3 Probems. 8 4 Proof of Perron-Frobenius. 9 5 Markov chains in a nutshe. 2 6 One parameter semi-groups of positive matrices. 2

2 Graphs. An undirected graph Γ = (N, E) consists of a set N (for us finite) and a subset E of the set of subsets of N of cardinaity two. We ca eements of N nodes or vertices and the eements of E edges. If e = {i, j} E we say that the edge e joins the vertices i and j or that i and j are adjacent. Notice that in this definition our edges are undirected : {i, j} = {j, i}, and we do not aow sef-oops. An exampe of a graph is the cyce A () with + vertices, so N = {0,, 2,..., } with 0 adjacent to and to, with adjacent to 0 and to 2 etc. The adjacency matrix A of a graph Γ is the (symmetric) 0 matrix whose rows and coumns are indexed by the eements of N and whose i, j-th entry A ij = if i is adjacent to j and zero otherwise. For exampe, the adjacency matrix of the graph A () 3 (see the figure beow) is We can think of A as foows: Let V be the vector space with basis given by the nodes, so we can think of the i-th coordinate of a vector x V as assigning the vaue x i to the node i. Then y = Ax assigns to i the sum of the vaues x j summed over a nodes j adjacent to i. As I hope to expain in cass, the Hücke theory of organic chemistry reduces considerations of stabiity of organic moecues to that of finding the eigenvaues of the adjacency matrix of a graph. A path of ength r is a sequence of nodes x i, x i2,..., x ir where each node is adjacent to the next. So, for exampe, the number of paths of ength 2 joining i to j is the i, j-th entry in A 2 and simiary, the number of paths of ength r joining i to j is the i, j-th entry in A r. The graph is said to be connected if there is a path (of some ength) joining every pair of vertices. In terms of the adjacency matrix, this means that for every i and j there is some r such that the i, j entry of A r is non-zero. In terms of the theory of non-negative matrices (see beow) this says that the matrix A is irreducibe. Notice that if denotes the coumn vector a of whose entries are, then is an eigenvector of the adjacency matrix of A (), with eigenvaue 2, and a the entries of are positive. In view of the Perron-Frobenius theorem to be stated beow, this impies that 2 is the maximum eigenvaue of this matrix. We modify the notion of the adjacency matrix as foows: We start with a connected graph Γ as before, but modify its adjacency matrix by repacing some of the ones that occur by positive integers a ij. If, in this repacement a ij >, we redraw the graph so that there is an arrow with a ij ines pointing towards the node i. For exampe, the graph abeed A () in Tabe Aff corresponds to 2

3 the matrix ( 0 ) ( which ceary has as an positive eigenvector with eigenvaue 2. ) Simiary, diagram A (2) 2 in Tabe Aff 2 corresponds to the matrix ( ) ( 2 which has as eigenvector with eigenvaue 2. In the diagrams, the coefficient ) next to a node gives the coordinates of the eigenvector with eigenvaue 2, and it is immediate to check from the diagram that this is indeed an eigenvector with eigenvaue 2. For exampe, the 2 next to a node with an arrow pointing toward it in C () satisfies 2 2 = etc. A the graphs so far have zeros aong the diagona. If we reax this condition, and aow for any non-negative integer on the diagona, then the ony new possibiities are those given in Figure 4. Indeed, et us ca a matrix symmetrizabe if A ij 0 A ji 0. The main content of this probem set wi be to show that the ists in the Figures -4 exhaust a irreducibe matrices with non-negative integer matrices, which are symmetrizabe and have maximum eigenvaue 2. 2 Perron-Frobenius. We say that a rea matrix T is non-negative (or positive) if a the entries of T are non-negative (or positive). We write T 0 or T > 0. We wi use these definitions primariy for square (n n) matrices and for coumn vectors = (n ) matrices. We et Q := {x R n : x 0, x 0} so Q is the non-negative orthant excuding the origin. Aso et C := {x 0 : x = }. So C is the intersection of the orthant with the unit sphere. A non-negative matrix square T is caed primitive if there is a k such that a the entries of T k are positive. It is caed irreducibe if for any i, j there is a k = k(i, j) such that (T k ) ij > 0. For exampe, as mentioned above, the adjacency matrix of a connected graph is irreducibe. If T is irreducibe then I + T is primitive. In this section we wi assume that T is non-negative and irreducibe. 3

4 < > A () A (), > 2 B () > < C () D () > > G () 2 F () E () E () E () 8 4 Figure : Aff.

