Linear Algebra II Tutorial Sheet no. 1
Summer term 2011
Prof. Dr. Otto April 13, 2011
Dr. Le Roux Dr. Linshaw
Discuss and compare as many different solution strategies as possible for the following two questions from your exam.
Exercise T1 (Exam problem 2)
LetB= (b1, . . . ,bn)be an ordered basis of ann-dimensionalF-vector spaceV. (a) LetB0be obtained by replacingbibyb0i=Pi
j=1bjfor1≤i≤n:
B0:= (b1,b1+b2,b1+b2+b3, . . . ,b1+· · ·+bn).
Determine whetherB0always also forms a basis ofV. (b) Forv∈V let
B−v:= (b1−v,b2−v, . . . ,bn−v).
Show that the set of thosev∈V for whichB−visnota basis ofV forms an affine subspace of dimensionn−1 (which contains, and is therefore spanned by, thebi).
Hint: turn the condition thatB−vadmits a non-trivial linear combination of0into a vector equation forv.
Solution:
a) Letϕ:V →V be the mapϕ(bi) = b0i. It is easy to check that the matrix¹ϕº
B
B ofϕ has1’s on and above the diagonal, and zeroes below the diagonal. The determinant of this matrix is1, soϕis invertible andB0also forms a basis ofV.
b) Suppose thatB−vis not a basis ofV. Then there are constantsλ1, . . . ,λn, not all zero, such that λ1(b1−v) +· · ·+λn(bn−v) =0.
Then
λ1b1+· · ·+λnbn=λv whereλ=Pn
i=1λn. First, we claim thatλ6=0; otherwise we would have a relation of linear dependence among the elements ofB. Hence we can divide both sides byλ, obtaining
µ1b1+· · ·+µnbn=v whereµi=λλi. ClearlyPn
i=1µi=1, and this is the only condition onv. Hence the set of allvsuch thatB−vis not a basis ofV is precisely the set of linear combinationsµ1b1+· · ·+µnbnsuch thatPn
i=1µi=1. This is an affine subset of dimensionn−1.
Exercise T2 (Exam Problem 4)
InV =R4, letϕ:R4→R4be the linear map with
ϕ((1, 0, 0, 1)) = (2, 0, 0, 1), ϕ((2, 0, 0, 1)) = (0, 1, 1, 0), ϕ((0, 1, 1, 0)) = (0, 1, 2, 0), ϕ((0, 1, 2, 0)) = (1, 0, 0, 1).
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(a) Check thatb1= (1, 0, 0, 1),b2= (2, 0, 0, 1),b3= (0, 1, 1, 0),b4= (0, 1, 2, 0)form a basisB= (b1,b2,b3,b4)ofR4 and determine the matrix representation¹ϕº
B Bofϕ.
Isϕinjective? Does it have an inverse?
(b) Let S = (e1,e2,e3,e4)be the standard basis. Derive the matrix representations ¹ϕºBS and ¹ϕºSS from ¹ϕºBB through a systematic application of suitable basis transformation matrices.
Solution:
a) It is easy to check that¹ϕº
B
Bhas the form
0 0 0 1
1 0 0 0
0 1 0 0
0 0 1 0
. This matrix is invertible (being a permutation matrix),
and its inverse is just the transpose
0 1 0 0
0 0 1 0
0 0 0 1
1 0 0 0
.
b) By definition, ¹ϕº
B
S is given by
2 0 0 1
0 1 1 0
0 1 2 0
1 0 0 1
. Similarly, ¹idº
B
S is given by
1 2 0 0
0 0 1 1
0 0 1 2
1 1 0 0
. Since
¹idº
S B¹idº
B
S = ¹idº
S
S, it follows that ¹idº
S
B is the inverse of ¹idº
B
S. By an easy computation, ¹idº
S
B is given
by
−1 0 0 2
1 0 0 −1
0 2 −1 0
0 −1 1 0
. It follows that
¹ϕºSS=¹ϕºSB¹idº
S B=
2 0 0 1
0 1 1 0
0 1 2 0
1 0 0 1
−1 0 0 2
1 0 0 −1
0 2 −1 0
0 −1 1 0
=
−2 −1 1 4
1 2 −1 −1
1 4 −2 −1
−1 −1 1 2
.
