Section 12.3 and 13.1.
Homework #8
Masaya Sato
Sec 12.3 2. Prove the if λ1, . . . , λn are the eigenvalues of the n × n matrix A then λ1k , . . . , λkn are the eigenvalues of Ak for any k ≥ 0.
Proof. By assumption for each λi there exists some nonzero vector vi such that Avi = λi vi
(i = 1, . . . , n).
Therefore for nonnegative k k times
k−1 times
Ak vi = (A · · · A) vi = (A · · · A)(Avi )
= (A · · · A)(λi vi ) = λi(A · · · A)vi
and thus Ak vi = λik vi . Hence λ1k , . . . , λkn are the eigenvalues of Ak . 17. Prove that any matrix A is similar to its transpose At .
Proof. Consider the Jordan canonical form J A of A. Then for an n × n invertible matrix
0 0 T = ... 0 1 Then T −1 = T and
0 ··· 0 ··· .. . 1 ··· 0 ···
0 1 .. .
1 0 .. . . 0 0 0 0
T −1 J A T = J At .
Moreover J At = J A , where J A is the Jordan canonical form of the transpose At . Henc Hencee A t and A are similar. similar. t
t
18. Determine all possible Jordan canonical forms for a linear transformation with characteristic teristic polynom p olynomial ial (x − 2)3 (x − 3)2 . Solution: Observe first that the characteristic polynomial ( x −2)3 (x −3)2 has the eigenvalues eigenvalues 2 and 3, with multiplicity 3 and 2, respectively. Let J i (λ) denote a Jordan block with respect to the eigenvalue λ with multiplicity i. Then the Jordan canonical J form is given by
(i) J 1 ∼ = J 3 (2) ⊕ J 2 (3) , (ii) J 2 ∼ = J 1 (2) ⊕ J 2 (2) ⊕ J 2 (3) , (iii) J 3 ∼ = J 1 (2) ⊕ J 1 (2) ⊕ J 1 (2) ⊕ J 2 (3) , (iv) J 4 ∼ = J 3 (2) ⊕ J 1 (3) ⊕ J 1 (3) , Abstract Abstract Algebra Algebra by Dummit and Foote 1
Section 12.3 and 13.1.
Homework #8
Masaya Sato
(v) J 5 ∼ = J 1 (2) ⊕ J 2 (2) ⊕ J 1 (3) ⊕ J 1 (3) , and (vi) J 6 ∼ = J 1 (2) ⊕ J 1 (2) ⊕ J 1 (2) ⊕ J 1 (3) ⊕ J 1 (3) . Therefore for (i) to (vi)
2 0 J 1 = 0 0 0
2 0 J 4 = 0 0 0
1 2 0 0 0
2 0 0 0 0
0 1 2 0 0
0 0 0 3 0
0 2 0 0 0
0 1 2 0 0
0 0 0 3 0
0 2 0 0 0 2 0 , J 3 = 0 0 1 0 0 3 0 0
0 1 2 0 0
0 0 0 3 0
0 2 0 0 0 2 0 , J 5 = 0 0 0 0 0 3 0 0
0 1 2 0 0
0 0 0 3 0
0 2 0 0 0 2 0 , and J 6 = 0 0 0 0 0 3 0 0
0 0 0 , J 2 1 3
1 2 0 0 0
0 0 2 0 0
0 0 0 3 0
0 0 0 , 1 3
0 0 2 0 0
0 0 0 3 0
0 0 0 . 0 3
20. Show that the following matrices are similar in M p (Fp ) (p × p matrices with entries from
Fp ):
0 1 0 .. .
0 0 1 .. .
0 ··· 0 ··· 0 ··· .. .
0 0 0 ···
0 0 0 .. .
1 0 0 .. .
1 0
and
1 0 0 .. .
1 1 0 .. .
