Let T1: P1 -> P2 be the linear transformation defined by:
T1(c0 + c1*x) = 2c0 - 3c1*x
Using the standard bases, B = {1, x} and B' = {1, x, x^2}, what is the transformation matrix
[T1]B',B
T(c0 + c1*x) = 2c0 - 3c1*x --->
T(1) = 2
T(x) = -3x
So, the matrix elements are:
T_{1,1} = 2
T_{1,2} = 0
T_{2,1} = 0
T_{2,2} = -3
T_{3,1} = 0
T_{3,2} = 0
To find the transformation matrix, [T1]B',B, we need to determine the images of the basis vectors of B under the linear transformation T1, and express those images as linear combinations of the basis vectors of B'.
So, let's start by applying T1 to each basis vector of B and representing the results as linear combinations of the basis vectors of B'.
For the first basis vector, 1, we have:
T1(1) = 2(1) - 3(0) * x = 2
So, the image of the first basis vector, 1, is represented as 2(1) + 0(x) + 0(x^2).
For the second basis vector, x, we have:
T1(x) = 2(0) - 3(1)*x = -3x
So, the image of the second basis vector, x, is represented as 0(1) - 3(x) + 0(x^2).
Now, we can construct the transformation matrix [T1]B',B using the coefficients obtained:
[T1]B',B = [2 0]
[-3 0]
[0 0]
This transformation matrix represents the linear transformation T1 in the standard bases B = {1, x} and B' = {1, x, x^2}.
Note: The subscripts of T in the matrix [T1]B',B represent the indices of the basis vectors in B' and B, respectively. So, for example, T_{1,1} corresponds to the coefficient of the first basis vector in B'.