hw5s(1) - MATH 4377, ADVANCED LINEAR ALGEBRA, SUMMER 2011,...

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Unformatted text preview: MATH 4377, ADVANCED LINEAR ALGEBRA, SUMMER 2011, Key to HW#5 1. Problem 5 (c), (d) on Page 85. Solution : (c) T 1 0 0 0 ! = tr 1 0 0 0 ! = 1 T 0 1 0 0 ! = tr 0 1 0 0 ! = 0 T 0 0 1 0 ! = tr 0 0 1 0 ! = 0 T 0 0 0 1 ! = tr 0 0 0 1 ! = 1 therefore [ T ] = 1 0 0 1 (d) T(1) = 1 T(x) = 2 T(x 2 ) = 4 therefore [ T ] = 1 2 4 2. (I wrote this problem on the whiteboard) Let T : M 2 2 ( R ) M 2 2 ( R ) be a linear transformation and let = ( 1 0 0 0 ! , 1 1 0 0 ! , 0 1 1 0 ! , 0 0 0 1 !) . Assume that [ T ] = 1 2 0 0 0 0 1 0 1 1 0 3 0 0 0 1 . For A = a b c d ! M 2 2 ( R ), find T ( A ). Solution : We use for formula on Page 91: [ T ( A )] = [ T ] [ A ] . 1 To do so, we first find [ A ] : Write a b c d ! = ( a- b + c ) 1 0 0 0 ! +( b- c ) 1 1 0 0 ! + c 0 1 1 0 ! + d 0 0 0 1 ! . ( remark : you can find above by soving a system of linear equations). So [ A ] = a- b + c b- c c d...
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hw5s(1) - MATH 4377, ADVANCED LINEAR ALGEBRA, SUMMER 2011,...

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