1. ## Re: Linear transformation formula

Originally Posted by Deveno
but suppose we want to use the basis B = {b1,b2} = {(1,0),(1,1)} instead.

in the basis B, the vector (1,0) = (1)(1,0) + (0)(1,1) = 1b1 + 0b2 = [1,0]B.
Do you mind

Hi Deveno,

Thank for you reply. Could you please do one more favour and just change that basis to say (2,3) and (5,7)? Because it's confusing me between the basis vector and standard vector. Thanks

2. ## Re: Linear transformation formula

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:

$P = \begin{bmatrix}4&3\\7&5 \end{bmatrix}$

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

$P^{-1} = \begin{bmatrix}-5&3\\7&-4 \end{bmatrix}$

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

$[T] = \begin{bmatrix}13&-7\\43&-24 \end{bmatrix}$.

it is straight-forward to verify that:

$\begin{bmatrix}13&-7\\43&-24 \end{bmatrix} \begin{bmatrix}4\\7 \end{bmatrix} = \begin{bmatrix}3\\4 \end{bmatrix}$

$\begin{bmatrix}13&-7\\43&-24 \end{bmatrix} \begin{bmatrix}3\\5 \end{bmatrix} = \begin{bmatrix}4\\9 \end{bmatrix}$

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:

$P^{-1}[T]P = \begin{bmatrix}-5&3\\7&-4 \end{bmatrix}\begin{bmatrix}13&-7\\43&-24 \end{bmatrix}\begin{bmatrix}4&3\\7&5 \end{bmatrix}$

$= \begin{bmatrix}-5&3\\7&-4 \end{bmatrix}\begin{bmatrix}3&4\\4&9 \end{bmatrix} = \begin{bmatrix}-3&7\\5&-8 \end{bmatrix}$.

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.

*************
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:

$Q = \begin{bmatrix}3&4\\4&9\end{bmatrix}$.

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

$P^{-1}Q = \begin{bmatrix}-3&7\\5&-8 \end{bmatrix}$.

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

$Q^{-1}P = \begin{bmatrix} \frac{8}{11}&\frac{7}{11}\\ \frac{5}{11}&\frac{3}{11} \end{bmatrix}$.

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

putting this all together, our new matrix is:

$Q^{-1}PM = \begin{bmatrix} \frac{8}{11}&\frac{7}{11}\\ \frac{5}{11}&\frac{3}{11} \end{bmatrix} \begin{bmatrix}-3&7\\5&-8 \end{bmatrix}$

$= \begin{bmatrix}1&0\\0&1 \end{bmatrix}$.

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([x1,x2]B) = T(x1b1+x2b2)

= x1T(b1) + x2T(b2) (by linearity)

= x1c1 + x2c2 = [x1,x2]C.

3. ## Re: Linear transformation formula

Good stuff. Thanks Deveno

4. ## Re: Linear transformation formula

u1 = (4,7) → Tu1 = u1’ = (3,4) &
u2 = (3,5) → Tu2 = u2’ = (4,9) in R2

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.

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:
4a11 +7a12=3, 4a21+7a22=4, 3a11+5a12=4, 3a21+5a22=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=d11u1+d21u2, Tu2=d12u1+d22u2

(3,4)=d11(4,7)+d21(3,5), (4,9)=d12(4,7)+d22(3,5) which gives:
3=4d11+3d21, 4=7d11+5d21, 4=4d12+3d22, 9=7d12+5d22 which gives
d11=6, d12=7, d21=-7, d22=-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

5. ## Re: Linear transformation formula

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’=x’e1+y’e2).
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’’ = Ax’of 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) it’s hard to keep track of things.

6. ## Re: Linear transformation formula

Hi Hartlw,

Thanks for your reply. I will write back after my exams are over and seek further clarifications

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