REU 2007 · Apprentice Class · Lecture 12

Instructor: L´aszl´oBabai Scribes: Courtney Morris and Gabriel Kerr July 11, 2007. Revised by instructor. Last updated July 11, 9:30 p.m.

A12.1 Rational roots of polynomials

Consider the polynomial f(x) = x4 + ax3 + bx2 + cx − 15, a, b, c ∈ Z. What can we say about the roots?

Claim A12.1.1. If f(m) = 0, m ∈ Z, then m ∈ {±1, ±3, ±5, ±15}. Proof. We have 15 = −m(m3 + am2 + bm + c), thus m divides 15.

m Claim A12.1.2. If f( n ) = 0, m, n ∈ Z, n 6= 0, gcd(m, n) = 1, then n = 1.

m m4 m3 4 Proof. We have 0 = f( n ) = n4 + a n3 + ··· + 15. Multiply through by n to get 0 = m4 + am3n + bm2n2 + cmn3 + 15n4. Moving m4 to the other side and factoring out an n we get m4 = −n(am3 + bm2n + cmn3 + 15n3). Thus n divides m4. Since we assumed gcd(m, n) = 1, n = ±1. Without any loss we can assume n = 1 (move the minus sign to the numerator.

k Exercise A12.1.3. Assume f(x) = a0 + a1x + ··· + akx ∈ Z[x] with a0 6= 0 and ak 6= 0. m Show that if f( n ) = 0 and gcd(m, n) = 1 then m | a0 and n | ak.

A12.2 Minimal graphs of girth 5 In this section we describe a mathematical gem due to Alan J. Hoﬀman and R. R. Singleton.

Deﬁnition A12.2.1. The girth of a graph G, girth(G), is the length of the shortest cycle.

If G is a tree (or a forest) then girth(G) is smallest member (the inf) of the empty set, i. e. +∞. With this convention, we have:

Observation A12.2.2. A graph has girth ≥ k exactly if it has no cycles of length < k.

Deﬁnition A12.2.3. A graph is r-regular if every vertex has degree r.

1 Lemma A12.2.4. If an r-regular graph has girth ≥ 5 then n ≥ r2 + 1, where n is the number of vertices. Remark A12.2.5. The Petersen graph shows that there exists a graph with r = 3 and n = 32 + 1 = 10. Proof of Lemma A12.2.4. We illustrate the proof on the case r = 3. To show that you cannot have fewer than 10 vertices when r = 3, start with one of your vertices, v0. It has three neighbors. Let S be the set consisting of v0 and its thre neighbors. Each of the 3 neighbors has 2 more neighbors. None of these additional neighbors belong to S (because then we would have a 3-cycle), and they are all distinct, because otherwise we would have a 4-cycle. Thus we have a minimum of 10 vertices. To show the inequality n ≥ r2 + 1 for general r, make the same argument, except replace 3 by r. You will see that it is necessary to have at least 1 + r + r(r − 1) = r2 + 1 vertices. Question A12.2.6. For which values of r is equality possible? If r = 1 then the graph with two vertices and one edge between them satisfies n = 12 + 1. If r = 2 then the pentagon satisfies n = 22 + 1. If n = 3 then the Petersen graph satisfies n = 32 + 1. For r = 4, 5, or 6, there does not exist a graph satisfying n = r2 + 1, however, for r = 7, there does exist one with n = 50. The graph is called the Hoffman-Singleton graph and just as the Petersen graph is a bunch of pentagons glued together in a clever way, the Hoffman-Singleton graph is a bunch of Petersen graphs glued together in a very clever way. (You can see pictures of this graph and explanations on Wolfram MathWorld.) The answer to the question is a theorem due to Hoffman-Singleton: Theorem A12.2.7 (Hoffman-Singleton Theorem). If there exists a graph of girth ≥ 5, degree r, and n = r2 + 1, then r ∈ {1, 2, 3, 7, 57}. Remark A12.2.8. It is unknown whether or not there exists a graph for r = 57. Theorem A12.2.9. (M. Aschbacher) If there exists a graph with r = 57 then it cannot be vertex-symmetrical. A graph is vertex-symmetrical if for each pair x, y of vertices there exists an automorphism that maps x to y. Definition A12.2.10. Let G be a graph with n vertices, numbered 1 to n. The adjacency matrix of G is an n × n (0, 1)-matrix, A = (aij), with aij = 1 if i ∼ j and 0 if i j, where i ∼ j if i is adjacent to j. Observation A12.2.11. A has zeros along the diagonal and is symmetric (AT = A). The number of nonzero entries is equal to twice the number of edges.

