Linear Algebra II
Exercise Sheet no. 15
Summer term 2011
Prof. Dr. Otto July 13, 2011
Dr. Le Roux Dr. Linshaw
Exercise 1 (Characteristic and minimal polynomials)
Find the characteristic and minimal polynomials of the following matrix
A=
1 1
1
1 11
0
2 1 2 1
2
0
2 44
Solution:
The characteristic polynomial is
pA(x) = (1−x)4(2−x)4(4−x)2, and the minimal polynomial is
qA(x) = (x−1)2(x−2)3(x−4).
Exercise 2 (Jordan normal form)
(a) Letϕbe an endomorphism of a ten-dimensionalF-vector spaceV. W.r.t. basisB= (b1, . . . ,b10)letϕbe represented by a Jordan normal form matrix with three Jordan blocks for the same eigenvalueλ∈F, of sizes2,3and5. Let ψ:=ϕ−λid. Complete the following table:
i 1 2 3 4 5 6 7 8 9 10
dim(kerψi) 10 10 10 10 10 10
In the notation of Lemma 1.6.4 of the notes: for whichv∈V does¹vºhave maximal dimension? Split the basisB in a way to obtain bases for the two invariant subspacesV =¹vº⊕V0(as in Claim 1.6.5). Ifϕ0is the restriction ofϕ toV0, what is the matrix representation ofϕ0with respect to this basis? Ifψ0=ϕ0−λid, how is the corresponding table forψ0related to the above?
(b) Now, letϕbe another endomorphism ofVwith characteristic polynomial(λ−X)10. Suppose we have the following data forψ=ϕ−λid:
i 1 2 3 4 5 6 7 8 9 10
dim(kerψi) 3 5 7 8 9 10 10 10 10 10
Determine the Jordan normal form representation ofϕfrom this data (up to permutation of Jordan blocks).
(c) (extra) In general, letϕ0andϕ1be two endomorphisms ofF-vector spacesV0andV1of the same finite dimension, with the same characteristic polynomial that splits into linear factors. Suppose moreover that for each eigenvalue λofϕ0andϕ1, the tables forψ0=ϕ0−λidandψ1=ϕ1−λidare the same.
Sketch a proof for the similarity ofϕ0andϕ1adapting the argument for the existence and uniqueness of the Jordan normal form. How can this be used to give a “different” proof for the similarity ofAandAt for any matrixA∈C(n,n)? 1
Solution:
a) Let
¹ϕº
B
B=A=
λ 1 0
λ 1 λ
λ 1 λ
λ 1 λ 1
λ 1 λ 1
0 λ
.
The entries of the table are:
i 1 2 3 4 5 6 7 8 9 10
dim(kerψi) 3 6 8 9 10 10 10 10 10 10
Forv=λb10withλ6=0, does¹vºhave maximal dimension 5. The natural choice for a basis forV0is(b1, . . . ,b5), and the representation ofϕ0with respect to this basis is the5×5-matrix in the top left corner ofA:
A=
λ 1 0
λ 1 λ
λ 1 λ
.
The table forψ0has the form:
i 1 2 3 4 5
dim(kerψ0i) 2 4 5 5 5
Theith entry of this table isiless than the entry of the previous table, until we reach the dimension of¹vº. b) The Jordan normal form is:
λ 0
λ 1 λ 1
λ λ 1 λ 1
λ 1 λ 1
λ 1
0 λ
.
Whenbi∈kerψj+1\kerψj, thenjdepends as follows oni:
i 1 2 3 4 5 6 7 8 9 10 j 1 1 2 3 1 2 3 4 5 6
c) Sketch: we can without loss of generality restrict ourselves to the case thatϕ0andϕ1haveλas sole eigenvalue.