0 ··· 0 0 1 ··· 0 0 1 ··· 0 0 .. .. .. . . . . 0 0 0 ··· 1 1 0 0 0 ··· 0 1
Proof. Suppose first that p = 2 since two matrices are not similar, i.e. the diagonal matrix
0 1 1 0
is not similar to another one
1 1 . 0 1
Let A denote the right above matrix, and let B denote denote the left above above matrix. Since A is a companion matrix, its characteristic polynomial pA (x) is given by det (xI − A) = xp + 0xp−1 + · · · + 0x − 1 = xp − 1. pA (x) = det Then since B is an upper triangular matrix, its characteristic polynomial pB (x) is given by pB (x) = det det (xI − B ) = (x − 1)p = xp − 1.
Moreover pA (x) and pB (x) are both minimal polynomial for A and B , respectively. Therefore two matrices A and B are similar. similar. Abstract Abstract Algebra Algebra by Dummit and Foote 2
Section 12.3 and 13.1.
Homework #8
Masaya Sato
21. Show that if A2 = A then A is similar to a diagonal matrix which has only 0’s and 1’s
along the diagonal. Proof. Consider an eigenvalue λ ∈ F for A. Then for the corresponding nonzero eigenvector v Av = λv
and λv = Av = A2 v = A(λv ) = λ2 v .
Therefore λ(λ − 1)v = 0
and hence λ = 0 or λ = 1. Now obverse that the Jordan canonical form J of an idempotent matrix A is also idempotent. In fact since A is similar to its Jordan canonical form J , there exists an invertible matrix T such that J = T −1 AT .
So J 2 = (T −1 AT )(T −1 AT ) = T −1 A2 T = T −1 AT = J .
Therefore J is of the form
0 0 .. .
0 ··· 0 ··· .. . . . . 0 0 ··· 0 0 ···
0 0 .. .
(0.1)
0 0 .. , . 1 0 0 1
a diagonal matrix whose entries are all equal to 0 or 1. Otherwise J would not be idempotent. 22. Prove that an n × n matrix A with entries from C satisfying A3 = A can be diagonalized. Is the same statement true over any field F ?
Proof. Note that every n × n matrix A over a field F is similar to some diagonal matrix if and only if the minimal polynomial of A has no repeat repeated ed roots. Since Since A3 = A by assumption, the minimal polynomial m(x) divides x3 − x = x(x − 1)(x + 1). 1). So m(x) has no repeated roots. Thus A can be diagonaliza diagonalizable. ble.
However, for a field F2 , the matrix A=
1 1 0 1
satisfies A3 = A. But A is not diagonalizab diagonalizable. le.
23. Suppose that A is an 2 × 2 matrix with entries from Q for which A3 = I but A = I . Write A in rational canonical form and in Jordan canonical form viewed as a matrix over C.
Abstract Abstract Algebra Algebra by Dummit and Foote 3
Section 12.3 and 13.1.
Homework #8
Masaya Sato
Proof. Observe first that the minimal matrix m(x) of A divides the polynomial x3 − 1 = (x − 1)(x2 + x +1). Moreover note that x2 + x +1 is irreducible over C since +1 or −1 would be a root by the Rational Root Theorem or Gauss’s Lemma . Since A = I , m(x) = x2 + x + 1. Therefore the rational canonical form for A is given by
0 −1 1 −1
and the Jordan canonical form is given by
√
1 2
3 2
− +i 0
0 √ . − − i 23 1 2
25. Determine the Jordan canonical form for the n × n matrix over Q whose entries are all
equal to 1. Solution: By applying finitely many row and column operations, the matrix is converted
to another of the form
1 0 .. .
0 0 .. .
0 ··· 0 0 ··· 0 .. . . .. , . . . 0 0 0 ··· 0
and the matrix is of the Jordan canonical form.
Determine the Jordan Jordan canonical canonical form for the n × n matrix over Fp whose entries are all 26. Determine equal to 1. Solution: Let A be an n × n matrix whose entries are all equal to 1 ∈ Fp . Then
A2 =
n n ··· n n ···
n n
n n ···
n
.. .