2 Exercise A12.2.12. If G is any graph and A is its adjacency matrix, then A = (bij), where bii = degi and bij = the number of common neighbors of i and j.

2 Exercise A12.2.13. If G is regular of degree r, has girth ≥ 5, and n = r2 + 1 vertices the number of common neighbors of a pair x, y of vertices is zero if x, y are adjacent and 1 if x, y are not adjacent.

Hint: let x = v0 in the proof of Lemma A12.2.4. Note that in particular, diam(G) = 2.

2 proof of Theorem A12.2.7. Let A = B = (bij) where A is the adjacency matrix. Combining Exx A12.2.12 and A12.2.13, we obtain that bii = r; and bij = 0 if i ∼ j and bij = 1 if i j. ¯ ¯ Let A = (¯aij) be the adjacency matrix of G. Thena ¯ii = 0 anda ¯ij = 1 − aij if i 6= j. We see that A + A¯ + I = J and that A2 = B = rI + A¯ = rI + J − A − I = (r − 1)I − A + J or

A2 + A − (r − 1)I = J. (1)

In general, P row i = degi. Thus

deg1  .  A · 1 =  .  , degn where 1 is the vector of all 1’s. In our case, A · 1 = r · 1, and we see that 1 is an eigenvector with eigenvalue r. From the Spectral Theorem (adjacency matrices are always real symmetric matrices) we know that A has an orthogonal eigenbasis, e1 = 1, e2,..., en. What can we say about the ei, i ≥ 2? Well, ei⊥1. It follows that for i ≥ 2 we have Jei = 0. Multiplying equation (1) by ei to the right (i ≥ 2), we see that

2 (A + A − (r − 1)I)ei = Jei = 0.

2 2 2 Now Aei = λiei and A ei = λi ei, thus λi ei + λiei − (r − 1)ei = 0 or

2 λi + λi − (r − 1) = 0, for i ≥ 2. Since this is a quadratic equation it has at most 2 roots. It follows that the matrix has 3 eigenvalues, call them r, µ1, and µ2 with multiplicities 1, m1 and m2, respectively. 2 µ1 and µ2 are roots of the equation t + t − (r − 1) = 0. Thus √ −1 ± p1 + 4(r − 1) −1 ± 4r − 3 µ = = . (2) 1,2 2 2 √ Letting s = 4r − 1 we have that s2 = 4r − 3, i. e.,

s2 + 3 r = . (3) 4 Note that s is the square root of an integer.

3 We can make the following table for the eigenvalues.

eigenvalue multiplicity r 1 µ1 m1 µ2 m2

Our unknowns are m1 and m2; we need two equations connecting them. Counting multi- plicites we have 1 + m1 + m2 = n. (4) Pn Consider the trace. We know that i=1 λi = Tr(A) = 0. Thus

1 · r + m1µ1 + m2µ2 = 0. (5)

Rewrite the ﬁrst equation as 2 m1 + m2 = r (6) and the second as 2r + m1(−1 + s) + m2(−1 − s) = 0 or 2r − (m1 + m2) + s(m1 − m2) = 0. (7) From this equation we ask whether s is rational or not. If it is irrational then clearly 2 m1 − m2 = 0 and then (after substituting r for m1 + m2) our equation reads

2r − r2 = 0 or r = 2 (since r 6= 0). This number occurs in the theorem. Henceforth we may assume that s is rational, and therefore (being the square root of an integer), s is an integer. 2 Again, substituting r for m1 + m2, our equation reads

2 r − 2r − (m1 − m2)s = 0.

s2 + 3 Putting r = into this equation and multiplying by 16 yields 4

2 2 2 (s + 3) − 8(s + 3) − 16(m1 − m2)s = 0.