Then we proceed by induction ondimV0=dimV1. By assumption, the tables forλare the same for bothψ0= ϕ0−λidandψ1=ϕ1−λid. This means the first jsuch thatdim(kerψ0j) =dimV0is also the first jsuch that dim(kerψ1j) =dimV1. Then there are vectorsv0∈kerψ0j+1\kerψ0j andv1∈kerψ1j+1\kerψ1j, allowing us to split upV0and V1asV0=¹v0º⊕V00 andV0 =¹v1º⊕V10. We letϕ00andϕ01be the restricitions ofϕ0andϕ1to V00 and V10 respectively. We know thatϕ00 andϕ10 also haveλas sole eigenvalue! , their tables are obtained in the same manner from those ofϕ0andϕ1, and thereforeϕ00 andϕ10 are represented by similar matrices by induction hypothesis. This can be extended to an isomorphism of V0 andV1showing thatϕ0 and ϕ1are represented by similar matrices, by sendingv0tov1,ϕ0v0toϕ1v1, etcetera. This completes the (sketch of the) proof.
SinceAandAt have the same characteristic polynomial, anddim(ker(A−λE)i) =dim(ker(At−λE)i)for every eigenvalueλandi∈N(for a matrix has the same rank as its transpose), we deduce that every matrix is similar to its transpose overC.
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Exercise 3 (Diagonalization using orthogonal matrices)
Letϕbe the endomorphism ofR3given in the standard basis by
A=
1 −1 −1/p 2
−1 1 1/p 2
−1/p 2 1/p
2 2
.
(a) Find an orthogonal matrixBsuch thatB−1ABis diagonal.
(b) Describe all subspaces ofR3which are invariant underϕ.
(c) Forx= (x1,x2,x3)∈R3, letQbe the quadratic form
Q(x) =x12+x22+2x23−2x1x2− 2
p2x1x3+ 2 p2x2x3.
Find the principle axes of the quadricX given byQ(x) =1.
Solution:
a) The characteristic polynomial ofAis−(x−1)(x−3)(x), and the normalized eigenvectors of eigenvalues3,1, and 0, respectively, are
v1=
−1/2 1/2 1/p 2
, v2=
1/2
−1/2 1/p
2
, v3=
1/p
2 1/p
2 0
.
Therefore the orthogonal matrixB=
−1/2 1/2 1/p 2 1/2 −1/2 1/p
2 1/p
2 1/p
2 0
satisfiesB−1AB=
3 0 0 0 1 0 0 0 0
.
b) The invariant subspaces are{0}, span(v1), span(v2), span(v3), span(v1,v2), span(v1,v3), span(v2,v3), andR3. c) Since the matrix representingQis preciselyA, the principal axes ofX lie along the vectorsv1,v2,v3.
Exercise 4 (Invariant planes inR4)
Letϕ be an orthogonal transformation ofR4which fixes a plane U1 pointwise, and acts by a nontrivial rotation on another planeU2. Prove thatU1andU2are the only invariant subspaces ofR4of dimension2.
Solution:
First, we haveU1∩U2={0}, since this is the only vector that is fixed pointwise under a nontrivial rotation. Therefore we have a direct sum decompositionR4=U1⊕U2. Suppose thatV is a two-dimensional invariant subspace ofR4, and V 6= U1and V 6=U2. Fix a non-null vectorv∈V; we have a unique decomposition v=u1+u2, withui∈Ui. Since V 6=U1, we may choosevso thatu26=0.
First, we claim that u1 6=0. Otherwise, we would have v ∈ U2, and hence V = U2 since v andϕ(v)are linearly independent. (This follows from the fact thatϕacts by a non-trivial rotation onU2). Then
ϕ(v) =ϕ(u1) +ϕ(u2) =u1+ϕ(u2).
Sinceu26=0, we havev−ϕ(v) =u2−ϕ(u2)6=0. SinceV is invariant underϕ, it follows thatu2−ϕ(u2)lies inV. Butu2−ϕ(u2)also lies inU2. By applyingϕ again, we see thatu2−ϕ(u2)andϕ(u2−ϕ(u2)) spanU2, so we have V =U2, which is a contradiction.
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