problemset6solutions - MAS 213 Linear Algebra II Problem...

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MAS 213: Linear Algebra II. Problem list for Week #6. Tutorial on 12th October. Problem 1: (Problems 2.3.3 in [FIS].) Consider two linear transformations: T : P 2 ( R ) P 2 ( R ) defined by T ( p ( x )) = (3 + x ) p ( x ) + 2 p ( x ). U : P 2 ( R ) R 3 defined by U ( a + bx + cx 2 ) = ( a + b, c, a b ). Let β and γ be the standard ordered bases of P 2 ( R ) and R 3 . (1) Check that the expected equation [ U T ] γ β = [ U ] γ β [ T ] β holds. (2) Let h ( x ) = 3 2 x + x 2 P 2 ( R ). Check that the expected equation [ U ( h ( x ))] γ = [ U ] γ β [ h ( x )] β holds. Solution. We need to calculate various ingredients for this problem. First, to determine [ T ] β : T (1) = 2 = (2) 1 + (0) x + (0) x 2 . T ( x ) = 3 + 3 x = (3) 1 + (3) x. T ( x 2 ) = 4 x 2 + 6 x = (0) 1 + (6) x + (4) x 2 . [ T ] β = [ T ] β β = 2 3 0 0 3 6 0 0 4 . Next, to determine [ U ] γ β : U (1) = (1 , 0 , 1) = (1) (1 , 0 , 0) + (0) (0 , 1 , 0) + (1) (0 , 0 , 1) . U ( x ) = (1 , 0 , 1) = (1) (1 , 0 , 0) + (0) (0 , 1 , 0) + ( 1) (0 , 0 , 1) . U ( x 2 ) = (0 , 1 , 0) = (0) (1 , 0 , 0) + (1) (0 , 1 , 0) + (0) (0 , 0 , 1) . Thus: [ U ] γ β = 1 1 0 0 0 1 1 1 0 . Finally, to calculate [ U T ] γ β : ( U T )(1) = U (2) = (2 , 0 , 2) = (2) (1 , 0 , 0) + (0) (0 , 1 , 0) + (2) (0 , 0 , 1) . ( U T )( x ) = U (3 + 3 x ) = (6 , 0 , 0) = (6) (1 , 0 , 0) + (0) (0 , 1 , 0) + (0) (0 , 0 , 1) . ( U T )( x 2 ) = U (6 x +4 x 2 ) = (6 , 4 , 6) = (6) (1 , 0 , 0)+(4) (0 , 1 , 0)+ ( 6) (0 , 0 , 1) . Thus: [ U T ] γ β = 2 6 6 0 0 4 2 0 6 . 1
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2 So, we can finally check that the required equation does indeed hold: [ U T ] γ β = 2 6 6 0 0 4 2 0 6 = 1 1 0 0 0 1 1 1 0 · 2 3 0 0 3 6 0 0 4 = [ U ] γ β [ T ] β . We move on to part 2 of this problem. There are two extra pieces we need, namely: [ h ( x )] β = 3 2 1 and [ U ( h ( x ))] γ = [(1 , 1 , 5)] γ = 1 1 5 . Now we can check the required equation: [ U ( h ( x ))] γ = 1 1 5 = 1 1 0 0 0 1 1 1 0 3 2 1 = [ U ] γ β [ h ( x )] β . Tick.
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3 Problem 2: (Problems 2.4.19 in [FIS].) Let T : M 2 × 2 ( F ) M 2 × 2 ( F ) be the linear transformation which sends a matrix to its transpose. Let β be the ordered basis β = { E 11 , E 12 , E 21 , E 22 } for M 2 × 2 ( F ) (where E ij is the 2 × 2 matrix which has a zero in every position except for a single 1 in the row i column j entry). Let A denote the matrix representation [ T ] β . Let M = [ 1 2 3 4 ] M 2 × 2 ( F ). Check that the expected equation L A ( ϕ β ( M )) = ϕ β ( T ( M )) holds. Solution Let’s go through the ingredients of this equation. It is straightforward to calculate that A = [ T ] β β = 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 . (Actually, we did precisely this example in the Week 5 problem sheet.) Next: ϕ β ( M ) = 1 2 3 4 . And: T ( M ) = [ 1 3 2 4 ] so that [ T ( M )] β = 1 3 2 4 . Putting the pieces together we can check the given equation: L A ( ϕ β ( M )) = β ( M ) = 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 1 2 3 4 = 1 3 2 4 = [ T ( M )] β , as required.
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4 Problem 3: (Problems 2.4.16 in [FIS].) Let B be an invertible n × n matrix. Define a function Φ : M n × n ( F ) M n × n ( F ) by Φ( A ) = B 1 AB . Prove that Φ is a linear transformation and is, more- over, an isomorphism.
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