Linear Algebra II

Linear Algebra II

Exercise Sheet no. 2

SS 2011

Prof. Dr. Otto April 20, 2011

Dr. Le Roux Dr. Linshaw

Exercise 1 (Further properties of eigenvalues/eigenspaces) Throughout this exercise,Ais a square matrix with entries inR.

(a) Prove or disprove thatAandA^t have the same eigenvalues. (A^t is the transpose ofA, see lecture notes for Linear Algebra I.)

(b) Prove or disprove thatAandA^thave the same eigenspaces.

(c) AssumeAis regular and letvbe an eigenvector ofAwith eigenvalueλ. Show thatvis also an eigenvector ofA⁻¹ with eigenvalue _λ¹.

(d) Letvbe an eigenvector of the matrixAwith eigenvalueλand lets be a scalar. Show thatvis an eigenvector of A−sEwith eigenvalueλ−s.

Solution:

a) The claim is correct: ifλis an eigenvalue ofA, then det(A−λE) =0. Since the determinant is invariant under transposition, also det(A−λE)^t =det(A−λE) =0. Making the following computation:(A−λE)^t=A^t−(λE)^t= A^t−λE, we see that det(A^t−λE) =0. Thereforeλis also an eigenvalue ofA^t. The other direction follows by symmetry ((A^t)^t=A).

b) This claim is incorrect, as can be seen by considering the following counterexample. WhenA:=

0 1 0 0

, the only eigenvalue ofAis 0. Its eigenspace isV₀=span(e₁) ={α

1 0

:α∈R}. 0 is also the only eigenvalue ofA^t, but its eigenspace isV₀⁰=span(e2) ={α

0 1

:α∈R}.

c) Letλbe an eigenvalue ofAandvbe one of its eigenvectors. ThenAv=λvholds. SinceAis invertible,A⁻¹exists.

By multiplying the previous equation byA⁻¹from the left, we obtainv=λA⁻¹v. Now we multiply by _λ¹ and get A⁻¹v= ¹_λv(λ6=0asAis regular). Thusvis an eigenvector ofA⁻¹with eigenvalue _λ¹.

d) CombiningAv=λvandsEv=svwe get:Av−sEv=λv−sv. This can be written as(A−sE)v= (λ−s)v, therefore vis an eigenvector ofA−sEwith eigenvalueλ−s.

Exercise 2 (Application of diagonalisation: Fibonacci Numbers, Golden Mean)

Recall that the sequence f₀,f₁,f₂, . . .of Fibonacci numbers is inductively defined as follows (cf Exercise H6.4 from LA I):

f₀ = 0, f₁ = 1, f_k+2 = f_k+1+f_k.

(a) We defineuk= f_k+1

f_k

∈R². Find a matrixAsuch thatuk+1=Aukfor allk∈N.

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(b) What are the eigenvalues ofA? Give an explicit formula for f_k.

Hint: Use the eigenvaluesλ1andλ2as abbreviations as long as possible.

(c) Compute the limita=lim_k→∞ ^fk+1_f

k .

The limit is called the Golden Mean, it divides a line segment of length1into two parts a and 1−a such that

1 a =₁₋^a_a. Solution:

a) u_k+1= f_k+2

f_k+1

=

f_k+1+f_k f_k+1

= 1 1

1 0 f_k+1

f_k

= 1 1

1 0

u_k.Therefore,

A= 1 1

1 0

.

As a consequence, for allk∈N,

u_k=A^ku₀=A^k 1

0

.

b) det(A−λE) =λ²−λ−1= (λ−¹⁺₂^p5)(λ−¹⁻₂^p5) =0.

Thus, the eigenvalues ofAare:

λ1= 1+p 5

2 and λ2=1−p 5 2 . Corresponding eigenvectors arev₁=

λ1

1

forλ1andv₂= λ2

1

forλ2. It follows thatA=S DS⁻¹, withS=

λ1 λ2

1 1

and D=

λ1 0 0 λ₂

. The conjugation map M 7→S M S⁻¹is an automorphism of the ring ofn×nmatrices (and in particular preserves sums and products of matrices), so we haveA^k= (S DS⁻¹)^k=S D^kS⁻¹. SinceS⁻¹=^p1₅

1 −λ2

−1 λ1

,

A^k = S D^kS⁻¹= 1 p5

λ₁ λ₂

1 1

‚λ₁^k 0 0 λ^k₂

Œ 1 −λ₂

−1 λ₁

= 1 p5

‚λ^k₁⁺¹ λ^k₂⁺¹ λ^k₁ λ₂^k

Œ 1 −λ2

−1 λ1

= 1 p5

‚λ^k₁⁺¹−λ₂^k+1 −λ₁^k+1λ2+λ^k₂⁺¹λ1

λ₁^k−λ₂^k −λ^k₁λ2+λ^k₂λ1

Π. It follows that

f_k+1 f_k

=u_k=A^k 1

0

= 1 p5

‚λ^k+1₁ −λ^k+1₂ λ^k₁−λ^k₂

Π,

hence

f_k= 1

p5(λ₁^k−λ₂^k) = 1 p5

‚1+p 5 2

Œk

−

‚1−p 5 2

Œk! .

c) We have that

f_k+1

f_k = ^λk+11_λk^−λ2k+1 1−λ₂^k = ^λ_λ^k+11k

1 ·

1−^λ k+1 2 λk+1

1 1−^λ k2 λk 1

=λ1·¹⁻⁽

λ2λ1)^k+1 1−(^λ2_λ1)^k and

λ2 λ1

=

1−p 5 1+p

5

=^pp5−1₅₊₁<1.

