MATH 5b: Solutions to Homework Set 7 March 2015

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1 MATH 5b: Solutions to Homework Set 7 March It is easy to see that the characteristic polynomial of the matrix is X 3 X = X(X + 1)(X 1). It is then clear from the factorization that this is the only possible invariant factor hence we have the rational canonical form of the matrix is the companion matrix of this invariant factor, i.e We want to find representatives for the similarity classes of 6 6 matrices over Q with minimal polynomial m(x) = (x + 2) 2 (x 1). We have that the minimal polynomial is the largest invariant factor. The product of the invariant factors has to be the characteristic polynomial which is a degree 6 polynomial. Then we can the representatives of the similarity classes to be the possible rational canonical forms. We have the following possibilities 1. x 1, x 1, x 1, (x + 2) 2 (x 1) and the matrix 2. x + 2, x + 2, x + 2, (x + 2) 2 (x 1) and the matrix 3. x + 2, (x + 2) 2, (x + 2) 2 (x 1) and the matrix

2 4. x 1, (x + 2)(x 1), (x + 2) 2 (x 1) and the matrix x + 2, (x + 2)(x 1), (x + 2) 2 (x 1) and the matrix (x + 2) 2 (x 1), (x + 2) 2 (x 1) and the matrix We want to find all rational canonical forms of matrices having characteristic polynomial x 2 (x 2 +1) 2. Suppose first that F = Q (or another field where X 2 +1 splits). There are four possibilities for the invariant factors 1. x 2 (x 2 + 1) 2 and the matrix 2. x, x(x 2 + 1) 2 and the matrix 3. x 2 + 1, x 2 (x 2 + 1) and the matrix

3 4. x(x 2 + 1), x(x 2 + 1) and the matrix 1 1 Now consider the possibility that F = C (or some other field in which X splits). The possible invariant factors are (besides the ones already listed above): 1. x + i, x 2 (x i) 2 (x + i) and the matrix 2. x i, x 2 (x i)(x + i) 2 and the matrix 3. x(x + i), x(x i) 2 (x + i) and the matrix 4. x(x i), x(x i)(x + i) 2 and the matrix i i 1 i i i i 1 i i i 1 i i i 4. Two matrices in GL 3 (F 2 ) are conjugate iff they have the same invariant factors. So we want to find all possible sets of invariant factors n 1 (x) n 2 (x) n k (x) in F 2 [x], with k 3, such that the companion matrices are all nonsingular, which is equivalent to the constant term of each polynomial being non-zero. We have the following possible cases: 3

4 1. deg(n 3 (x)) = 1, which implies that n 1 (x) = n 2 (x) = n 3 (x) = x + 1. Then the rational canonical form of the matrices in the corresponding conjugacy class is 1 2. deg(n 2 (x)) = 2. Thus n 1 (x) = x + 1 and n 2 (x) = (x + 1) 2. The rational canonical form of the matrices in the corresponding conjugacy class is 1 3. n 1 (x) = x 3 + ax 2 + bx + 1 for a, b F 2. Thus we obtain the following rational canonical forms: (a) a = b =, so n 1 (x) = x and (b) a = b = 1, so n 1 (x) = x 3 + x 2 + x + 1 and (c) a = 1, b =, so n 1 (x) = x 3 + x and (d) a =, b = 1, so n 1 (x) = x 3 + x + 1 and Note that by Proposition 14 and Theorem 15, GL 3 (F 2 ) has the following conjugacy classes: the identity (case 1), two classes of elements of order 7, each of which contains 24 elements (cases 3(c) and (d)), one class of elements of order 3 containing 56 elements (case 3(a)), one class of elements of order 4 containing 42 elements (case 3(b)) and one class of elements of order 2 containing 21 elements (case 2). 5. We have m T (x) (x 3 + x 2 1) in Q[x], where m T (x) denotes the minimal polynomial of T. Clearly x 3 +x 2 1 is irreducible in Q[x] because it has no rational roots. Then we must have m T (x) = x 3 +x 2 1. Let n = dim(v ). Then we have m T (x) p T (x), where p T is the characteristic polynomial of T and deg(t ) = n. We have then 3 n. If n = 3, then m T = p T and the only invariant factor if m T. Since all linear transformations of V with the same set of invariant factors are similar, all such transformations T are similar. 6. This problem depends on F and the factorization of the polynomials in F. Let s first take F = Q (or some other field in which X is irreducible). 4

