# Linear transformation formula

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• Oct 23rd 2012, 07:20 PM
M.R
Linear transformation formula
Hi,

I am trying to understand and solve the following problem:

Let T: $R^2 -> R^2$ be a linear transformation. Find a formula for $T(x_{1}, x_{2})$ given that

$T(4,7) = (3,4)$ and $T(3,5) = (4,9)$

My attempt:

Now the basis $B={(4,7), (3,5)}$ is the basis for the domain. And we are trying the find the basis for the codomain? Is this correct?

Now,

$T(x_{1}(4,7) + x_{2}(3,5)) = x_{1} T(4, 7) + x_{2} T(3,5) = x_{1} (, 3, 4) + x_{2} (4, 9) = 3x_{1}+4x_{2} , 4x_{1}+9x_{2}$

But this is not correct, because if i plug in the value $(4,7)$ the formula doesn't give me $(3,4)$. But I don't understand where I went wrong?
• Oct 23rd 2012, 07:48 PM
chiro
Re: Linear transformation formula
Hey M.R.

A linear operator taking R^2 to R^2 will have 4 elements. If this operator is [a, b; c, d] in matrix form then you have

You should get ax1 + by1 = r0, cx1 + dy1 = r1, ax2 + by2 = r2, cx2 + dy2 = r3 which gives four equations in four unknowns. You know x1,y1,x2,y2,r0,r1,r2,r3 so you can solve this system with standard linear algebra techniques.
• Oct 23rd 2012, 07:52 PM
TheEmptySet
Re: Linear transformation formula
Quote:

Originally Posted by M.R
Hi,

I am trying to understand and solve the following problem:

Let T: $R^2 -> R^2$ be a linear transformation. Find a formula for $T(x_{1}, x_{2})$ given that

$T(4,7) = (3,4)$ and $T(3,5) = (4,9)$

My attempt:

Now the basis $B={(4,7), (3,5)}$ is the basis for the domain. And we are trying the find the basis for the codomain? Is this correct?

Now,

$T(x_{1}(4,7) + x_{2}(3,5)) = x_{1} T(4, 7) + x_{2} T(3,5) = x_{1} (, 3, 4) + x_{2} (4, 9) = 3x_{1}+4x_{2} , 4x_{1}+9x_{2}$

But this is not correct, because if i plug in the value $(4,7)$ the formula doesn't give me $(3,4)$. But I don't understand where I went wrong?

Here is the way I thought of it.

$(4,7)=4e_1+7e_2$ and $(3,5)=3e_1+5e_2$

We know that

$T(4e_1+7e_2)=3e_1+4e_2 \iff 4T(e_1)+7T(e_2)=3e_1+4e_2$

and

$T(3e_1+5e_2)=4e_1+9e_2 \iff 3T(e_1)+5T(e_2)=4e_1+9e_2$

This gives a system of equations for $T(e_1) \text{ and } T(e_2)$

Solving for each (work not shown) gives

$T(e_1)=13e_2+43e_2 \quad T(e_2)=-7e_1-24e_2$

This gives the matrix

$T=\begin{bmatrix}13 & -7 \\ 43 & -24 \end{bmatrix}$
• Oct 24th 2012, 04:43 AM
Hartlw
Re: Linear transformation formula
Quote:

Originally Posted by M.R
Hi,

I am trying to understand and solve the following problem:

Let T: $R^2 -> R^2$ be a linear transformation. Find a formula for $T(x_{1}, x_{2})$ given that

$T(4,7) = (3,4)$ and $T(3,5) = (4,9)$

My attempt:

Now the basis $B={(4,7), (3,5)}$ is the basis for the domain. And we are trying the find the basis for the codomain? Is this correct?

Now,

$T(x_{1}(4,7) + x_{2}(3,5)) = x_{1} T(4, 7) + x_{2} T(3,5) = x_{1} (, 3, 4) + x_{2} (4, 9) = 3x_{1}+4x_{2} , 4x_{1}+9x_{2}$

But this is not correct, because if i plug in the value $(4,7)$ the formula doesn't give me $(3,4)$. But I don't understand where I went wrong?

Doing it your way: Let u1 = (4,2) and u2 = (3,5) be a basis.

