Linear Algebra II Tutorial Sheet no. 4
Summer term 2011
Prof. Dr. Otto May 2, 2011
Dr. Le Roux Dr. Linshaw
Exercise T1 (Algebraic and geometric multiplicity)
Letϕ be an endomorphism on a finite dimensionalF-vector spaceV and λ∈Fan eigenvalue ofϕ with geometric multiplicitydand algebraic multiplicitys. Show thatd≤s.
Hint: Choose a basisBofV that containsdeigenvectors ofϕwith eigenvalueλ.
Solution:
We choose a basisv1,· · ·,vd for the eigenspaceVλand extend it to a basisv1,· · ·,vnforV, wherenis the dimension of V. The matrix ofϕwith respect to this basis is a block upper tringular matrix of the form
A=¹ϕº
B B=
λEd B
0 C
,
where Ed is thed×d-unity matrix,B∈F(d,n−d)andC∈F(n−d,n−d). By successive expansion w.r.t. first, second, .. d-th column we findpϕ= (λ−x)ddet(C−x E) = (λ−x)dpC. Indeed
A−x En=
λ−x 0
... B
0 λ−x
0 C−x E
Hence,(λ−x)d is a divisor of the characteristic polynomialpϕ and we conclude thats≥d, wheresis the algebraic multiplicity of the eigenvalueλ.
Exercise T2 (Upper triangle shape)
Find a real upper triangular matrix similar to
A=
3 0 −2
−2 0 1
2 1 0
.
Solution:
Consider A1 = Aas the representation of a linear endomorphism ϕ1 of R3 with respect to the standard basis B1 = (e1,e2,e3). The corresponding characteristic equation
pϕ1=pA1=det
3−λ 0 −2
−2 −λ 1
2 1 −λ
=−λ3+3λ2−3λ+1= (1−λ)3=0,
hasλ1=1as only solution. A corresponding eigenvector can be found by solving the homogeneous system of equations (A1−E3)v=0. A possible result is
v1=
1
−1 1
.
1
We extendv1to a new basis, for exampleB2= (v1,e2,e3). The transition matrix and its inverse are given by
S1=
1 0 0
−1 1 0
1 0 1
and S1−1=
1 0 0
1 1 0
−1 0 1
.
We obtain
A2=S1−1A1S1=
1 0 −2 0 0 −1
0 1 2
,
as the representation ofϕ1with respect to the new basisB2. Our next step is to consider the endomorphismϕ2of the subspace ofR3spanned by(e2,e3)represented by the submatrix
A02=
0 −1
1 2
with respect to that basis. From the characteristic equationpϕ2=pA0
2=λ2−2λ+1= (λ−1)2=0, we getλ2=1. From the homogeneous system(A2−E2)v2=0we obtain
v2= 1
−1
.
After extending(v1,v2=e2−e3)to a new basis, for example: B3= (v1,v2,e3), we compute the transition matrix and its inverse:
S2=
1 0 0
−1 1 0 1 −1 1
and S2−1=
1 0 0 1 1 0 0 1 1
.
We derive
A3=S2−1A1S2=
1 2 −2 0 1 −1
0 0 1
,
whereA3has the desired upper triangle form.
Exercise T3 (Ideals)
Recall that a non-empty subset I of a commutative ringRis called an ideal, if it is closed under addition and under multiplication with arbitrary ring elements. The principal idealIagenerated by a fixed elementa∈Ris defined by
Ia={r a: r∈R}
as the set of all multiples ofa(see Definition 1.2.16 on page 22 of the notes).
(a) Verify thatIais the smallest (⊆-minimal) ideal containinga.
(b) LetI andJ be two ideals in a commutative ringR. Prove that
I+J={i+j:i∈I,j∈J}
is again an ideal, in fact, the smallest ideal containing bothIandJ.
(c) Prove that every ideal overZis principal. Is the same true in the ringsZn(n∈Z)? (As already discussed in H3.3 from LA I 2010/11.)
(d) For two elementsm,n∈Z, the setIm+Inis an ideal overZ, hence principal. This means thatIm+In=Ikfor some elementk∈Z. Expresskin terms ofmandn.
(e) For any two idealsI andJ in a commutative ringR, find an expression forI∧J, the largest ideal contained in both IandJ. Over the ringZ, how does one determine for any pairm,n∈Zthek∈Zsuch thatIm∧In=Ik?
Solution:
2
a) a∈Ia,r a+sa= (r+s)aands(r a) = (sr)a, soIa is an ideal. Since ideals have to be closed under multiplication by arbitrary ring elements,Ia⊆I for any idealI containinga.
b) I+J is non-empty, since bothI and J are, and therefore I+J is an ideal by the equalities(i+j) + (i0+j0) = (i+i0) + (j+j0)ands(i+j) =si+s j. Since ideals are closed under addition,I+J ⊆Kfor any idealKcontaining bothIandJ.
c) LetI be a ideal overZcontaining elements other than0. Sincei ∈I implies−i ∈I, leta be the least positive element ofI, and let j∈I. By the division algorithm, there exist integersk,lsuch that j=ka+lwith0≤l<a.
Sincel=j−ka∈I and by definition ofa, we havel=0.
Also for every idealI inZn, define
J:={k∈Z:kmodn∈I}.
J is easily seen to be an ideal inZ, so it is generated by somea∈Z. ThereforeIis generated byamodn.
d) From a previous OWO lecture or Exercise H3.3 from LA I 2010/11, it should be known thatIm+In={am+bn: a,b∈Z}is precisely the set of all multiples of the greatest common divisor ofmandn.
e) For any two idealsI andJ in a commutative ring, their intersectionI∩J is again an ideal, which is then clearly the largest contained in both.
Form,n∈Z, let Im and In be the corresponding ideals. Recall that the least common multiple lcm(m,n)is an integerkcharacterized by the following properties:
1. m|kandn|k.
2. Ifais any integer for whichm|aandn|a, thenk|a.
We claim thatIm∧In=Ik, wherek=lcm(m,n). ClearlyIm∧In=Ik for somek, and we havem|kandn|ksince Ik⊆ImandIk⊆In. Suppose thatn|aandm|a for some integera. Thena∈In∩Im, soa∈Ik by definition, and hencek|a. It follows thatk=lcm(m,n).
3
