Linear Algebra II Tutorial Sheet no. 5
Summer term 2011
Prof. Dr. Otto May 6, 2011
Dr. Le Roux Dr. Linshaw
Exercise T1 (Polynomials of matrices and linear maps)
In the following p,qstand for polynomials inF[X],ϕ for an endomorphism of an n-dimensionalF-vector spaceV, A,Bforn×nmatrices overF. Which of these claims are generally true, which are false in general (and which are plain nonsense)?
(a) p(AB) =p(A)p(B)(?)
(b) (pq)(A) =p(A)q(A) =q(A)p(A)(?) (c) (p(ϕ))(v) =p(ϕ(v))(?)
(d) ¹p(ϕ)ºBB=p(¹ϕºBB)(?) (e) Aregular⇒p(A)regular (?)
(f) A∼B⇒p(A)∼p(B)(?)
(g) ϕ(v) =λv⇒(p(ϕ))(v) =p(λ)v(?) (h) p(A)q(A) =0⇒(p(A) =0∨q(A) =0)(?)
(i) ϕandp(ϕ)have the same invariant subspaces (?)
(j) U⊆V an invariant subspace ofϕ ⇒ U invariant underp(ϕ)(?)
(k) U⊆V an invariant subspace ofϕ ⇒ (p(ϕ))(v+U) = (p(ϕ))(v) +U (?) (ϕviewed as a map on subsets of V.) (l) U ⊆V an invariant subspace ofϕ0 ⇒ (p(ϕ0))(v+U) = (p(ϕ))(v) +U (?) (ϕ0the induced endomorphism of
V/U.) Solution:
a) False in general, even ifAandBcommute.
b) True,p7→p(A)is a ring homomorphism.
c) Nonsense!
d) True.
e) False, for instancepA(A) =0is not regular.
f) True.
g) True.
h) False, considerp=q=X andA= 0 1
0 0
. i) False, considerpϕ.
j) True.
k) False, considerϕ=0.
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l) True.
Exercise T2 (Eigenvectors) Consider the matricesA:=
1 2 3 2 1 3 3 3 6
andB:=
2 1 0 0
0 2 0 0
0 0 1 1
0 0 −2 4
(a) Determine the characteristic and minimal polynomials ofAandB.
(b) For the matrix B:
i. Show thatv1= (1, 0, 0, 0)andv2= (0, 0, 1, 1)are eigenvectors with eigenvalue 2.
ii. Determine an eigenvectorv4with eigenvalue3.
iii. Check thatv3= (0, 1, 0, 0)is a solution of(B−2E4)2x=0and thatBv3=2v3+v1. iv. Determine the matrix that representsϕBw.r.t. the basis(v1,v3,v2,v4).
Solution:
a) ForAwe obtainpA=
1−X 2 3
2 1−X 3
3 3 6−X
=−X(X+1)(X−9).
Since this polynomial splits into distinct linear factors, the minimal polynomial ofAis equal to−pAby Proposition 1.5.2 on page 33 of the notes. (Recall that the minimal polynomial is normalised.)
ForBwe obtainpB=
2−X 1 0 0
0 2−X 0 0
0 0 1−X 1
0 0 −2 4−X
=
2−X 1 0 2−X
1−X 1−2 4−X
= (2−X)3(X−3).
The minimal polynomialqBhas the same linear factors as the characteristic polynomialpB. SoqBhas to be one of the following:(X−2)(X−3), or(X−2)2(X−3), orpB.
As(B−2E4)(B−3E4)6=0and(B−2E4)2(B−3E4) =0, we conclude thatqB= (X−2)2(X−3).
b) For the matrix B:
i. SinceB−2E4=
0 1 0 0
0 0 0 0
0 0 −1 1 0 0 −2 2
we have
(B−2E4)
1 0 0 0
=0and(B−2E4)
0 0 1 1
=0.
ii. We are looking for a non-trivial solution of the equation(B−3E4)v4=0:
−1 1 0 0 0
0 −1 0 0 0
0 0 −2 1 0
0 0 −2 1 0
−1 0 0 0 0
0 −1 0 0 0
0 0 −2 1 0
0 0 0 0 0
So one solution isv4= (0, 0, 1, 2).
iii. We obtain(B−2E4)2=
0 0 0 0
0 0 0 0
0 0 −1 1 0 0 −2 2
.