It follows that

k→∞lim f_k+1

f_k =λ₁·

1−lim_k→∞λ

2 λ1

k+1

1−lim_k→∞λ

2 λ1

k =λ₁.

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Exercise 3 (Euclidean algorithm and recursive functions)

The greatest common divisord=gcd(a,b)of two non-zero natural numbersaandbis a natural number characterised by the following property:

• d|bandd|a.

• Ifr|aandr|b, thenr|d

The Euclidean algorithm is a procedure for determining the greatest common divisor of two numbersaandb.

Step0: Swap a and b ifa<b.

Euclid (a,b): IFb=0THEN returna, ELSE return Euclid(b,amodb).

[After initialising: d₁:=min{a,b},d₀:=max{a,b}, we divide with remainder in each step: d_k−1=q_kd_k+d_k+1with 0≤d_k+1<d_k, and ends ifd_k+1=0. We getd_k=gcd(a,b).]

(a) Prove that the gcd is well defined, that is, uniquely characterised by the above properties.

(b) Prove that gcd(a,b) =gcd(b,amodb)for0<b.

(c) Assume thatb<aand that(a,b)→^E (a⁰,b⁰)→^E (a⁰⁰,b⁰⁰)are two steps in the Euclidean algorithm. Prove thata⁰⁰< ^a₂ andb⁰⁰<₂^b.

(d) Deduce that Euclid(a,b)returns gcd(a,b).

Solution:

a) 1divides bothaandbso the set of common divisors is non-empty. Letdandd⁰satisfy the second condition. Then d|d⁰andd⁰|d, sod=d⁰.

b) Supposea=q b+rwhere0≤r<b, sor=amodb. Sincee|aande|biffe|r=a−q bande|b. Letd=gcd(a,b) andd⁰=gcd(b,r). We haved|d⁰andd⁰|d, sod=d⁰.

c) Ifb≤ ^a₂, then b⁰< b≤ ^a₂; if ^a₂ <b, thenq=1so b⁰ =a−b< ^a₂. So in every case, b⁰< ^a₂. Similarly b⁰⁰< ^a₂⁰. Moreovera⁰=banda⁰⁰=b⁰, soa⁰⁰<^a₂ andb⁰⁰< ^b₂.

d) Every step(a,b)→^E (a⁰,b⁰)of the algorithm preserves the gcd, in the sense that gcd(a,b) =gcd(a⁰,b⁰). Moreover the algorithm terminates due to part (c), so we obtain(gcd(a,b), 0)after finitely many steps.

Exercise 4 (Diagonalization and recursive sequences)

Leta_k be the sequence of real numbers defined recursively as follows: a₀=0,a₁=1, anda_k+2= ¹₂(a_k+1+a_k). In other words, each term in the sequence is the average of the two previous terms.

(a) As we did with the Fibonacci sequence, we want to study this sequence using diagonalization of matrices. For k≥0, let

u_k= a_k+1

a_k

.

Using the equationsa_k+2= ¹₂(ak+1+a_k)anda_k+1=a_k+1, find a2×2matrixAsuch thatu_k+1=Au_k. (b) Find the eigenvalues and eigenvectors ofA, and find a matrixSand a diagonal matrixDwithD=S⁻¹AS.

(c) Find a formula fora_k, and calculatelim_k→∞a_kif it exists.

Solution:

a) A=

‚ ₁

2 1 2

1 0

Π.

b) The eigenvalues ofAare1and−¹₂. Forλ=1, the eigenspaceV₁is spanned by

1 1

. Forλ=−¹₂, the eigenspaceV_−1/2is spanned by 1

−2

. So S=

1 1 1 −2

andD=

‚ 1 0 0 −¹₂

Π.

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c) First, we haveu_k=A^ku₀=S D^kS⁻¹u₀, whereu₀= 0

1

, anda_kis just the second component of the vectoru_k. We calculate

S⁻¹=

2/3 1/3 1/3 −1/3

, S⁻¹u₀=

1/3

−1/3

, D^k=

‚ 1 0 0 (−¹₂)^k

Π.

It follows easily thata_k=¹₃+²₃ −¹₂k

. It is clear from this formula thatlimk→∞a_k=¹₃.

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