Let v be any vector, then v = x1u1 +x2u2

if Tu1 = (3,4) and Tu2 = (4,9)

Tv = x1Tu1 + x2Tu2 = x1(3,4) + x2(4,9)

If v = u1, v has coordinates x1=1 and x2 = 0 so Tv = (3,4)
If v = u2, v has coordinates x1= 0 and x2 = 1 so Tv = (4,9)
• Oct 24th 2012, 02:38 PM
Hartlw
Re: Linear transformation formula

Just another way of looking at it in the e1 = (1,0), e2 = (0,1) basis.
Principle:
Given v = x1e1 + x2e2, Tv = x1Te1 + x2Te2.
We know Tu1 and Tu2, so if we find e1 and e2 in terms of u1 and u2 we have the transformation.
e1 = x11u1 + x12u2
e2 = x21u1 + x22u2
(1,0) = x11(4,7) + x12(3,5)
(0,1) = x21(4,7) + x22(3,5)
equating first and second components gives:
e1 = -5u1 + 7u2
e2 = 3u1 -4u2

Te1 = -5(3,4) +7(4,9) = (13,43)
Te2 = 3(3,4) -4(4,9) = (-7,-24)
• Oct 24th 2012, 04:03 PM
M.R
Re: Linear transformation formula
Quote:

Originally Posted by Hartlw
Doing it your way: Let u1 = (4,2) and u2 = (3,5) be a basis.

Let v be any vector, then v = x1u1 +x2u2

if Tu1 = (3,4) and Tu2 = (4,9)

Tv = x1Tu1 + x2Tu2 = x1(3,4) + x2(4,9)

If v = u1, v has coordinates x1=1 and x2 = 0 so Tv = (3,4)
If v = u2, v has coordinates x1= 0 and x2 = 1 so Tv = (4,9)

But how was my way correct? Given that the answer is $T=\begin{bmatrix}13 & -7 \\ 43 & -24 \end{bmatrix}$
• Oct 25th 2012, 08:36 AM
Hartlw
Re: Linear transformation formula
MR wrote: "But how was my way correct? Given that the answer is http://latex.codecogs.com/png.latex?... \end{bmatrix}"

Given u1 = (4,7), u2 = (3,5) and Tu1 = (3,4), Tu2 = (4,9)

u1 and u2 are not THE basis for the domain, they are A basis. You want to work with the basis u1 and u2? Fine.

Now you want to know, given a vector (x1,x2), what is it transformed to under T? Since you chose u1, u2 as your basis, you have to assume the vector (x1,x2) has coordinates wrt this basis, ie,

v =x1u1 + x2u2.

Then T(x1,x2) = x1Tu1 + x2Tu2

Now there are 2 options:

Option 1) Express result in e1,e2 basis, which is what you did, to get:
So you have answered the question: If I have a vector with coordinates in the u1,u2 basis, what are the coordinates of the transformed vector in the e1,e2 basis.

If x1,x2 are the coordinates of the starting vector in the e1,e2 system, to use your procedure, you must first express it in terms of coordinates wrt the u1,u2 system.

If I want to test my transformation on u1, then u1 = (1)u1 + (0)u2. So you are asking, if (x1,x2) = (1,0), what is T(x1,x2) and the answer is (3,4)

Option 2) Express result in u1, u2 basis, ie, work exclusiveley in the u1,u2 basis. To do this, you have to express Tu1 and Tu2 in the u1,u2 basis knowing what they are in the e1,e2 basis.
Tu1 = a11u1 + a12u2
Tu2 = a21u1 + a22u2, where u1 = (4,7), u2 = (3,5) and Tu1 = (3,4), Tu2 = (4,9).
The result would have the form: T(x1,x2) = x1k1u1 + x2k2u2.
So now, starting with a vector expressed in the u1,u2, basis, you have found the transformed vector in the u1,u2 basis.

Edit: The answer you refer to, given by TheEmptySet, works exclusiveley in the e1,e2 coordinate system, and assumes (x1,x2) are coordinates in e1,e2.