5 2 < A (2) < < < <... > < A (2) 2 2 A (2) 2 3 D (2) +, 2 E (2) 6 Figure 2: Aff 2 2 < D (3) 4 Figure 3: Aff 3 5

6 L L 2 2 >..... LC <..... LB LD 2 Figure 4: Loops aowed 6

7 Theorem 2.. Perron-Frobenius.. T has a positive (rea) eigenvaue λ max such that a other eigenvaues of T satisfy λ λ max. 2. Furthermore λ max has agebraic and geometric mutipicity one, and has an eigenvector x with x > Any non-negative eigenvector is a mutipe of x. 4. More generay, if y 0, y 0 is a vector and µ is a number such that then y > 0, T y µy and µ λ max with µ = λ max if and ony if y is a mutipe of x. 5. If 0 S T, S T then every eigenvaue σ of S satisfies σ < λ max. 6. In particuar, a the diagona minors T (i) obtained from T by deeting the i-th row and coumn have eigenvaues a of which have absoute vaue < λ max. I wi present a concise proof of this theorem (for those of you who have not seen it in a inear agebra course) ater on in this handout. Let me make one carification as to the ast two assertions of the theorem. The matrix T (i) is usuay thought of as an (n ) (n ) matrix obtained by striking out the i-th row and coumn. But we can aso consider the matrix T i obtained from T by repacing the i-th row and coumn by a zeros. If x is and n-vector which is an eigenvector of T i, then the n vector y obtained from x by omitting the (0) i-th entry of x is then an eigenvector of T (i) with the same eigenvaue. Conversey, if y is an eigenvector of T (i) then inserting 0 at the i-th position wi give an n-vector which is an eigenvector of T i with with the same eigenvaue as that of y. More generay, suppose that S is obtained from T by repacing a certain number of rows and the corresponding coumns by a zeros. Then we may appy item 5) of the theorem to this n n matrix, S, or the compressed version of S obtained by eiminating a these rows and coumns. We wi want to appy this to the foowing specia case. A subgraph Γ of a graph Γ is the graph obtained by eiminating some nodes, and a edges emanating from these nodes. Thus, if A is the adjacency matrix of Γ and A is the adjacency matrix of A, then A is obtained from A by striking out some rows and their corresponding coumns. Thus if Γ is irreducibe, so that we may appy the Perron Frobenius theorem to A, and if Γ is a proper subgraph (so we have actuay deeted some rows and coumns of A to obtain A ), then the maximum eigenvaue of A is stricty ess than the maximum eigenvaue of A. Simiary, if an entry A ij is >, the matrix A obtained from A by decreasing this entry whie sti keeping it positive wi have a stricty smaer maxima eigenvaue. 7

