If you find any mistakes, please make a comment! Thank you.

Chapter 8 Exercise C


1. Solution: Because
4=dimC4=dimG(3,T)+dimG(5,T)+dimG(8,T), it follows that the multiplicities of the eigenvalues of T are at most 2. Thus p(z)=(z3)2(z5)2(z8)2 is a polynomial multiple of the characteristic polynomial. Therefore, we must have p(T)=0.


2. Solution: We have
1dimG(5,T),dimG(6,T)n1. Hence p(z)=(z5I)n1(z6I)n1 is a polynomial multiple of the characteristic polynomial. Therefore, p(T)=0.


3. Solution: Define TL(C4) by
T(z1,z2,z3,z4)=(7z1,7z2,8z3,8z4). Since E(7,T)G(7,T),E(8,T)G(8,T) and C4=E(7,T)E(8,T), it follows that E(7,T)=G(7,T) and E(8,T)=G(8,T). We have dimE(7,T)=dimE(8,T)=2. Hence, the characteristic polynomial of T is
(z7)2(z8)2.


4. Solution: A=(0000000000000000) Let
A=(1100051000500005). Then
(A5I)2(AI)=(4100001000000000)2(0100041000400004)=(4100001000000000)(0010004000000000)=(0000000000000000). Define TL(C4) by
T(z1,z2,z3,z4)=(z1+z2,5z2+z3,5z3,5z4). Then M(T)=A and the eigenvalues of T are thus 1 and 5 (the entries on the diagonal). Now 8.36 implies that the minimal polynomial of T is a polynomial multiple of (z5)(z1). The previous work shows that (T5I)(TI)0 and (T5I)2(TI)=0. Hence (z5)2(z1) is the minimal polynomial of T. By Exercise 11 in section 8B, the multiplicity of 1 is 1 and of 5 is 3. Thereby the characteristic polynomial of T is (z1)(z5)3.


5. Solution: Let
A=(0000011000100003). Then
(AI)2(A3)A=(1000001000000002)2(3000021000200000)(0000011000100003)=(1000001000000002)2(0000021000200000)=(1000001000000002)(0000002000000000)=(0000000000000000). Define TL(C4) by
T(z1,z2,z3,z4)=(0,z2+z3,z3,3z4). Then M(T)=A. The rest of this exercise is almost the same as the previous one.


6. Solution: Define TL(C4) by
T(z1,z2,z3,z4)=(0,z1,z2,3z4). Then, the standard basis of C4 consists of eigenvectors of T corresponding to the eigenvalues 0,1,1,3. Applying T(TI)(T3) to each of these basis vectors shows that T(TI)(T3)=0. Hence z(z1)(z3) is the minimal polynomial of T. We have
M(T)=(0000010000100003). Thus, by Exercise 11 in section 8B, the characteristic polynomial of T is z(z1)2(z3).


7. Solution: By Exercise 4 in section 5B, we have
(1)V=nullPrangeP. It is easy to check that if vrangeP then Pv=v. Thus
nullPG(0,T) and rangePG(1,T). (1) and 8.26 then imply that these inclusions are actually equalities, which gives the desired result.


8. Solution: Let p denote the minimal polynomial of T. We have
T is invertible 0 is not an eigenvalue of Tp(0)0the consant term of p is nonzero, where the equivalence between the first and second lines follows from 8.49.


9. Solution: Since
4+5T6T27T3+2T4+T5=0, multiplying both sides by T5 we get
4T5+5T46T37T2+2T1+I=0. Hence 4z5+5z46z37z2+2z+1 is a polynomial multiple of the minimal polynomial of T1 (by 8.46). As it turns out, this is actually the minimal polynomial of T1 (with the coefficients multiplied by 4). To see this, suppose by contradiction that it is not. Hence, the minimal polynomial of T has degree at most 4. This means that
a0I+a1T1+a2T2+a3T3+a4T4=0 for some a0,a1,a2,a3,a4F, not all equal 0. Multiplying both sides of the equation above by T4, we get
a0T4+a1T3+a2T2+a3T+a4I=0, which is a contradiction, because the minimal polynomial of T has degree 5.


