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:
$\displaystyle P = \begin{bmatrix}4&3\\7&5 \end{bmatrix}$
the matrix that takes standard coordinates to B-coordinates is:
$\displaystyle P^{-1} = \begin{bmatrix}-5&3\\7&-4 \end{bmatrix}$
in your original problem, the matrix for T(x,y) = (13x-7y,43-24y) is (as has been discussed to death):
$\displaystyle [T] = \begin{bmatrix}13&-7\\43&-24 \end{bmatrix}$.
it is straight-forward to verify that:
$\displaystyle \begin{bmatrix}13&-7\\43&-24 \end{bmatrix} \begin{bmatrix}4\\7 \end{bmatrix} = \begin{bmatrix}3\\4 \end{bmatrix}$
$\displaystyle \begin{bmatrix}13&-7\\43&-24 \end{bmatrix} \begin{bmatrix}3\\5 \end{bmatrix} = \begin{bmatrix}4\\9 \end{bmatrix}$
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:
$\displaystyle P^{-1}[T]P = \begin{bmatrix}-5&3\\7&-4 \end{bmatrix}\begin{bmatrix}13&-7\\43&-24 \end{bmatrix}\begin{bmatrix}4&3\\7&5 \end{bmatrix}$
$\displaystyle = \begin{bmatrix}-5&3\\7&-4 \end{bmatrix}\begin{bmatrix}3&4\\4&9 \end{bmatrix} = \begin{bmatrix}-3&7\\5&-8 \end{bmatrix}$.
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.
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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:
$\displaystyle Q = \begin{bmatrix}3&4\\4&9\end{bmatrix}$.
and we can change standard coordinates to B-coordinates using P^{-1}, so to change C-coordinates to B-coordinates we use:
$\displaystyle P^{-1}Q = \begin{bmatrix}-3&7\\5&-8 \end{bmatrix}$.
it thus stands to reason that to "change back to C (from B)" we use: (P^{-1}Q)^{-1} = Q^{-1}P,
$\displaystyle Q^{-1}P = \begin{bmatrix} \frac{8}{11}&\frac{7}{11}\\ \frac{5}{11}&\frac{3}{11} \end{bmatrix}$.
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:
$\displaystyle Q^{-1}PM = \begin{bmatrix} \frac{8}{11}&\frac{7}{11}\\ \frac{5}{11}&\frac{3}{11} \end{bmatrix} \begin{bmatrix}-3&7\\5&-8 \end{bmatrix}$
$\displaystyle = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$.
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}.
u1 = (4,7) → Tu1 = u1 = (3,4) &
u2 = (3,5) → Tu2 = u2 = (4,9) in R^{2}
1) First part of MR post (e1,e2 basis) was answered by Chiro. Given x,y in e1,e2 basis, what is the matrix representation of T when T(x,y) is in e1,e2 basis?
2) Second part of MR post was answered by MR: Given x,y in the u1,u2 basis, what is the formula for T(x,y) in the e1,e2 basis?
T(x,y) = x(3,4) + y(4,9).
3) Given x,y in the u1,u2 basis, what is the formula for T(x,y) in the Tu1,Tu2 basis?
T(x,y) = xTu1 + yTu2.
Other questions one could ask:
4) Given x,y in the u1,u2 basis, what is the matrix representation of T when T(x,y) is in the u1,u2 basis?
5) Given x,y in the e1,e2 basis, what is the matrix representation of T when T(x,y) is in the u1,u2 basis?
6) Given x,y in the Tu1,Tu2 basis, what is T when T(x,y) is in the u1,u2 basis?
7) Given x,y in any basis, what is T when T(x,y) is in any basis? This is general case of matrix representation of Linear Transformations.
Examples:
1) Let A be matrix representation of case 1). Then:
A(4,7)^{T}=(3,4)^{T}
A(3,5)^{T}=((4,9)^{T}, which in component form is:
4a_{11} +7a_{12}=3, 4a_{21}+7a_{22}=4, 3a_{11}+5a_{12}=4, 3a_{21}+5a_{22}=9
These 4eqs in 4 unknowns give A = (13,-7 ; 43,-24) which is the matrix representation of T(x,y) in the e1,e2 → e1,e2 bases.
2) Let B be matrix representation of case 2)
B(x,y)^{T} = (3,4 ; 4,9)(x,y)^{T} = (3x+4y, 4x+9y)^{T} = x(3,4)+ y(4,9)
B =(3,4 ; 4,9) is the matrix representation of T(x,y) in the u1,u2 → e1,e2 bases
3) Let C be matrix representation of case 3)
u1=(1,0) , Tu1=(1,0). u2=(0,1), Tu2=(0,1). C=I
C=(1,0 ; 0,1) is the matrix representation of T(x,y) in the u1,u2 → Tu1, Tu2
4) Let D be matrix representation of case 4). Need Tu1 and Tu2 in terms of u1 and u2:
Tu1=d_{11}u1+d_{21}u2, Tu2=d_{12}u1+d_{22}u2
(3,4)=d_{11}(4,7)+d_{21}(3,5), (4,9)=d_{12}(4,7)+d_{22}(3,5) which gives:
3=4d_{11}+3d_{21}, 4=7d_{11}+5d_{21}, 4=4d_{12}+3d_{22}, 9=7d_{12}+5d_{22} which gives
d_{11}=6, d_{12}=7, d_{21}=-7, d_{22}=-8 which gives:
u1=(1,0), u2=(0,1)
Tu1=6u1-7u2=(6,-7)
Tu2=7u1-8u2=(7,-8)
D(1,0)^{T}=(6,-7)^{T}
D(0,1)^{T}=(7,-8)^{T}
D=(6,7 ; -7,-8)
or, given (x,y) coordinates of v in u1,u2 basis, then D(x,y)T are coordinates of T(x,y) in u1,u2 basis. Or, D is matrix representation of T(x,y) in the u1,u2 → u1,u2 bases.
etc, etc
PROBLEM
Given a mapping of two vectors u1,u2 → Tu1,Tu2 in basis e1,e2. Let g1,g2 and h1,h2 be arbitrary LI vectors. If x = (x,y) is given in g1,g2 basis, what is T(x) in h1,h2 basis?
Assumption
It is known how to go from representation of a vector in one basis to representation of same vector in another basis.
NOTATION
The element X in R2 is mapped into the element T(X) in R2.
x = R (X; g1,g2), x is the representation of X in the basis g1,g2 (x=xg1+yg2).
x= R (X; e1,e2), x is the representation of X in the basis e1,e2 (x=xe1+ye2).
x= R (T(X); e1,e2), x is the representation of the vector T(X) in the basis e1,e2.
x = R (T(X); h1,h2), x is the representation of the vector T(X) in the basis h1,h2.
A = R (T: e1,e2; e1,e2), A is the matrix representation x = Axof the mapping x = T(x). (Chiro, post #2).
SOLUTION
x = Ax
x = Px
x = P*x = P*Ax = P*APx
T(x,y) = P*AP(x.y)T
P*AP = R (T: g1,g2; h1,h2)
Note: Without the R notation (Mirsky) its hard to keep track of things.