8 3 Probems. Using Perron-Frobenius, these probems wi show that that the (generaized) graphs occurring in Figures - 4 are the ony irreducibe graphs with maxima eigenvector 2. Here is how to get started. We have aready checked that A (2) 2 has maxima eigenvector 2. Any graph B which has an entry 4 must contain A (2) 2 as a subgraph, by the condition A ij 0 A ji 0. The graph A (2) 2 can not be a proper subgraph of B, because this wi impy that the maxima eigenvaue of B is stricty > 2. This impies that A (2) 2 is the ony graph whose matrix contains a 4. Simiary, A () is the ony graph for which there is a pair i, j with both A ij and A ji >. We have verified that A, 2 has 2 as maxima eigenvaue. This impies that no other graph with maxima eigenvaue 2 can contain a cyce, and that the straight ine graph with no doube or tripe edges or branches has maxima eigenvector < 2 an so is not a candidate.. Write out the matrices corresponding to G () 2 and to D (3) 4 and verify that the given vectors are eigenvectors with eigenvaue 2. Concude that these are the ony graphs with a tripe edge. 2. Write out the matrices corresponding to F () 4 and E (2) 6 and verify that the given vectors are eigenvectors with eigenvaue 2. Concude that these are the ony graphs with a doube edge which does not terminate at an end point of the graph. In other words, any graph with maxima eigenvaue 2 and which has a doube edge must have the property that one of the nodes of the doube edge is not adjacent to any other node. 3. Write out the matrix corresponding to D () 4. This is a graph with five nodes, a star with the four branches emanating from the center node and connecting to four end nodes. Verify that the given vector is an eigenvector with eigenvaue 2. Concude that this is the ony graph with maxima eigenvaue 2 which has four or more branches emanating from any node (and with no oops). 4. Verify that the maxima eigenvaue of D () is 2, and concude that these are the ony graphs with maxima eigenvaue 2 which has two or more branch points (that is a node with more than two adjacent nodes). Aso concude that a graph with a branch node connecting to two end nodes and no other branch nodes, oops, or doube edges has maxima eigenvaue stricty ess than Write out the matrix corresponding to E () 6. Even without writing out the matrix, notice that the given vector is an eigenvector with eigenvaue 2, since the sum of the vaues at the adjacent nodes is twice the vaue at any given node. 8

9 This graph has three branches (emanating from the centra node) and each of the branches has two nodes not counting the centra node. Concude that this is the ony graph with maxima eigenvaue 2 with a branching node that has two or more nodes on each branch (not counting the branching node). So any other graph with a branch has at east one branch with ony one additiona node on it if the graph has maxima eigenvaue The graph E () 7 has a branch nodes with one additiona node on one of the branches and three additiona node on the two others. Verify that E () 7 has maxima eigenvaue 2, and concude that any graph other than E () 6 and E () 7 with a branch node and with maxima eigenvaue 2, has at most one additiona node on one of its branches and at most two additiona nodes on a second of its branches. 7. Verify that E () 8 has maxima eigenvaue 2, and concude that outside of the three E () graphs, = 6, 7, 8 any graph with maxima eigenvaue 2, and a branch point, must have two of the branches with ony additiona node on each, and the other end of the graph must be a doube edge or a oop. Concude that these possibiities are exhausted by B (), A (2) 2 and LD. 8. The matrix of L 0 is (2). So this is the ony graph with maxima eigenvaue 2 with two (or more) oops emanating from a singe node. Draw the graph LD 2 which has three nodes, write out its matrix. Verify that its maxima eigenvaue is 2. Concude that any other graph with oops and with maxima eigenvaue 2 has its oop(s) emanating from an end node. The possibiity of a oop at each end is accounted for by L,. We are thus eft to consider graphs with no branches and either doube edges at both ends or a doube edge at one end and a (singe) oop at the other end. In the case of a doube edge at both ends, there are there possibiities as to the pacement of the arrows. They can both face in, as in C () they can both face out, as in D (2) +or one can face in and the other out as in A(2) 2. We have thus accounted for a graphs with maxima eigenvaue 2 and with doube edges at both ends. If there is a doube edge at one end and a oop at the other, there are two possibiities, the arrow can face in as in LC or can face out as in LB, and both these possibiities occur. 4 Proof of Perron-Frobenius. Let P := (I + T ) n 9