Exercise T3 (Complex numbers)
Recall that complex numbers are represented by expressions of the form z=a+bi
witha,b∈R,i6∈Ra new constant. Identifyinga∈R with the complex numbera+0iand the new constant iwith 0+1i, one may introduce addition and multiplication as the natural extensions of addition and multiplication inRbased on associativity, commutativity, distributivity and the identityi2=−1.Rthus becomes a subfield of the field of complex numbers.
(a) Letz1=3+4iandz2=5+12ibe complex numbers. Compute
z−11 , z−12 , z12, z22, and z1z2,
and draw them in the complex plane. Find the complex square roots of i,z1 and z2, i.e., solve the equations x2=i,x2=z1,x2=z2overC.
(b) Define forϕ∈R,
eiϕ:=cosϕ+isinϕ.
Show thateiϕeiψ=ei(ϕ+ψ)and(eiϕ)n=einϕfor every natural numbern.
(c) Show that every complex numberz∈C\{0}can be represented as:
z=r eiϕ,
withr∈R>0. Prove that this representation is unique in the following sense:
z=seiψwiths>0impliesr=sandϕ≡ψmod2π.
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(d) Use the representation from (c) to
i. give a geometric description of complex multiplication in terms of rotations and rescalings (i.e., dilations or contractions) inR2.
ii. find all complex solutions ofz5=1and draw these in the complex plane. In general, find all solutions to zn=wforw∈C\{0},n∈N.
Solution:
a)
z1−1= 1
3+4i = 3−4i
(3+4i)(3−4i)=3−4i
25 , z2−1= 1
5+12i =5−12i 169 z12=−7+24i z22=−119+120i, and z1z2=−33+56i Forx2=i:
x1= p2
2 + p2
2 i and x2=− p2
2 − p2
2 i.
Forx2=z1:
x1=2+i and x2=−2−i.
Forx2=z2:
x1=3+2i and x2=−3−2i.
b)
eiϕeiψ = (cosϕ+isinϕ)(cosψ+isinψ)
= (cosϕcosψ−sinϕsinψ) + (cosϕsinψ+sinϕcosψ)i
= cos(ϕ+ψ) +sin(ϕ+ψ)i
= ei(ϕ+ψ),
using the trigonometric formulas forcos(ϕ+ψ)andsin(ϕ+ψ).
The equality(eiϕ)n=einϕthen follows by induction onn.
c) Letz=a+bi6=0. Trying to find a representationz=r eiϕ=rcosϕ+ (rsinϕ)imeans solvinga=rcosϕand b=rsinϕ. One findsrby observing that
a2+b2=r2(cos2ϕ+sin2ϕ) =r2, so r=p
a2+b2>0and ris uniquely determined (it is themodulusofz). Furthermore,ϕ has to be an angle such that
cosϕ= a p
a2+b2
and sinϕ= b p
a2+b2 .
This has a unique solutionϕ0∈[0, 2π), called theargumentofz. The argument ofzis just the angle between the positive part of thex-axis and the vectorzin the complex plane. Furthermore, the set of all solutions is given by {ϕ0+2kπ:k∈Z}.
d) 1. Letz∈C. Ifz=0, thenwz=0for allw∈C. Assume thatz6=0, so by (iii), it has the formz=r eiϕwith r>0andϕ∈[0, 2π). Then, for any complex numberw, multiplication ofwbyzis equivalent to a rotation ofwthrough angleϕfollowed by a rescaling using the modulusrofz.
2. z=0is certainly not a solution ofz5=1, so assumez is of the formr eiϕwithr>0. We have to solve the equation:
(r eiϕ)5=1, that is r5ei5ϕ=1ei0.
By the uniqueness of the representation, this implies r5=1and 5ϕ =0mod2π. So r=1, since r>0.
Then by solving5ϕ=2πkfor every integerkwith0≤k<5, we find solutionsϕk= 2π5k ∈[0, 2π). This means we have found five different solutions (viz, eiϕk with0≤ k <5), which must be all, since a fifth degree equations can have at most five solutions.
For generalw=eiϕ∈C\ {0}, the equationzn=r eiϕhas n solutions zk=pn
r eiϕk, with ϕk=2πk
n , k=0, . . . ,n−1.
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