what we need to do, is find the matrix for T with respect to these two bases, and then find the inverse of that matrix.

let's give the bases names, to keep them straight. so B =

and C = .

so to find the matrix for T, we're looking for a 3x3 matrix. now we know the columns of T will be the C-coordinate images of the B basis vectors (in B-coordinates the elements of B are just (1,0,0), (0,1,0) and (0,0,1)).

the trouble is, we aren't given the images of the B-basis vectors in C-coordinates, but instead in the standard basis of P2. now, this might seem "backwards" but what we are going to do, is find the matrix P that changes C-coordinates, into "standard coordinates", because this is very easy. i am going to use {1,x,x^2} as the standard basis, if your text uses {x^2,x,1} you'll have to reverse the order of some things.

P([1,0,0]C) = x^2 = 0(1) + 0(x) + 1(x^2) so the first column of P is (0,0,1)^T.

P([0,1,0]C) = x+1 = 1(1) + 1(x) + 0(x^2), so the 2nd column of P is (1,1,0)^T.

P([0,0,1]C) = 1-x = 1(1) + (-1)(x) + 0(x^2), so the 3rd column of P is (1,-1,0)^T.

so P (the "change of basis matrix" from C-->standard) is

but what we want to do is go from standard--->C, so we need to find . verify that is indeed:

so the C-coordinates of 2x-2, which is (-2,2,0) in the standard basis, are: [0,0,-2]C,

the C-coordinates of x^2+x+1, which is (1,1,1) in the standard basis, are: [1,1,0]C

the C-coordinates of 3x^2 which is (0,0,3) in the standard basis, are: [3,0,0]C

(you could have just figured these out directly, but i wanted to show you "how it's done in general").

so our matrix for T, relative to the bases B and C is:

now that we have a matrix for T, we need to find a matrix for .

verify that the matrix for is:

so now we have a matrix for relative to C and B. the thing is, our input will probably be in "standard form", so to compute explicitly what is (and not just the matrix for it), we need to first apply to p(x), or what is equivalent, to find the images of the basis elements 1,x, and x^2 under and THEN apply the matrix for to those.

the C-coordinates of 1 = 1 + 0x + 0x^2 = (1,0,0) are [0,1/2,1/2]C

the C-coordinates of x = 0 + 1x + 0x^2 = (0,1,0) are [0,1/2,-1/2]C

the C-coordinates of x^2 = 0 + 0x + 1x^2 = (0,0,1) are [1,0,0]C

applying our matrix for to these C-coordinate vectors, we get:

1-->[-1/4,1/2,-1/6]B

x-->[1/4,1/2,-1/6]B

x^2-->[0,0,1/3]B

almost done....we still want to express these B-coordinates as symmetric matrices. do so, and express f(x) as a + bx + cx^2, to find T^-1(f) as a 2x2 symmetric matrix. check your answer by verifying that T^-1(2x-2) gives you the first element of B, and so on.