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M2PM2 Algebra II: Solutions To Problem Sheet 8

This document contains solutions to problems from an algebra II problem sheet. It first shows that the relation A ∼ B if there exists an invertible matrix P such that B = P-1AP is an equivalence relation on matrices. It then works through routine linear algebra problems, including finding change of basis matrices and determining the rank, injectivity, and surjectivity of linear transformations. Finally, it examines properties of the general linear group GL(V), showing that it forms a group, that the determinant map is a surjective group homomorphism, and that GL(V) is isomorphic to the general linear group of matrices with respect to a fixed basis.

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98 views2 pages

M2PM2 Algebra II: Solutions To Problem Sheet 8

This document contains solutions to problems from an algebra II problem sheet. It first shows that the relation A ∼ B if there exists an invertible matrix P such that B = P-1AP is an equivalence relation on matrices. It then works through routine linear algebra problems, including finding change of basis matrices and determining the rank, injectivity, and surjectivity of linear transformations. Finally, it examines properties of the general linear group GL(V), showing that it forms a group, that the determinant map is a surjective group homomorphism, and that GL(V) is isomorphic to the general linear group of matrices with respect to a fixed basis.

Uploaded by

sticker592
Copyright
© Attribution Non-Commercial (BY-NC)
We take content rights seriously. If you suspect this is your content, claim it here.
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KMB, 29th November 2012

M2PM2 Algebra II: Solutions to Problem Sheet 8


1. Dene A B if P invertible such that B = P 1 AP . Then A A as A = I 1 AI . And A B B = P 1 AP A = P BP 1 B A. Finally A B, B C B = P 1 AP, C = Q1 BQ C = Q1 P 1 AP Q = (P Q)1 A(P Q) A C . Hence is an equivalence relation. 2. Routine rst year stu: P = [v ]F = (5a + 2b, 3a b)T , [T ]E = 1 2 5 2 , Q = , [v ]E = (a, b)T , 3 5 3 1 30 48 0 2 . , [ T ]F = 18 29 3 1

3. (i) The determinant must be 0 because T is not surjective (one spots (x1 x2 + 2x3 ) + (x1 3x3 ) + (x2 + x3 ) = 0) and hence, by rank-nullity, cant be injective either. Alternatively, just bash it out. (ii) The matrix of T w.r.t the usual basis 1, x, x2 , x3 is triangular with diagonal entries all 1, so has determinant 1. 1 0 0 0 0 1 0 0 , , , is A = (iii) Matrix of T w.r.t. basis 0 0 1 0 0 0 0 1 1 2 0 0 1 4 0 0 2 0 0 1 2 , which has determinant equal to (det(M )) = 36. 0 0 1 4 4. The T of Q3(ii) satises T (1) = 1, T (x) = x, T (x2 ) = x2 + 4x 1, and T (x3 ) = x3 + 9x 2, so matrix of T w.r.t. basis 1, x, x2 , x3 is 1 0 1 2 0 1 4 9 0 0 1 0 0 0 0 1 The only eigenvalue is 1, and basis for 1-eigenspace of T is checked to be 1, x without too much trouble. There is hence no basis of evectors: g (1) < a(1). The T of Q3(iii) has matrix A as in the solution above. The characteristic 1 2 polynomial of the matrix is (x 2)(x 3) so this matrix has distinct 1 4 evalues so can be diagonalised, say by a 2 2 matrix P . Then the 4 4 matrix P 0 2a 2b a b diagonalises A. The eigenspaces are and . 0 P a b a b 5. (a)(i) Characteristic polynomial is (x + 1)2 (x 2), so eigenvalues are 1, 2 with algebraic multiplicities 2,1 respectively. The geometric multiplicity of the evalue 1 is dimension of the 1 eigenspace, which is easily checked to be 1; the geometric

multiplicity of 2 must also be 1 (as 1 g (2) 1). Since the geom multiplicity of 1 is less than the algebraic multiplicity, there is no basis of eigenvectors.
2 2 2 (ii) T sends 1 0, x 3x, x x + 6x , so matrix of T wrt basis 1, x, x is 0 0 0 0 3 1 . This has distinct eigenvalues 0,3,6, all with algebraic and geometric 0 0 6 multiplicity 1, and there is a basis of eigenvectors.

(b) The char poly is (x + 1)2 (x 1), so (as in part (a)(i) above) A is diagonalisable i the 1 eigenspace has dimension 2. This eigenspace consists of solutions 0 a b to the system 0 2 c v = 0, so it is 2-dimensional i ac 2b = 0. 0 0 0 6. (i) Well-denedness of multiplication (i.e. closure): S, T GL(V ) implies that ST is a linear transformation, and it is invertible as (ST )1 = T 1 S 1 , so ST GL(V ). Associativity: follows from associativity of composition. Identity: is identity map I (v ) = v v V . Inverse: exists by detion. Hence GL(V ) is a group. (ii) If T, U GL(V ) then det(T U ) = det(T )det(U ) by lectures, so det is a homomorphism. Let B = {v1 , . . . , vn } be a basis of V . For R , dene T : V V to be the linear map which sends v1 v1 , v2 v2 , . . . , vn vn . Then det(T ) = . Hence det is surjective. (iii) Fix a basis B of V . Then the map T [T ]B is an isomorphism from GL(V ) to GL(n, R).

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