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Brock Byrd 2022-07-16 Answered
Let T : M 2 x 2 R 3 have matrix [ T ] B , A = [ 1 2 0 1 0 1 1 0 1 1 1 1 ] relative to A = { [ 2 0 0 0 ] , [ 0 3 0 0 ] , [ 0 0 5 0 ] , [ 0 0 0 6 ] } and β = { ( 1 , 1 , 1 ) , ( 1 , 2 , 3 ) , ( 1 , 4 , 9 ) }. Find the matrix of T relative to the bases A = { [ 1 0 0 0 ] , [ 0 4 0 0 ] , [ 0 0 2 0 ] , [ 0 0 0 7 ] } and β = { ( 1 , 1 , 1 ) , ( 1 , 0 , 0 ) , ( 1 , 1 , 0 ) }
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Answers (1)

Kaya Kemp
Answered 2022-07-17 Author has 18 answers
First of all, notice that M 2 x 2 R 4 with [ a b c d ] ( a , b , c , d ).
Your first step is the right way to go, but I would be careful with the notation. Let α be the standard basis auf R 3 , then
T ( [ 2 0 0 0 ] ) = ( 1 , 0 , 1 ) B = ( 2 , 5 , 10 ) α = ( 10 , 3 , 5 ) B
T ( [ 0 3 0 0 ] ) = ( 2 , 1 , 1 ) B = ( 4 , 8 , 9 ) α = ( 9 , 4 , 1 ) B
T ( [ 0 0 5 0 ] ) = ( 0 , 1 , 1 ) B = ( 2 , 5 , 12 ) α = ( 12 , 3 , 7 ) B
T ( [ 0 0 0 6 ] ) = ( 1 , 0 , 1 ) B = ( 0 , 3 , 8 ) α = ( 8 , 3 , 5 ) B
Thus, we have
[ T ] B , A = [ 10 9 12 8 3 4 3 3 5 1 7 5 ]
Now we are almost done. The new basis A is good to handle because every new basis vector is a scalar multiple of a vector of A. For example the first basis vector of A is half of its counterpart of A (So ( 2 , 0 , 0 , 0 ) A = ( 1 , 0 , 0 , 0 ) A ). So if you map ( 1 , 0 , 0 , 0 ) A with our wanted matrix, it will have the same result as if we map ( 1 2 , 0 , 0 , 0 ) A into [ T ] B , A (because they both map into the vector space generated by B ) Thus, the first column of [ T ] B , A . You can get the other columns with the same method. The solution is:
[ T ] B , A == [ 5 9 4 3 12 2 5 8 7 6 3 2 4 4 3 3 2 5 3 7 6 5 2 1 4 3 7 2 5 5 7 6 ]
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