OK, i will write [a,b] to indicate "coordinates in the basis B", where B = {(4,7),(3,5)}, and (a,b) to indicate standard basis coordinates.

so [1,0] = 1(4,7) + 0(3,5) = (2,3), and [0,1] = 0(4,7) + 1(3,5) = (3,5).

it is often helpful to know what the standard basis vectors (1,0) and (0,1) are in B-coordinates:

(1,0) = [a,b] = a(4,7) + b(3,5) = (4a+3b,7a+5b) gives:

7a+5b = 0 (the second coordinate), so b = (-7/5)a, and since:

4a+3b = 1 (the first coordinate), we have 4a + 3(-7/5)a = 1:

4a - (21/5)a = 1

(20/5 - 21/5)a = 1

(-1/5)a = 1

a = -5, and thus b = (-7/5)(-5) = 7.

so (1,0) = [-5,7]. these are both describing "the same vector in the plane", but "using different coordinate systems" (so the NUMBERS are different).

a similar calculation shows that (0,1) = [3,-4].

the matrix that takes B-coordinates to standard coordinates is:

the matrix that takes standard coordinates to B-coordinates is:

in your original problem, the matrix for T(x,y) = (13x-7y,43-24y) is (as has been discussed to death):

.

it is straight-forward to verify that:

we can ask: what would the matrix of T be in "B-coordinates"?

let's do this 2 ways:

first we calculate P^{-1}[T]P:

.

now let's do it by using what we're given about T:

we know T(4,7) = (3,4). let's call the B-coordinate matrix version of T, M. in B-coordinates, (4,7) = [1,0].

in B-coordinates, (3,4) = [a,b] = P^{-1}(3,4) = [-3,5]:

-3(4,7) + 5(3,5) = (-12,-21) + (15,25) = (3,4), so the first column of M is (-3,5)^{T}.

similarly, T takes [0,1] (what (3,5) is in B-coordinates) to (4,9), which in B-coordinates is [7,-8]:

7(4,7) - 8(3,5) = (28,49) - (24,40) = (4,9), so the second column of M is (7,-8)^{T}. same answer either way.

*************

what YOU did (in your original working) is use the basis B = {(4,7),(3,5)} for the domain, and use the basis C = {(3,4),(4,9)} for the image of T.

to make THAT work out, we need to know how to form a matrix we haven't looked at before: one that takes B-coordinates as input, and spits out C-coordinate vectors as output.

so we need to know how to change B-coordinates to C-coordinates.

well, we can change (3,4) = [1,0]_{C}and (4,9) = [0,1]_{C}to "standard coordinates" easily, the matrix that does THAT is:

.

and we can change standard coordinates to B-coordinates using P^{-1}, so to change C-coordinates to B-coordinates we use:

.

it thus stands to reason that to "change back to C (from B)" we use: (P^{-1}Q)^{-1}= Q^{-1}P,

.

now we can start with B-coordinates, and apply M (which takes B-coordinates to B-coordinates), and then apply Q^{-1}P, which will turn the B-coordinates into C-coordinates.

putting this all together, our new matrix is:

.

THIS is what Hartlw was talking about, the linearity of T tells us that the matrix for T relative to B and C (= T(B)) is the identity matrix:

T([x_{1},x_{2}]_{B}) = T(x_{1}b_{1}+x_{2}b_{2})

= x_{1}T(b_{1}) + x_{2}T(b_{2}) (by linearity)

= x_{1}c_{1}+ x_{2}c_{2}= [x_{1},x_{2}]_{C}.