.. .
...
.. . .
Then consider the following two cases. hen A2 = 0. So the minima minimall polynomia polynomiall m(x) is of the form m(x) = x2 . Case 1: p|n. Then Therefore the Jordan canonical form of A is given by A∼ = B ⊕ C ,
where B=
0 1 0 0
Abstract Abstract Algebra Algebra by Dummit and Foote 4
Section 12.3 and 13.1.
Homework #8
Masaya Sato
and C is the (n − 2) × (n − 2) zero matrix. relatively ly prime. So A2 = A and the minimal polynomial Case 2 : p n. Then p and n are relative of A is m(x) = x2 − x. Therefore the Jordan canonical form of A is given by J ∼ = B ⊕ C , where B = 1, 1, a 1 × 1 matrix, and C is the (n − 1) × (n − 1) zero matrix. 31. Let N be an n × n matrix with coefficients in the field F . The matrix N is said to be Prove that that any any nilpotent if some power of N is the zero matrix, i.e., N k = 0 for some k . Prove
nilpotent matrix is similar to a block diagonal matrix whose blocks are matrices with 1’s along the superdiagonal and 0’s elsewhere. Proof. Let u be a nonzero vector such that N k−1 u = Observe first first that that {u , N u , . . . , N k −1 u} 0. Observe is linearly independent. So for a0 , a1 , . . . , ak−1 ∈ F it is immediate to show that a0 u + a1 (N u) + · · · + ak−1 (N k−1 u) = 0
has only the trivial solution a0 = 0, a1 = 0, and ak−1 = 0 by applying N k−1 , . . . , N 2 , N to the equation. Now consider the following two cases. Case 1: k ≥ n. The matrix representation for N with respect to the basis {N n−1u , . . . , N u , u} is given by 0 1 0 ··· 0 0 0 1 ··· 0 .. .. .. . . .. , . . . . . 0 0 0 ··· 1 0 0 0 ··· 0
which consists of a single Jordan block whose entries are 1 along the superdiagonal and 0’s elsewhere. there is a unique pair of positive positive integers integers q and r such that n = qk + r, Case 2 : k < n. Since there where 0 ≤ r ≤ k − 1 consider a k × k submatrix Q contained in N . Then its representation for K with respect to a basis {u , N u , . . . , N k −1 u} is given by
1 0 .. .
1 0 .. .
0 0 .. .
0 ··· 1 ··· .. . . . . 0 0 0 ··· 0 0 0 ···
0 0 .. . . 1 0
Now let R be an r × r matrix. matrix. Then the represent representation ation matrix matrix for R with respect to the basis {N k−1 u , . . . , N k −r−1 u, N k−r u} is given by 0 0 .. .
0 ··· 1 ··· .. . . . . 0 0 0 ··· 0 0 0 ···
0 0 .. , . 1 0
Abstract Abstract Algebra Algebra by Dummit and Foote 5
Section 12.3 and 13.1.
Homework #8
Masaya Sato
and thus N ∼ = Q ⊕ · · · ⊕ Q ⊕ R, where Q ⊕ · · · ⊕ Q is q copies of a k × k submatrix Q. Therefore N is similar to a block diagonal matrix whose entries are 1’s on the superdiagonal and 0’s elsewhere. 32. Prove that if N is an n × n nilpotent matrix then in fact N n = 0.
Proof. Using the result from Exercise 31, consider the two cases. Case 1: k ≥ n. Then N is similar to M , where M is an n × n matrix whose entries are 1’s on the superdiagonal superdiagonal and 0’s elsewhere. elsewhere. So there exists an n × n invertible matrix T such 1 − that N = T M T . N n = T −1 M n T
and moreover M n = 0. Therefore N n = 0. hen N is similar to a block diagonal matrix M , where each block has Case 2 : k < n. Then entries 1’s on the superdiagonal and 0’s elsewhere. Note that M k = 0 since since M k = Qk ⊕ · · · ⊕ Qk ⊕ Rk = 0 ⊕ · · · ⊕ 0 ⊕ 0.