10. Solution: Let λ1,,λm denote the distinct eigenvalues of T and d1,,dm their corresponding multiplicities. Then
p(z)=(zλ1)d1(zλm)dm. The eigenvalues of T1 are 1λ1,,1λm and by Exercise 3 in section 8A they have multiplicities d1,,dm. Thus
q(z)=(z1λ1)d1(z1λm)dm=zd1(11λ1z)d1zdm(11λmz)dm=zd1++dm(11λ1z)d1(11λmz)dm=zdimV(11λ1z)d1(11λmz)dm=zdimV1λ1d1(λ11z)d11λmdm(λm1z)dm=zdimV1λ1d1λmdm(λ11z)d1(λm1z)dm=zdimV(1)d1++dmλ1d1λmdm(1zλ1)d1(1zλm)dm=zdimV(1)d1++dmλ1d1λmdmp(1z)=zdimV1p(0)p(1z).


11. Solution: Let zm++a1z+a0 be the minimal polynomial of T. Then
a0I=Tma1T. Multiplying both sides of the equation above by 1a0T1 gives the desired result (note that a0 is nonzero by Exercise 8).


12. Solution: Suppose V has a basis consisting of eigenvectors of T. Let λ1,,λm denote the distinct eigenvalues of T. Then, it is easy to see that
(Tλ1I)(TλmI)=0 by applying the left side of the equation to each of the basis vectors, because the parentheses commute. By 8.46, (zλ1)(zλm) is a polynomial multiple of the minimal polynomial of T. This polynomial has no repeated zeros. Hence the minimal polynomial has no repeated zeros.

Conversely, suppose the minimal polynomial of T, call it p, has no repeated zeros. By 8.23, T has a basis of generalized eigenvectors. Let v be one vector in this basis. Then vG(λ,T) for some eigenvalue λ of T. By 8.49, λ is zero of p. Thus, we can write p(z)=(zλ)q(z) for some polynomial q with q(λ)0. We have
0=p(T)v=q(T)(Tλ)v. Note that (Tα)v^0 for all nonzero v^G(λ,T) and all αF with αλ. This implies that (Tλ)v=0, since G(λ,T) is invariant under every polynomial of T and so (TλI)v0 implies q(T)(TλI)0 (because we can factor q(T) and λ is not a zero of q). Therefore v is an eigenvector of T and the basis of V consisting of generalized eigenvectors of T actually consists of eigenvectors of T.


13. Solution: If F=C, this follows directly from the previous exercise from the Complex Spectral Theorem (7.24).


14. Solution: As you can see in the solution to Exercise 10, the constant term in the characteristic polynomial, which equals p(0), is is 1 or 1 times the product of the eigenvalues of S raised to the their respective multiplicities. The eigenvalues of S all have absolute value 1 (see 7.43 (b)), hence |p(0)|=1.


15. Solution:

(a) Just repeat the proof of 8.40 replacing I with v, Tj with Tjv and n2 with n.

(b) This is essentially the same as the proof of 8.46.
Let q denote the minimal polynomial of T. Then we also have q(T)v=0. Furthermore, degpdegq. By the Division Algorithm for Polynomials (4.8), there exist s,rP(F) such that p=sq+r and degr<degq. This implies that
0=p(T)v=s(T)q(T)v+r(T)v=r(T)v. By the same reasoning used in the proof, it follows that r=0.


16. Solution: We have
a0I+a1T+a2T2++am1Tm1+Tm=0. Taking the adjoint of it side yields
a0I+a1T+a2(T)2++am1(T)m1+(T)m=0 and we see that the minimal polynomial of T is
a0+a1z+a2z2++am1zm1+zm. To see this, suppose by contradiction this is not the minimal polynomial of T. Let p denote the minimal polynomial T. Then degp<m. Because p(T)=0, taking the adjoing of each side as we did above shows that p(T)=0, where p equals p with conjugated coefficients. But degp<m, which is a contradiction because the minimal polynomial of T has degree m.


17. Solution: The characteristic polynomial of T, call it q, has degree dimV (see 8.36) and is a polynomial multiple of the minimal polynomial of T (see 8.48), call it p. Because q is a multiple of q, it follows that q=ps for some polynomial s and, because degq=degp, it follows that degs=0. Hence s is a constant. Since they’re both monic, s=1 and they should be equal.


19. Solution: This follows directly from Exercise 11 in section 8B.


20. Solution: This is a bit obvious if you realize that
(2)dimG(λ,T)=dimG(λ,T|V1)++G(λ,T|Vm) for every λF. This is true because the subspaces on the right side of the equation are contained in G(λ,T) and because when we write each Vj as a direct sum of generalized eigenspaces of T|Vj in the equation
V=V1Vm the only way the dimensions will fit is if (2) is true.


Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
Close Menu
Close Menu