Exercise Sheet no. 3

Linear Algebra II

Exercise Sheet no. 3

SS 2011

Prof. Dr. Otto April 29, 2011

Dr. Le Roux Dr. Linshaw

Exercise 1 (Warm-up: Multiple Zeroes) For a polynomialp=Pn

i=0a_iXⁱ∈F[X]define itsformal derivativep⁰by p⁰:=

n

X

i=1

ia_iXⁱ⁻¹.

(a) Check that the usual product rule for differentiaton applies to the formal derivative of polynomials considered here!

(b) Letαbe a zero ofP. Show the equivalence of the following:

i. αis a multiple zero ofp. (In other words,(X−α)²dividesp.) ii. αis a zero ofp⁰.

iii. αis a zero of gcd(p,p⁰).

Solution:

a) The mapp7→p⁰is linear by definition, so it suffices to check that the claim holds for monomialsp=X^kandq=X^l. On the one hand(pq)⁰= (X^k+l)⁰= (k+l)X^k+l−1; on the other handp⁰q+q⁰p=kX^k−1X^l+l X^l−1X^k= (k+l)X^k+l−1. b) Letαbe a zero ofp. Thenp= (X−α)^rqfor someq∈F[X]not divisible byX−α. Then

p⁰= (X−α)^rq⁰+r(X−α)^r−1q.

To see that (i) and (ii) are equivalent, note thatαis a multiple root ofpiffr≥2, which is clearly equivalent to (ii).

To see that (ii) and (iii) are equivalent, note thatαis a zero of bothpandp⁰iff(X−α)is a divisor of gcd(p,p⁰).

Exercise 2 (Commutative subrings of matrix rings)

LetA∈F^(n,n) be ann×nmatrix over a fieldF. LetR_A⊆F^(n,n)be the subring generated byA, which consists of all linear combinations of powers ofA.

(a) Prove thatR_Ais a commutative subring ofF^(n,n).

(b) Consider the evaluation map˜:F[X]→R_Adefined by ˜p=Pn

i a_iAⁱ forp=Pn

i a_iXⁱ. Show that this map is a ring homomorphism. Is it surjective? Injective?

Hint: By forgetting about the multiplicative structure, we may regardF[X]andR_Aas vector spaces overF, and we may regard˜as a vector space homomorphism. DoF[X]andR_Ahave the same dimension asF-vector spaces?

Solution:

a) We haveA^kA^l=A^k+l=A^lA^kfor all0≤k,land since every element of the ring is a linear combination of powers ofA, the claim follows.

b) It is straightforward to check that it is a ring homomorphism, and it is surjective by definition. Since F^(n,n)is a finite-dimensional vector space overFandR_Ais a subspace ofF^(n,n),R_Ais also a finite-dimensional vector space overF. On the other handF[X]is an infinite-dimensional vector space overF. (See exerciseT3.1.) Therefore the map cannot be injective. (Note that this is consistent with the Cayley-Hamilton Theorem.)

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Exercise 3 (The Euclidean algorithm revisited)

Recall the Euclidean algorithm from Exercise Sheet 2. In particular, given natural numbersa,b, we normalise so that d₁=min{a,b}andd₀=max{a,b}. In each step, we divide with remainder, obtainingd_k−1=q_kd_k+d_k+1. At the end of this procedured_k+1=0, andd_k=gcd(a,b).

(a) Letkbe the number of steps needed to compute gcd(a₀,b₀)in this way. Consider the matrixM∈Z^(2,2)given by M=

0 1 1 q₁

0 1

1 q₂

· · · 0 1

1 q_k

.

Show thatMis regular and thatM⁻¹is again a matrix overZ. ComputeM⁻¹ d₁

d₀

. (b) Interpret the entries in second row ofM⁻¹in terms of gcd(d₀,d₁).

(c) Recall that theleast common multiplelcm(d₀,d₁)is an integerzcharacterized by the following properties:

i. d₀|zandd₁|z.

ii. Ifais any integer for whichd₀|aandd₁|a, thenz|a.

Interpret the entries in the first row ofM⁻¹in terms of lcm(d₀,d₁).

Solution:

a) Each matrix 0 1

1 q_k

has determinant−1. Therefore det(M) = (−1)^k, soM is regular. It follows that

M⁻¹=

m₁₁ m₁₂ m₂₁ m₂₂

−1

= (−1)^k

m₂₂ −m₁₂

−m₂₁ m₁₁

which clearly has integer entries. Using the above notation we have 0 1

1 q_l

d_l+1 d_l

= d_l

d_l−1

. It follows that

M⁻¹ d₁

d₀

= 0 1

1 q_k −1

0 1

1 q_k−1 −1

. . . 0 1

1 q₂ −1

0 1

1 q₁ −1

d₁ d₀

= 0 1

1 q_k −1

0 1

1 q_k−1 −1

. . . 0 1

1 q₂ −1

d₂ d₁

=

0 gcd(d₀,d₁)

.

b) LetM⁻¹= p l

m n

. Somd₁+nd₀=gcd(d₀,d₁). Thereforemandnare the coefficients used to write gcd(d₀,d₁) as an integer linear combination ofd₀andd₁.

c) We havepd₁+l d₀=0sopd₁=−l d₀. It follows thatd₀|pd₁andd₁|pd₁. Letz=lcm(d₀,d₁). Thenzalso divides pd₁, that ispd₁=rzfor somer. Moreoverdet(M⁻¹) =±1, sopandlare relatively prime. Sincepd₁=−l d₀=rz and sincez=lcm(d₀,d₁), it follows thatr dividespandl. Sor=±1and|pd₁|=|l d₀|=lcm(d₀,d₁). (See the OWO Lecture Notes from 2008/09.)

Exercise 4 (Polynomial factorisation and diagonalisation) Consider the following polynomials inF[X]forF=Q,RandC:

p₁=X³−2, p₂=X³+4X²+2X, p₃=X³−X²−2X+2.

(a) Which of these polynomials are irreducible inF[X]?

(b) Which of these polynomials decompose into linear factors overF[X]?

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(c) Supposep_iis the characteristic polynomial of a matrixA_i∈F^(3,3). Which of theA_iis diagonalisable overF? Solution:

Over the complex numbers these polynomials decompose as p₁ = (X−p³

2)(X−p³

2ω)(X−p³ 2ω²), p₂ = X(X+2+p

2)(X+2−p 2), p₃ = (X−1)(X+p

2)(X−p 2),

withω=e^23πi.

a) Sincep₁has no rational roots, it is irreducible overQ, whilep₂andp₃are not.

None of these polynomials is irreducible overRorC(every third degree polynomial is reducible overRandC).

b) None of the above polynomials decompose into linear factors overQ. p₂andp₃decompose into linear factors overR, whilep₁does not.

All polynomials inC[X]decompose into linear factors overC, especially so dop₁,p₂andp₃. c) Applying Propositions 1.1.15 and 1.3.1, it follows that

1. the matricesA_iare diagonalisable overC; 2. A₂andA₃, but notA₁, are diagonalisable overR; 3. none of theA_iare diagonalisable overQ.

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