problemset9solutions - MAS 213: Linear Algebra II. Problem...

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Unformatted text preview: MAS 213: Linear Algebra II. Problem list for Week #9. Solutions Problem 1: (Problem 5.1.4, various parts, from [FIS].) In the parts below you are given a vector space V and a linear operator T : V → V . Determine the eigenvalues of T , and find an ordered basis of V with respect to which the operator has a diagonal matrix representation. Solution to Part (i). In this part the vector space is V = P 1 ( R ) and the operator is T ( ax + b ) = ( − 6 a + 2 b ) x + ( − 6 a + b ) . To determine the eigenvalues we first need to calculate the characteristic polynomial of T . This is defined to be the characteristic polynomial of any matrix representation of T . The matrix representation of T with respect to α = { 1 ,x } , the standard ordered basis of P 1 ( R ), is: [ T ] α = [ 1 − 6 2 − 6 ] . The characteristic polynomial of this matrix is: det([ T ] α − tI 2 ) = det ([ 1 − 6 2 − 6 ] − t [ 1 0 0 1 ]) = (1 − t )( − 6 − t ) − ( − 6)(2) = t 2 + 5 t − 6 + 12 = ( t + 3)( t + 2) . Thus the eigenvalues of T are − 3 and − 2. Next we’ll determine the corresponding eigenspaces of T . According to the general theory, we determine these eigenspaces by first finding the corresponding eigenspaces of a matrix representation, such as [ T ] α , and then mapping the result from R 2 back to V via the inverse of the co-ordinate vector isomorphism ϕ α . 1 So that is the next thing we’ll do. To determine ( R 2 ) − 2 , the eigenspace of the matrix [ T ] α corresponding to the eigenvalue − 2, we have to solve for the space of solutions to the equation: ([ 1 − 6 2 − 6 ] − ( − 2) [ 1 0 0 1 ])[ x y ] = [ ] . We can find the solution space by using the usual row reduction algorithm, which proceeds as follows: [ 3 − 6 2 − 4 ] → [ 1 − 2 ] . Thus: ( R 2 ) − 2 = {[ 2 u u ] : u ∈ R } ⊂ R 2 . According to the general theory, ( P 1 ( R )) − 2 , the eigenspace of T corre- sponding to this eigenvalue is the image of ( R 2 ) − 2 under ( ϕ α ) − 1 . α is the standard ordered basis of P 1 ( R ), so: ( ϕ α ) − 1 ([ 2 u u ]) = 2 u + ux. Thus: ( P 1 ( R )) − 2 = span( { 2 + x } ) ⊂ P 1 ( R ) . Next we turn to the determination of the eigenspace ( P 1 ( R )) − 3 . As usual, we first solve the corresponding matrix problem. The corresponding eigenspace ( R 2 ) − 3 is the space of solutions to the equation: ([ 1 − 6 2 − 6 ] − ( − 3) [ 1 0 0 1 ])[ x y ] = [ ] . This is solved via row reduction algorithms in the following way: [ 4 − 6 2 − 3 ] → [ 2 − 3 ] . We deduce: ( R 2 ) − 3 = {[ 3 u 2 u ] : u ∈ R } . Mapping that back to P 1 ( R ), we deduce: ( P 1 ( R )) − 3 = span( { 3 + 2 x } ) ⊂ P 1 ( R ) ....
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problemset9solutions - MAS 213: Linear Algebra II. Problem...

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