b) If x is an eigenvector for both T1 and T2, prove that x is an eigenvector for aT1
+bT2 and find the eigenvalues of aT1 +bT2 in terms of the eigenvalues.
Problem Set 7 Solutions MATH 110: Linear Algebra Each problem is worth 5 points. PART 1 1. Curtis p. 192 3. 2. Curtis p. 192 5. 3. Curtis p. 215 1(a,b,c,d,e).
Solutions in book. PART 2 Problem 1(10) a) If T : V → V has an eigenvalue λ, prove that aT has the eigenvalue aλ. Proof: Let x be an eigenvector for the eigenvalue λ. (aT )(x) = a(T (x)) = a(λx) = λ(ax) so ax is an eigenvector of aT with eigenvalue λ. b) If x is an eigenvector for both T1 and T2 , prove that x is an eigenvector for aT1 + bT2 and find the eigenvalues of aT1 + bT2 in terms of the eigenvalues of T1 and T2 . Proof: Let λ be the eigenvalue of x for T1 and let µ be the eigenvalue of x for T2 . (aT1 + bT2 )(x) = (aT1 )(x) + (bT2 )(x) = (λa + µb)x. c) Suppose that x is an eigenvector of T with eigenvalue λ. Show that x is an eigenvector of T 2 with eigenvalue λ2 . Proof: T 2 (x) = T (T (x)) = T (λx) = λ(T (x)) = λ(λx) = λ2 x. d) Let P be a polynomial. Show that if x is an eigenvector of T with eigenvalue λ then x is an eigenvector of P (T ) with eigenvalue P (λ). Proof: By induction on the degree of P (we will denote the degree with the variable n). Suppose n = 0: the polynomial is a constant, and certainly x is an eigenvector of P (T ) (every vector actually is) with eigenvalue P (λ). Now suppose the theorem is true for all polynomials of degree less than or equal to n. Consider P of degree n + 1. If P has no constant term, then P (T ) = T Q(T ) where Q has degree n. By induction x is an eigenvector of Q with eigenvalue Q(λ). Therefore, T Q(T )(x) = T (Q(λ)x) = Q(λ)T (x) = Q(λ)(λx) = λQ(λ)x. Thus x is an eigenvector of P with eigenvalue λQ(λ) = P (λ). Now, if P has a constant term c, then P − c is a polynomial of degree 1
n + 1 with no constant term. By the above argument x is an eigenvector of P − c with eigenvalue (P − c)(λ). Notice that x is also an eigenvector of c with eigenvalue c (proved for the base case). By part (b) x is an eigenvector of the polynomial P − c + c = P with eigenvalue (P − c)(λ) + c = P (λ). Problem 2(10) If T : V → V has the property that T 2 has a nonnegative eigenvalue λ2 , prove that at least one of λ or −λ is an eigenvalue for T . (Hint: T 2 − λ2 I = (T + λI)(T − λI)). Proof: Let x be an eigenvector of T 2 with the eigenvalue λ2 . (T 2 − λ2 I)(x) = 0. Therefore, (T + λI)(T − λI)(x) = 0. If (T − λI)(x) = 0 then x is an eigenvector of T with eigenvalue λ. Otherwise, (T − λI)(x) = y where y is some nonzero vector. In this case, (T + λI)(y) = 0, which means that y is an eigenvector of T with eigenvalue −λ. Thus, either λ or −λ is an eigenvalue. Problem 3 (10) Let V be the linear space of all real convergent sequences {xn }. Define T : V → V as follows: if x = {xn } is a convergent sequence with limit a, let T (x) = {yn } where yn = a − xn for n ≥ 1. Prove that T has only two eigenvalues λ = 0 and λ = −1 and determine the eigenvectors belonging to each such λ. Solution: Let λ be an eigenvalue of T . Then a − xn = λxn for each n, a (assuming that λ 6= −1). Taking the limit, as which means that xn = λ+1 a n → ∞, we see that λ+1 = a which implies that λ = 0. Thus λ is either 0 or −1. If λ = 0 this means that xn is a constant sequence, thus the eigenvectors corresponding to eigenvalue 0 are the constant sequences. If λ = −1 then we see that there is no restriction on xn , other than that a = 0. Thus, the sequences with limit 0 are the eigenvectors with eigenvalue −1. Problem 4 (20) Let V be the vector space of sequences {an } over the real numbers. The shift operator S : V → V is defined by S((a1 , a2 , . . .) = (a2 , a3 , a4 , . . .). Find the eigenvectors of S, and show that the subspace W consisting of the sequences {xn } satisfying xn+2 = xn+1 + xn is a two dimensional, S-invariant subspace of V . Also, find an explicit basis for W . Using these results, find an explicit formula for the nth Fibonacci number fn where fn+2 = fn+1 + fn and f1 = f2 = 1. 2
Solution: Suppose that (a1 , a2 , . . .) is an eigenvector with eigenvalue λ. Then S((a1 , . . .)) = λ(a2 , a3 , . . .) which means that a2 = λa1 , a3 = λa2 , . . .. Thus, the eigenvectors are of the form a1 (1, λ, λ2 , . . .). Now suppose that x = (x1 , x2 , . . .) ∈ W . x is completely determined by its two initial values x1 , x2 . Therefore, the dimension of W is at most 2. If an element of W is an eigenvector, it must be associated with an eigenvalue λ2 = λ + 1, which gives the two possible eigenvalues √ √ 1+ 5 1− 5 λ1 = , λ2 = . 2 2 A basis for W is therefore b1 = (λ1 , λ21 , λ31 , . . .) and b2 = (λ2 , λ22 , . . .) which is clearly invariant under S. In order to express the Fibonacci sequence in terms of this basis, we need to find constants k1 , k2 such that 1 = k 1 λ1 + k 2 λ2 1 = k1 λ21 + k2 λ22 . Solving for k1 , k2 , we find that 1 fn = √ 5
√ !n √ !n ! 1+ 5 1− 5 . − 2 2
Problem 5∗ (10) Prove that the eigenvalues of an upper triangular matrix are its diagonal entries. Proof: Let A be the upper triangular matrix. Consider det(λI − A). This is just n Y
(λ − aii )
i=1
where aii is the ith diagonal entry. This expression equals zero if, and only if, λ = aii for some i. Thus, the diagonal entries of the matrix are exactly the eigenvalues. PART 3 - Optional Problem Recall that a graph is consists of two sets: a set of vertices, and a set of edges consisting of pairs of vertices. The adjacency matrix of a graph on n vertices is an n × n graph with the ijth entry equal to 1 if vertex i is adjacent to vertex j and zero otherwise. 3
Let f (x) = xn +c1 xn−1 +c2 xn−2 +. . .+cn be the characteristic polynomial of the adjacency matrix of a graph. Show that c1 = 0, −c2 is the number of edges in the graph, and −c3 is twice the number of triangles in the graph.
4
