problemset8solutions

# problemset8solutions - MAS 213 Linear Algebra II Problem...

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MAS 213: Linear Algebra II. Problem list for Week #8. Solutions. Problem 1: In this problem you are given a linear operator T ∈ L ( V ) on some vector space V , and a basis β for the vector space. The problem is to calculate the matrix representation of the operator with respect to the given basis, and so determine whether the basis is a basis of eigenvectors for the operator. Solution to Problem 1, part (i). In this case, V = P 1 ( R ), T ( a + bx ) = (6 a 6 b ) + (12 a 11 b ) x , and β = { 3 + 4 x, 2 + 3 x } . We calculate the matrix representation in the usual way, expressing the values of the given basis vectors back in terms of the basis: T (3 + 4 x ) = 6 8 x = ( 2)(3 + 4 x ) + (0)(2 + 3 x ) . T (2 + 3 x ) = 6 9 x = (0)(3 + 4 x ) + ( 3)(2 + 3 x ) . The matrix representation is obtained from these formulas by assembling the co-eﬃcients we have found into the columns of the matrix: [ T ] β = [ 2 0 0 3 ] . This matrix representation is diagonal, so in this case: yes, β is a basis of eigenvectors for T . Solution to Problem 1, part (ii). In this part V = R 3 , T a b c = 3 a + 2 b 2 c 4 a 3 b + 2 c c , and the given basis is β = 0 1 1 , 1 1 0 , 1 0 2 . The matrix representation cor- responding to this choice of basis is calculated following the standard pro- cedure: T 0 1 1 = 0 1 1 = ( 1) 0 1 1 . T 1 1 0 = 1 1 0 = (1) 1 1 0 . T 1 0 2 = 1 0 2 = ( 1) 1 0 2 . 1

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2 So the matrix representation with respect to this ordered basis is: [ T ] β = 1 0 0 0 1 0 0 0 1 and we can confirm that the given basis is a basis of eigenvectors for T . Solution to Problem 1, part (iii). Here V = M 2 × 2 ( R ), T ([ a b c d ]) = [ 7 a 4 b + 4 c 4 d b 8 a 4 b + 5 c 4 d d ] , and the given basis is β = {[ 1 0 1 0 ] , [ 1 2 0 0 ] , [ 1 0 2 0 ] , [ 1 0 0 2 ]} . We calculate: T ([ 1 0 1 0 ]) = [ 3 0 3 0 ] = ( 3) [ 1 0 1 0 ] . T ([ 1 2 0 0 ]) = [ 1 2 0 0 ] = (1) [ 1 2 0 0 ] . T ([ 1 0 2 0 ]) = [ 1 0 2 0 ] = (1) [ 1 0 2 0 ] . T ([ 1 0 0 2 ]) = [ 1 0 0 2 ] = (1) [ 1 0 0 2 ] . So the matrix representation corresponding to this choice of basis is: [ T ] β = 3 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 , and we confirm that the given basis is indeed a basis of eigenvectors.
3 Problem 2: (Problem 5.1.3, parts (a) through (c), from [FIS].) In this problem you are given a square matrix A , and you are told what field F its entries are understood to lie in, A M n × n ( F ). In each case: (i) Determine all the eigenvalues of A . (ii) For each eigenvalue determine the set of eigenvectors that correspond to that eigenvalue. (iii) If it is possible, determine a basis for the vector space F n consisting of eigenvectors of the matrix A . (iv) In the cases that such a basis exists, find an invertible matrix Q and a diagonal matrix D such that D = Q - 1 AQ .

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• Fall '10
• ANDREWKRICKER

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