Hence(B−2E4)2
0 1 0 0
=0. FurthermoreBv3=
1 2 0 0
=2v3+v1.
2
iv. We obtain
2 1 0 0
0 2 0 0
0 0 2 0
0 0 0 3
Exercise T3 (Complexification)
ForA∈R(2,2)consider the associated endomorphismsϕARandϕCA, which are represented by A w.r.t. the standard bases ofR2and ofC2, respectively.
Let the characteristic polynomialpAbe irreducible inR[X].
(a) Show thatpAhas a pair of complex conjugate zeroes. (Recall that the complex conjugate ofz=α+iβis¯z=α−iβ.) (b) Show thatC2has a basisB= (v, ¯v)of eigenvectors ofϕAconsisting of a vectorvwith eigenvalueλ, and its complex
conjugate¯v, which has eigenvalueλ.¯
(c) Letb1=12(v+¯v)andb2=2i1(v−¯v), which lie inR2. i. Show thatB0={b1,b2}is a basis forR2.
ii. Determine the matrix representation ofϕAR w.r.t. basisB0 and discuss the similarity ofAwith a matrix that would suggest the interpretation as "rotation followed by dilation"
Solution:
a) The characteristic polynomial is a quadratic of the form x2+t x+d, where t =tr(A)and d = det(A). By the quadratic formula,x= −t±p
t2−4d
2 . SincepA(x)is irreducible inR[X]we must havet2−4d<0, so it is clear that the roots are distinct and occur as a complex conjugate pair.
b) Letλandλ¯be the eigenvalues ofϕA, and letvbe an eigenvector ofϕAwith eigenvalueλ, so thatAv=λv. Taking complex conjugates of both sides and noting thatA¯=AsinceAis real, we haveA¯v=λ¯¯v. Hence¯vis eigenvector ofϕA with eigenvalueλ. Finally, let¯ B ={v, ¯v}. The fact thatB is a basis forC2is clear from the fact that the corresponding eigenvaluesλandλ¯are distinct.
c) i. First we regardb1andb2as elements ofC2. It is clear that they form a basis ofC2because they are related tovand¯vvia the invertible matrix
¹idº
B0 B =
1
2 1 1 2i 2 −2i1
.
Sinceb1andb2are linearly independent overC, they must be linearly independent overRas well. Hence if we regard them as elements ofR2, they form a basis ofR2.
ii. We have¹ϕAº
B0
B0=¹idº
B
B0◦¹ϕAº
B B◦¹idº
B0
B. Also, note that¹ϕAº
B B=
λ 0 0 λ¯
and¹idº
B
B0= (¹idº
B0 B)−1= 1 1
i −i
. We therefore obtain
¹ϕº
B0 B0=
Re(λ) Im(λ)
−Im(λ) Re(λ)
=r
cos(θ) sin(θ)
−sin(θ) cos(θ)
,
whereλ=r eiθ.
Exercise T4 (Simultaneous diagonalisation and polynomials)
LetA∈R(n,n) be a matrix withndistinct real eigenvalues, and letB∈R(n,n)be an abitrary matrix such thatAandB are simultaneously diagonalisable. Show that there exists a polynomialp∈R[X]such thatB=p(A).
Hint. Recall that, last semester in Linear Algebra I, we have shown in exercise (E14.2) that, given ndistinct real numbersa1, . . . ,an∈Randnarbitrary real numbersb1, . . . ,bn∈R, there exists a polynomialpof degreen−1such that p(ai) =bifor alli.
Solution:
By assumption, there exists a matrixCsuch thatD:=C−1ACandH:=C−1BCare both diagonal matrices. Suppose that D=
λ1
...
λn
andH=
µ1
...
µn
.
Note that λ1, . . . ,λnare the eigenvalues of A. Since these are all distinct, we can use the hint to find a polynomial p such that p(λi) = µi . It follows that p(D) = H. Since(C DC−1)k = C DkC−1 it follows that p(A) = p(C DC−1) = C p(D)C−1=C H C−1=B.
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