Therefore N n = T −1 M n T = T −1 M k M n−k T = 0
as desired. 34. Prove that the trace of a nilpotent n × n matrix is 0.
Proof. Let M be an n × n matrix given above. Since a nilpotent matrix N is similar to M , N and M share the same characteristic polynomial p(x) = det det (xI − N ) = det det (xI − M ) = xn .
Observe that the coefficient of xn−1 for p(x), which is the trace of M , is equal to 0. Therefore the trace of N is 0. Sec 13.1 3. Prove that x3 + x + 1 is irreducible over F2 and let θ be a root. Compute the powers of θ in F2 (θ ).
Proof. Suppose by contradiction that x3 + x +1 is reducible, i.e. there exists some polynomial p(x) and q (x) in F2 [x] such that x3 + x + 1 = p(x)q (x).
Observe that one of p(x) and q(x) is a polynomial of degree 1, so let p(x) be of degree 1 for simplicity. Then p(x) is of the form p(x) = x or p(x) = x + 1.
Abstract Abstract Algebra Algebra by Dummit and Foote 6
Section 12.3 and 13.1.
Homework #8
Masaya Sato
It is obvious that p(x) = x. So p(x) = x + 1 and thus p(1) = 0. However x3 + x + 1 ⇒ 13 + 1 + 1 = 1 = 0.
This contradicts that p(x) divided x3 + x + 1. Therefore x3 + x + 1 is irreducib irreducible le over over F2 . 3 Now let θ is a root. Then θ + θ + 1 = 0. Moreover θ4 θ5 θ6 θ7 θ8
= θ2 + θ, = θ 3 + θ 2 = θ2 + θ + 1, = θ 3 + θ 2 + θ = θ2 + 1, = θ 3 + θ = 1, and = θ.
Therefore for a nonnegative integer k θ 7k+1 θ 7k+2 θ 7k+3 θ 7k+4 θ 7k+5 θ 7k+6
= θ, = θ2 , = θ + 1, = θ2 + θ, = θ 2 + θ + 1, and = 1.
5. Suppose that α is a rational root of a monic polynomial in Z[x]. Prove Prove that that α is an
integer. Proof. Since α ∈ Q is a root of some polynomial p(x) ∈ Z[x] of the form p(x) = xn + an−1 xn−1 + · · · + a1 x + a0 ,
Moreove verr α ∈ Z Root Theorem Theorem or Gauss’s Lemma . Moreo α = ±a0 or α = ±1 by the Rational Root since a0 ∈ Z. 7. Prove that x3 − nx + 2 is irreducible over Q for n = −1, 3, 5.
Proof. Suppose that there is a pair of monic polynomials p(x) and q (x) of degrees 1 and 2
respectively such that x3 − nx + 2 = p(x)q (x).
Then p(x) is of the form p(x) = x − r, where r ∈ Q. r is a root of the polynomial x3 − nx + 2 and moreover r = 2, r = −2, r = −1, or r = 1 by the Rational Root Theorem or Gauss’s Lemma . If r = 2, then x3 − nx + 2 ⇒ 23 − 2n + 2 = 10 − 2n.
If r = −2 or r = 1, then x3 − nx + 2 ⇒ (−2)3 − (−2)n + 2 = −6 + 2 n, or
Abstract Abstract Algebra Algebra by Dummit and Foote 7
Section 12.3 and 13.1.
Homework #8
Masaya Sato
x3 − nx + 2 ⇒ (1)3 − (1)n + 2 = 3 − n.
Moreover if r = −1, then x3 − nx + 2 ⇒ (−1)3 − (−1)n + 2 = 1 + n.
Therefore x3 − nx + 2 is reducible reducible over over Q if n = −1, n = 3, or n = 5. Otherwise x3 − nx + 2 is irreducible over Q.
Abstract Abstract Algebra Algebra by Dummit and Foote 8
