1. Solution: Because
it follows that the multiplicities of the eigenvalues of are at most . Thus is a polynomial multiple of the characteristic polynomial. Therefore, we must have .
2. Solution: We have
Hence is a polynomial multiple of the characteristic polynomial. Therefore, .
3. Solution: Define by
Since and , it follows that and . We have . Hence, the characteristic polynomial of is
4. Solution: Let
Then
Define by
Then and the eigenvalues of are thus and (the entries on the diagonal). Now 8.36 implies that the minimal polynomial of is a polynomial multiple of . The previous work shows that and . Hence is the minimal polynomial of . By Exercise 11 in section 8B, the multiplicity of is and of is . Thereby the characteristic polynomial of is .
5. Solution: Let
Then
Define by
Then . The rest of this exercise is almost the same as the previous one.
6. Solution: Define by
Then, the standard basis of consists of eigenvectors of corresponding to the eigenvalues . Applying to each of these basis vectors shows that . Hence is the minimal polynomial of . We have
Thus, by Exercise 11 in section 8B, the characteristic polynomial of is .
7. Solution: By Exercise 4 in section 5B, we have
It is easy to check that if then . Thus
and 8.26 then imply that these inclusions are actually equalities, which gives the desired result.
8. Solution: Let denote the minimal polynomial of . We have
where the equivalence between the first and second lines follows from 8.49.
9. Solution: Since
multiplying both sides by we get
Hence is a polynomial multiple of the minimal polynomial of (by 8.46). As it turns out, this is actually the minimal polynomial of (with the coefficients multiplied by ). To see this, suppose by contradiction that it is not. Hence, the minimal polynomial of has degree at most . This means that
for some , not all equal . Multiplying both sides of the equation above by , we get
which is a contradiction, because the minimal polynomial of has degree .
10. Solution: Let denote the distinct eigenvalues of and their corresponding multiplicities. Then
The eigenvalues of are and by Exercise 3 in section 8A they have multiplicities . Thus
11. Solution: Let be the minimal polynomial of . Then
Multiplying both sides of the equation above by gives the desired result (note that is nonzero by Exercise 8).
12. Solution: Suppose has a basis consisting of eigenvectors of . Let denote the distinct eigenvalues of . Then, it is easy to see that
by applying the left side of the equation to each of the basis vectors, because the parentheses commute. By 8.46, is a polynomial multiple of the minimal polynomial of . This polynomial has no repeated zeros. Hence the minimal polynomial has no repeated zeros.
Conversely, suppose the minimal polynomial of , call it , has no repeated zeros. By 8.23, has a basis of generalized eigenvectors. Let be one vector in this basis. Then for some eigenvalue of . By 8.49, is zero of . Thus, we can write for some polynomial with . We have
Note that for all nonzero and all with . This implies that , since is invariant under every polynomial of and so implies (because we can factor and is not a zero of ). Therefore is an eigenvector of and the basis of consisting of generalized eigenvectors of actually consists of eigenvectors of .
13. Solution: If , 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 , is is or times the product of the eigenvalues of raised to the their respective multiplicities. The eigenvalues of all have absolute value (see 7.43 (b)), hence .
15. Solution:
(a) Just repeat the proof of 8.40 replacing with , with and with .
(b) This is essentially the same as the proof of 8.46.
Let denote the minimal polynomial of . Then we also have . Furthermore, . By the Division Algorithm for Polynomials (4.8), there exist such that and . This implies that
By the same reasoning used in the proof, it follows that .
16. Solution: We have
Taking the adjoint of it side yields
and we see that the minimal polynomial of is
To see this, suppose by contradiction this is not the minimal polynomial of . Let denote the minimal polynomial . Then . Because , taking the adjoing of each side as we did above shows that , where equals with conjugated coefficients. But , which is a contradiction because the minimal polynomial of has degree .
17. Solution: The characteristic polynomial of , call it , has degree (see 8.36) and is a polynomial multiple of the minimal polynomial of (see 8.48), call it . Because is a multiple of , it follows that for some polynomial and, because , it follows that . Hence is a constant. Since they’re both monic, 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
for every . This is true because the subspaces on the right side of the equation are contained in and because when we write each as a direct sum of generalized eigenspaces of in the equation
the only way the dimensions will fit is if is true.