Linear Algebra II
Exercise Sheet no. 1
SS 2011
Prof. Dr. Otto April 11, 2011
Dr. Le Roux Dr. Linshaw
Exercise G1 (Warm-up)
In R3, let g be a line through the origin and E be a plane through the origin such that g is not in E. Determine (geometrically) the eigenvalues and eigenspaces of the following linear maps:
(a) reflection in the planeE.
(b) central reflection in the origin.
(c) parallel projection in the direction ofgontoE.
(d) rotation aboutgthrough 1
3πfollowed by rescaling in the direction ofgwith factor6.
Which of these maps admit a basis of eigenvectors?
Solution:
a) Two eigenvalues:1, with eigenspace E, and−1, whose corresponding eigenspace is the orthogonal complement ofE.
b) One eigenvalue−1with eigenspaceR3.
c) Two eigenvalues: 0, with eigenspaceg, and 1, with eigenspaceE.
d) One eigenvalue: 6, with eigenspaceg.
We have a basis of eigenvectors in cases (i), (ii) and (iii).
Exercise G2 (Warm-up)
(a) Suppose thatϕ:V →V is a linear map over an arbitrary field, and such that all vectorsv∈Vare eigenvectors of ϕ. Show thatϕmust have exactly one eigenvalueλ, and thatϕis preciselyλ·id, where id is the identity map.
(b) Letψ:R4→R4be the map defined by
ϕ
x y z w
=
x y
−w z
.
Find the (real) eigenvalues ofϕand their multiplicity, and find bases for the corresponding eigenspaces.
Solution:
a) Suppose thatλ1andλ2are distinct eigenvalues ofϕ. Letv1andv2be (non-zero) vectors in the corresponding eigenspaces, so thatϕ(v1) =λ1v1andϕ(v2) =λ2v2. Clearlyv1,v2are linearly independent. Thenϕ(v1+v2) = λ1v1+λ2v2. Since every vector inV is an eigenvector, this must be equal toλ(v1+v2)for some scalarλ. Then
(λ1−λ)v1+ (λ2−λ)v2=0,
which implies thatλ1=λ=λ2, by linear independence. This is a contradiction, so there is only one eigenvalue λ. It is immediate thatϕ=λ·id.
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b) The only eigenvalue is1, and a basis for the corresponding eigenspace consists of
1 0 0 0
and
0 1 0 0
.
Exercise G3 (Fixed points of affine maps)
Recall that an affine map is a functionϕ:R2→R2of the formϕ(x) =ϕ0(x) +bwhereϕ0is a linear map andb∈R2 is a vector. In this exercise we are interested in the question of whether such a mapϕhas afixed point,i.e., a pointx such thatϕ(x) =x.
(a) Prove thatϕhas a fixed point, provided that1is not an eigenvalue ofϕ0.
(b) Letϕbe a rotation through the angleαabout a pointc. Give a formula forϕw.r.t. the standard basis, i.e., find functions f and gsuch thatϕ(x,y) = (f(x,y),g(x,y)).
(c) Let%α:R2→R2be a rotation through the angleα(about the origin) and letτc:x7→x+cbe the translation byc.
Using (ii), show that the compositionτc◦%α◦τ−cis a rotation throughαabout the pointc.
(d) Suppose that the linear mapϕ0is a rotation through an angleα6=0. Prove that the affine mapϕ:x7→ϕ0(x) +b has a fixed pointcand thatϕ=τc◦%α◦τ−c, i.e.,ϕis a rotation throughαaboutc.
(Bonus question: how can you find the centrecgeometrically(i.e., without computation)?)
(e) Give an example of an affine mapϕ(x) =ϕ0(x) +bwithout fixed points such thatϕ0is not the identity map.
Solution:
a) A pointxis a fixed point ofϕifϕ0(x) +b=x. We can rewrite this equation as (ϕ0−id)(x) =−b.
We claim that the mapϕ0−id is invertible. Since we are in a finite dimensional vector space it is sufficient to show thatϕ0−id is injective, i.e., thatker(ϕ0−id) =0. Suppose thatv∈ker(ϕ0−id). Then0= (ϕ0−id)(v) = ϕ0(v)−v. Hence,ϕ0(v) =v. Sinceϕ0does not have the eigenvalue1, the only solution to this equation isv=0.
Hence,ker(ϕ0−id) =0, as desired.
It follows thatϕ0−id is invertible and
x= (ϕ0−id)−1(−b) is a fixed point.
b) The rotation throughαaround the origin is the function
%α(x,y) = xcosα−ysinα, xsinα+ycosα . If we rotate aroundc= (cx,cy), we obtain
ϕ(x,y) = cx+ (x−cx)cosα−(y−cy)sinα, cy+ (x−cx)sinα+ (y−cy)cosα . c) Direct computation shows that
(τc◦%α◦τ−c)(x,y) = cx+ (x−cx)cosα−(y−cy)sinα, cy+ (x−cx)sinα+ (y−cy)cosα
. Hence, the result follows by (ii).
d) To show thatϕhas a fixed pointcit is sufficient, by (i), to show thatϕ0does not have the eigenvalue1. Hence, suppose thatv is a vector withϕ0(v) =v. Sinceϕ0is a rotation through an angleα6=0it only fixes the zero vector. Hence,v=0is the only solution to this equation.
It remains to prove thatϕ=τc◦%α◦τ−c. Note thatϕ0=%α. Hence, we have (τc◦%α◦τ−c)(x) =%α(x−c) +c
=%α(x)−%α(c) +c
=%α(x)−%α(c)−b+c+b
=%α(x)−ϕ(c) +c+b
=%α(x) +b
=ϕ(x).
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e) For instance, the map
ϕ(x,y) = (−x, 1+x+y)
has no fixed point. (It must have eigenvalue1. What is the corresponding eigenvector?) Exercise G4 (Eigenvalues and eigenvectors)
Consider the real2×2matrixA=
−2 6
−2 5
and the linear mapϕ=ϕAgiven byAw.r.t. the standard basis.
(a) Calculate the eigenvalues ofAby expanding det(A−λE)and find the zeroes/roots of the characteristic polynomial.
(b) For each eigenvalueλidetermine the eigenspaceVλ
i.
(c) Find a basisB ofR2that only consists of eigenvectors ofϕ and find the matrix of the mapϕwith respect to the basisB.
Solution:
a) We have
det(A−λE) = (−2−λ)(5−λ) +12=λ2−3λ+2= (λ−1)(λ−2).
Thus the characteristic polynomial splits into linear factors corresponding to rootsλ1=1andλ2=2.
b) In order to determine the kernels ofA−λiE, we perform Gauss–Jordan elimination.
A−λ1E=
−3 6
−2 4
−3 6
0 0
A−λ2E=
−4 6
−2 3
−4 6
0 0
We may choosev1= 2
1
such thatspan(v1) =ker(A−λ1E)andv2= 3
2
withspan(v2) =ker(A−λ2E).
c) The vectorsv1andv2form a basisBofR2, since they are linearly independent. W.r.t. to this basisϕis represented by the matrix
1 0 0 2
.
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