Expanding and simplifying gives

4 2 s − 2s − 16(m1 − m2)s − 15 = 0,

s2 + 3 so that s must divide 15, i. e., s = ±1, ±3, ±5, ±15. Again using r = we obtain 4 r = 1, 3, 7, 57, as required.

4 A12.3 Helly’s Theorem We state Helly’s theorem:

n Theorem A12.3.1. If C1,...,Cm are convex sets in R and every n + 1 of them intersect then all do, i. e., there is a point contained in all the Ci. T We outline a proof of this for m = n + 2. First take a pi ∈ j6=i Cj. Then if we replace Ci with conv({pj : j 6= i}) the result will still hold (since conv({pj : j 6= i}) ⊆ Ci (where conv(S) denotes the convex hull of S ⊆ Rn).

n Lemma A12.3.2. (Radon’s Lemma) If p1, . . . , pn+2 are distinct points in R then it is possible to divide them into two disjoint subsets A and B such that A ∪ B = {p1, . . . , pn+2} and conv(A) ∩ conv(B) 6= ∅.

Exercise A12.3.3. Prove Radon’s Lemma.

Exercise A12.3.4. Use Radon’s Lemma to prove Helly’s Theorem in the special case m = n + 2.

Exercise A12.3.5. Use the preceding exercise to Helly’s Theorem for general m. Hint: induction on m.

Now we consider a combinatorial duality theorem on the real line.

Exercise A12.3.6. Given nonempty intervals I1,...,Im in R show that the minimum number of points for which every interval contains at least one point is equal to the maximum number of disjoint intervals.

Note that min ≥ max is trivial; your task is to prove min ≥ max.

We say that a family of sets A1,...,Am can be “nailed with s points” if there exusts a set S of size s which intersects all the Ai. The following exercise recasts the preceding one as a Helly-type theorem.

Exercise A12.3.7. Infer from the previous exercise: if I1,...,Im are intervals in R and every s + 1 of them can be “nailed” with s points then all of them can.

We now ask for the smallest value of N as a function of r, s that makes the following Helly-type statement true:

Let A1,...,Am be sets such that |Ai| = r. If every N of them can be nailed by s points then all of them can be nailed by s points.

r+s Exercise A12.3.8. Show that N = r . Hint: Use a theorem proved in class.

5 A12.4 Omissions: the minimal polynomial This is the last lecture of this series (the two remaining classes will be problem sessions). The series was meant to give an introduction to linear algebra, while illustrating the power of this traditional subject on highly non-traditional, and often striking, applications. The tradeoﬀ was that some of the traditional concepts and results of linear algebra had to be omitted. The minimal polynomial of a matrix A has been one of the victims. We at least give the deﬁnition and the basic properties, including one important result.

Deﬁnition A12.4.1. Let A be an n × n matrix over a ﬁeld F . We call mA ∈ F [x] the minimal polynomial of A if mA is a monic polynomial of minimum degree for which mA(A) = 0.

Exercise A12.4.2. Show that if f ∈ F [X] satisﬁes f(A) = 0 then mA divides f. This also proves the uniqueness of mA. Hint: show that the set {f ∈ F [x]: f(A) = 0} is an ideal in F [x]. Recall that a matrix is diagonalizable if and only if it is similar to a diagonal matrix if and only if it has an eigenbasis.

Exercise* A12.4.3. Let A be an n × n matrix over the ﬁeld F = C (or any ﬁeld that contains all the eigenvalues of A). Show that A is diagonalizable if and only if mA has no multiple roots.

One direction of the “if and only if” statement is easy. Which one?