5 1. Q[x]/((x + 1) 2 ) 2. Q[x]/((x 1)(x 2 + 1) 2 ) = Q[x]/(x 1) Q[x]/((x 2 + 1) 2 ) 3. Q[x]/(x 4 1) = Q[x]/(x 1) Q[x]/(x + 1) Q[x]/(x 2 + 1) 4. Q[x]/((x + 1)(x 2 1)) = Q[x]/((x + 1) 2 ) Q[x]/(x 1). So, the list of elementary divisors of V is x 1, x 1, x 1, x + 1, (x + 1) 2, (x + 1) 2, x 2 + 1, (x 2 + 1) 2. According to the textbook page 494, the largest invariant factor is the product of the largest of the distinct prime powers among the elementary divisors, the next largest invariant factor is the product of the largest of the distinct prime powers among the remaining elementary divisors etc. So we have a 3 (x) = (x 1)(x + 1) 2 (x 2 + 1) 2 (this is the minimal polynomial) a 2 (x) = (x 1)(x + 1) 2 (x 2 + 1) a 1 (x) = (x 1)(x + 1). Now consider the case F = C (or some other field in which X splits). In this case, all polynomials factor into linear terms. 1. C[x]/((x + 1) 2 ) 2. C[x]/((x 1)(x 2 + 1) 2 ) = C[x]/(x 1) C[x]/((x + i) 2 ) C[x]/((x i) 2 ) 3. C[x]/(x 4 1) = C[x]/(x 1) C[x]/(x + 1) C[x]/(x + i) C[x]/(x i) 4. C[x]/((x + 1)(x 2 1)) = C[x]/((x + 1) 2 ) C[x]/(x 1). The elementary divisors of V are x 1, x 1, x 1, x + 1, (x + 1) 2, (x + 1) 2, (x + i), (x + i) 2, (x i), (x i) 2. The invariant factors are a 3 (x) = (x 1)(x + 1) 2 (x i) 2 (x + i) 2 a 2 (x) = (x 1)(x + 1) 2 (x + i)(x i) a 1 (x) = (x 1)(x + 1). 7. (1) Suppose that α(u) U, α(w ) W. Let B = M Y (α U ) and C = M Z (α W ). For all u U and w W, we have Bu U and Cw W. The matrix A maps U W U W, (u, v) (Bu, Cv). We have A = (B, C), B(u) U, C(w) W, U W = {} and B C = {}. Hence we get that A = B C. Conversely, suppose that A = B C. Let A = (a i,j ) i,j, with a i,j = if i > m, j m or i m, j > m. Now let u U, so u i = for i > m. Then the entry j > m of Au is n i=1 a j,iu i = m i=1 a j,iu i + n i=m+1 a j,iu i =. Thus Au U, so α(u) U. Similarily, we can show that α(w ) W. (2) Using the fact that ( B det C ) = det(b) det(c) we immediately get that p A (x) = p B (x) p C (x), where p A (x) is the characteristic polynomial of A. If m A (x) denotes the minimal polynomial of A, then note that m B (x) and m C (x) are generators of Ann(U) and Ann(W ) respectively. We have V = U W, Ann(V ) = Ann(U) Ann(W ). Then the generator m A (x) of Ann(V ) is the least common multiple of m B (x) and m C (x). (3) We know the fact that a matrix is diagonalizable iff its minimal polynomial splits completely over F and if its roots are all distinct. Suppose A is diagonalizable. Then m A (X) split completely over F and has distinct roots. Then since m A (X) = lcm(m B (X), m C (X)), it must be the case that m B (X) and m C (X) also split completely and each have distinct linear factors. Hence B and C are also diagonalizable. Conversely, suppose B and C are diagonalizable. Then m B (X) and m C (X) split completely into distinct linear factors. Then m A (X) = lcm(m B (X), m C (X)) also splits completely into distinct linear factors. Hence A is diagonalizable. 5

Solution. That ϕ W is a linear map W W follows from the definition of subspace. The map ϕ is ϕ(v + W ) = ϕ(v) + W, which is well-defined since

Solution. That ϕ W is a linear map W W follows from the definition of subspace. The map ϕ is ϕ(v + W ) = ϕ(v) + W, which is well-defined since MAS 5312 Section 2779 Introduction to Algebra 2 Solutions to Selected Problems, Chapters 11 13 11.2.9 Given a linear ϕ : V V such that ϕ(w ) W, show ϕ induces linear ϕ W : W W and ϕ : V/W V/W : Solution.