Help : Linear Transformations
In a vector space R^3, is given a transformation A with a subscript A(x1,x2,x3)=(2*x1+x2, x1+x2+2*x3, x2+x3).
Linear transformation B has in the basis; (1,1,1), (1,0,1), (1,1,0) a matrix T: [1 2 3] [ 1 1 0] [ 0 1 1] Write down a matrix which belongs to the transformation AB in the standard basis of the vector space R^3. I just can't solve this problem. Please help! 
Hmm, the standard basis is (1,0,0) (0,1,0) (0,0,1)
You will need first to find the transformation matrix for A. Then you will need transform T, to a matrix to change from the standard basis to the standard basis. You do this by multiplying it by the matrix to change from the given basis to the standard basis, and then multiplying the result by the matrix to change from the standard basis to the given basis. B = G * T * S G: Matrix to change from the given basis to the standard basis S: Matrix to change from the standard basis to the given basis All you need to then is to multiply the 2 matrix. If you need more complete explanition on any step, let me know. 
Yes, a more complete explanation would be very helpful.
Thanks, for the reply! 
Well, at which step does the problem lie?

Quote:
Quote:
I actually have problems with transforming a matrix from one basis to another. 
To find a transformation matrix:
you calculate the image of the vectors of the basis: so for example A(e1) = (2,1,0) and then write it as the product of the vectors of the basis you transfering to: (2,1,0) = 2e1 + e2 + 0 e3 And last write it as a matrix, where this would fill a row, in the matrix. You will get a 3x3 Matrix, because you have the 3 basis vectors. To find the Matrix to transform from a basis to another, you do the same, but just express the vectors of the matrix you want to transform to as vectors of the matrix you transform from. 
I think i got it now. So the transformation matrix A is : [2 1 0 ]
[1 1 1] [0 2 1] Is that correct? Quote:

All times are GMT 5. The time now is 02:18 PM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums