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Hi, i'm currently going through past papers to get my head into shape, and i'm struggling on this certain transformations question as it hasn't been covered properly in my book. Is there any chance anyone can do it for me so I can get a clear idea what to do? :)
http://img703.imageshack.us/img703/7...lequestion.png
I haven't seen the notation b(T)b' before. From context, it must be the matrix associated with the linear transformation T taking input with respect to b and giving output with respect to b'.
We have, in terms of bases B and B',
T(1, 2, 3, 4) = (-1, -1, -26)
T(1, -1, 0, 1) = (2, 5, -2)
T(3, 1, 2, 1) = (0, 1, 2)
T(1, 0, 3, 5) = (0, 6, -29)
In terms of bases b and B',
T(1, 0, 0, 0) = (-1, -1, -26)
T(0, 1, 0, 0) = (2, 5, -2)
T(0, 0, 1, 0) = (0, 1, 2)
T(0, 0, 0, 1) = (0, 6, -29)
We just need to convert the vectors on the right from basis B' to basis b'. In general this problem is simply that of creating a change of basis matrix. Changing from basis b' to B' is extremely easy and can be done with the matrix
[1 1 0]
[0 2 1]
[1 3 2]
Computing the inverse by a variety of methods, we can change from B' to b' via
[1/2 -1 1/2]
[1/2 1 -1/2]
[-1 -1 1]
It follows that...
base B' -> base b'
(-1, -1, -26) -> 1/2 * (-25, 23, -48)
(2, 5, -2) -> (-5, 7, -9)
(0, 1, 2) -> (0, 0, 1)
(0, 6, -29) -> 1/2 * (-41, 41, -70)
Thus, in terms of bases b and b',
T(1, 0, 0, 0) = (-25/2, 23/2, -24)
T(0, 1, 0, 0) = (-5, 7, -9)
T(0, 0, 1, 0) = (0, 0, 1)
T(0, 0, 0, 1) = (-41/2, 41/2, -35)
T's matrix in these bases is then just
[-25/2 -5 0 -41/2]
[23/2 7 0 41/2]
[-24 -9 1 -25]
which is the final answer (barring any computational mistakes; you should double check my calculations).