Problem Set 7 SolutionsMATH 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 2Problem 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 eigenvaluesof T1 and T2.

Proof: Let λ be the eigenvalue of x for T1 and let µ be the eigenvalue ofx 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 xis an eigenvector of T 2 with eigenvalue λ2.

Proof: T 2(x) = T (T (x)) = T (λx) = λ(T (x)) = λ(λx) = λ2x.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 certainlyx 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 orequal to n. Consider P of degree n + 1. If P has no constant term, thenP (T ) = TQ(T ) where Q has degree n. By induction x is an eigenvector ofQ with eigenvalue Q(λ). Therefore, TQ(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

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n + 1 with no constant term. By the above argument x is an eigenvector ofP − c with eigenvalue (P − c)(λ). Notice that x is also an eigenvector of cwith eigenvalue c (proved for the base case). By part (b) x is an eigenvectorof 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 − λ2I =(T + λI)(T − λI)).

Proof: Let x be an eigenvector of T 2 with the eigenvalue λ2. (T 2 −λ2I)(x) = 0. Therefore, (T + λI)(T − λI)(x) = 0. If (T − λI)(x) = 0 then xis an eigenvector of T with eigenvalue λ. Otherwise, (T − λI)(x) = y wherey is some nonzero vector. In this case, (T + λI)(y) = 0, which means thaty is an eigenvector of T with eigenvalue −λ. Thus, either λ or −λ is aneigenvalue.

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 twoeigenvalues λ = 0 and λ = −1 and determine the eigenvectors belonging toeach such λ.

Solution: Let λ be an eigenvalue of T . Then a − xn = λxn for each n,which means that xn = a

λ+1(assuming that λ 6= −1). Taking the limit, as

n →∞, we see that aλ+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 eigenvectorscorresponding to eigenvalue 0 are the constant sequences. If λ = −1 thenwe see that there is no restriction on xn, other than that a = 0. Thus, thesequences 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 consistingof 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 numberfn where fn+2 = fn+1 + fn and f1 = f2 = 1.

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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 byits two initial values x1, x2. Therefore, the dimension of W is at most 2. Ifan element of W is an eigenvector, it must be associated with an eigenvalueλ2 = λ + 1, which gives the two possible eigenvalues

λ1 =1 +

√5

2, λ2 =

1−√

5

2.

A basis for W is therefore b1 = (λ1, λ21, λ

31, . . .) and b2 = (λ2, λ

22, . . .) which

is clearly invariant under S. In order to express the Fibonacci sequence interms of this basis, we need to find constants k1, k2 such that

1 = k1λ1 + k2λ2

1 = k1λ21 + k2λ

22.

Solving for k1, k2, we find that

fn =1√5

((1 +

√5

2

)n

−(

1−√

5

2

)n).

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 justn∏

i=1

(λ− aii)

where aii is the ith diagonal entry. This expression equals zero if, and onlyif, λ = aii for some i. Thus, the diagonal entries of the matrix are exactlythe eigenvalues.

PART 3 - Optional ProblemRecall 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 nvertices is an n×n graph with the ijth entry equal to 1 if vertex i is adjacentto vertex j and zero otherwise.

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Let f(x) = xn+c1xn−1+c2x

n−2+ . . .+cn be the characteristic polynomialof the adjacency matrix of a graph. Show that c1 = 0, −c2 is the number ofedges in the graph, and −c3 is twice the number of triangles in the graph.

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