Linear Algebra: Change of basis

mccoy1
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(a) Let A (matrix) =c1= [1,2,1], c2 = [0,1,2], c3 = [3,-2,-1] be a matrix (c1,c2,c3 refer to the columns of the matrix A, which is a 3x3 matrix) expressed in the standard basis and let w1 = (0,0,1)T, w2 = (0,1,2)T , w3 =(3,0,2)T , find the vector AUE
in w basis.
(b). Referring to problem (a), find the transformation Uw = BUw, corresponding to VE = AUE.

The Attempt at a Solution



(a) Ans = W^-1A UE...no problem.
(b), I think I'm either stupid or the question is (lol well you know what i mean)...what do they mean by 'Uw = BUw, corresponding to VE = AUE'?
Thanks fellows.
 
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What do you think guys?
 
A is the matrix representing a linear transformation in one basis, the "standard" basis, {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, B is the matrix representing the same linear transformation in another basis, {(0, 0, 1), (0, 1, 2), (3, 0, 2)}.

That's why this was titled "change of basis".
 
HallsofIvy said:
A is the matrix representing a linear transformation in one basis, the "standard" basis, {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, B is the matrix representing the same linear transformation in another basis, {(0, 0, 1), (0, 1, 2), (3, 0, 2)}.

That's why this was titled "change of basis".

Thanks for your help. I have no problem with change of basis, so to speak and that's why i did the first question (part a)...part a involved a change of basis. What I do have problem with though is what they meant by transformation Uw = BUw corresponding to VE = AUE. Your explanation makes sense, I'm starting to wonder why they just didn't simply us to find the matrix that transform vector u (in w basis) to the same point/that A mapped vector U(basis E, i.e the std basis). From your explanation, I think they are asking for a similarity transformation. If that's true, then I hate the use of the term correspond because it doesn't mean anything (in my opinion)..
Thanks.
 
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