10 Reca that Q denotes the positive orthant minus {0} and that C denotes the intersection of the unit sphere with the positive orthant. For any z Q et L(z) := max{s : sz T z} = (T z) i min. i n,z i 0 z i By definition L(rz) = L(z) for any r > 0, so L(z) depends ony on the ray through z. If z y, z y we have P z < P y. Aso P T = T P. So if sz T z then sp z P T z = T P z so L(P z) L(z). Furthermore, if L(z)z T z then L(z)P z < T P z. So L(P z) > L(z) uness z is an eigenvector of T. Consider the image of C under P. It is compact (being the image of a compact set under a continuous map) and a of the eements of P (C) have a their components stricty positive (since P is positive). Hence the function L is continuous on P (C). Thus L achieves a maximum vaue, L max on P (C). Since L(z) L(P z) this is in fact the maximum vaue of L on a of Q, and since L(P z) > L(z) uness z is an eigenvector of T, we concude that L max is achieved at an eigenvector, ca it x of of T and x > 0 with L max the eigenvaue. Since T x > 0 and T x = L max x we have L max > 0. We wi now show that this is in fact the maximum eigenvaue in the sense of the theorem. So et y be any eigenvector with eigenvaue λ, and et y denote the vector whose components are y j, the absoute vaues of the components of y. We have y Q and from T y = λy and the triange inequaity we concude that λ y T y. Hence λ L( y ) L max. So we may use the notation since we have proved that λ max := L max λ λ max. We have proved item in the theorem. Suppose that 0 S T. Then sz Sz and Sz T z impies that sz T z so L S (z) L T (z) for a z and hence L max (S) L max (T ). We may appy the same argument to T to concude that it aso has a positive maximum eigenvaue. Let us ca it η. (We sha soon show that η = λ max.) This means that there is a vector w > 0 such that w T = ηw. 0

11 Reca the x > 0 denotes the eigenvector with maximum eigenvaue λ max of T. We have w T x = ηw x = λ max w x impying that η = λ max since w x > 0. Now suppose that y Q and T y µy. Then λ max w y = w T y µw y impying that λ max µ, again using the fact that a the components of w are positive and some component of y is positive so w y > 0. In particuar, if T y = µy then then µ = λ max. Furthermore, if y Q and T y µy then µ 0 and so 0 < P y = (I + T ) n y ( + µ) n y y > 0. This proves the first two assertions in item 4. If µ = λ max then w (T y λ max y) = 0 but T y λ max y 0 and so w (T y λ max y) = 0 impies that T y = λ max y. Then the ast assertion of item 4) wi then foow from item 2) once we prove item 2), since we have shown that y must be an eigenvector with eigenvaue λ max. Suppose that 0 S T and Sz = σz, z 0. Then so T z S z σ z σ L max (T ) = λ max, as we have aready seen. But if σ = λ max then L( z ) = L max (T ) so z > 0 and z is aso an eigenvector of T with the same eigenvaue. But then (T S) z = 0 and this is impossibe uness S = T since z > 0. Repacing the i-th row and coumn of T by zeros give an S 0 with S < T since the irreducibiity of T precudes a the entries in a row being zero. This proves items 5) and 6). From inear agebra we know that d dλ det(λi T ) = i det(λi T (i) ) and each of the matrices λ max I T (i) has stricty positive determinant by what we have just proved. This shows that the derivative of the characteristic poynomia of T is not zero at λ max, and therefore the agebraic mutipicity and hence the geometric mutipicity of λ max is one. This proves 2) and hence the remaining assertions. QED

12 5 Markov chains in a nutshe. A non-negative matrix M is a stochastic matrix if each of the row sums equa. Then the coumn vector a of whose entries equa is an eigenvector with eigenvaue. So if M is irreducibe is the maxima eigenvaue since has a positive entries. If M is primitive, then we know from the genera theory that π π 2 π n M k π π 2 π n.... π π 2 π n where p := (π, π 2,, π n ) is the unique vector whose entries sum to one and satisfies pm = p. 6 One parameter semi-groups of positive matrices. We are going to be interested (in the infinite dimensiona case) in one parameter semi-groups of the form e ta and wi want to know when this is non-negative in for a t > 0. If A = (a) is a one-by-one matrix then e at > 0 no matter what a is. Suppose that an n n matrix A with n > has the property that e ta is non-negative for a t > 0. Then the matrix I e ta has non-positive off diagona matrices. Dividing by t > 0 and etting t 0 shows that the off diagona entries of A are non-positive. If B is a matrix with a of its entries non-positive, then ceary has a of its entries non-negative. e tb = I tb + 2 t2 B 2 9. Show that if the off-diagona entries of A are non-positive then e ta has non-negative entries for a t > 0.[Hint: Use Lie s formua which says that ( e t(d+b) = im e t n D e t D) n n. n We wi prove a usabe infinite dimensiona version of this ater which is known as the Trotter product formua. At the moment, take Lie s formua for granted.] 2