Edit: There is one interpretation I completeley missed. If (x1,x2) are the coordinates of v in the u1,u2 basis, then (x1,x2) are also the coordinates of Tv in the Tu1,Tu2 basis.
• Oct 25th 2012, 10:07 AM
Hartlw
Re: Linear transformation formula
There is an exquisiteley simple interpretation to the original post problem, which I saw in last edit of previous post and repeat here so it isn't missed:

If v=(x1,x2) in the u1,u2 basis, then Tv = (xl,x2) in the Tu1,Tu2 basis:

v =x1u1 + x2u2
Tv = x1Tu1 +x2Tu2,
• Oct 25th 2012, 10:20 AM
Deveno
Re: Linear transformation formula
my totally naive approach:

the formula for T (provided we CAN find one) is going to be something like:

T(x,y) = (ax+by,cx+dy) <---general linear function from R2 to R2.

so what we really want to do is find a,b,c and d.

now T(4,7) = (3,4). to me this means:

4a+7b = 3
4c+7d = 4

that by itself isn't enough to say what a,b,c and d are, but it's a start.

we also know T(3,5) = (4,9) so:

3a+5b = 4
3c+5d = 9

looking at just a&b, we have the two equations:

4a+7b = 3
3a+5b = 4

so:

12a+21b = 9
-12a-20b = -16

thus b = -7 (and then from 4a+7b = 3, we get: 4a - 49 = 3, so 4a = 52, thus a = 13).

similarly:

12c+21d = 12
=12c-20d = -36

so d = -24 (and then from 4c+7d = 4, we get: 4c - 168 = 4, so 4c = 172, so c = 43).

so our formula for T is:

T(x,y) = (13x-7y,43x-24y)

yes, you CAN consider a basis B, and then find the matrix of T relative to that basis. but since we're in R2, and using coordinates (x,y) to denote elements of R2, we can by-pass that altogether, and just deal with "coordinate functions":

T1(x,y) = ax+by (the first coordinate of T(x,y))
T2(x,y) = cx+dy (the second coordinate of T(x,y)).

again, for emphasis: linear functions are linear in EACH coordinate.

*******

NOTE: there is nothing wrong with any of the previous answers. realize that if you are going to find a FORMULA for a linear transformation T in terms of COORDINATES, one really should say "which basis the coordinates are IN". writing (x,y) usually tacitly assumes that the basis is B = {(1,0),(0,1)}, but the matrix for T will have "different numbers" (i.e., a different formula), if you use some OTHER basis (in the domain, range, or both).
• Oct 25th 2012, 10:23 AM
Deveno
Re: Linear transformation formula
Quote:

Originally Posted by Hartlw
There is an exquisiteley simple interpretation to the original post problem, which I saw in last edit of previous post and repeat here so it isn't missed:

If v=(x1,x2) in the u1,u2 basis, then Tv = (xl,x2) in the Tu1,Tu2 basis:

v =x1u1 + x2u2
Tv = x1Tu1 +x2Tu2,

one caveat to this way of looking at things: you need to show that {Tu1,Tu2} actually IS a basis, after all: T might be singular.
• Oct 25th 2012, 12:41 PM
Hartlw
Re: Linear transformation formula
Tu1 = (4,7), Tu2 = (3,5). They are obviously Linearly Independent, as are u1 and u2, so T is non-sigular.

But that's not why I write. The point of the original post is as follows:

Example: Imagine u1 and u2 plotted on a piece of paper. I have a vector v whose coordinates I only know in terms of u1 and u2. Now imagine I have a transformation which rotates u1 and u2 through 30degs, which become the transformed basis. If I apply the same transformation to v, what are the new coordinates of the transformed vector? Since the vector rotated with u1 and u2, it still has the same components wr to the transformed basis. They all rotated together.

Edit: In the more general case for vectors drawn on a piece of paper, the theory becomes quite interesting. Suppose the transformation takes u1 to an arbitrary vector Tu1 and u2 to an arbitrary vector Tu2. Now what are the components of v after this transformation? The theory says it still has the same components with respect to the transformed vectors. Now that is not obvious.
• Oct 27th 2012, 02:06 PM
M.R
Re: Linear transformation formula
Quote:

Originally Posted by Hartlw
Edit: The answer you refer to, given by TheEmptySet, works exclusiveley in the e1,e2 coordinate system, and assumes (x1,x2) are coordinates in e1,e2.

Edit: There is one interpretation I completeley missed. If (x1,x2) are the coordinates of v in the u1,u2 basis, then (x1,x2) are also the coordinates of Tv in the Tu1,Tu2 basis.

Thanks for that explanation. If you look at my original question, I was confused if we needed to apply a change of basis, before and after the transformation. So the answer by TheEmptySet makes sense, since he applied a change of basis before the transformation (domain) and left the answer in the standard basis after transformation.
• Oct 27th 2012, 02:11 PM
M.R
Re: Linear transformation formula
Quote:

Originally Posted by Deveno
my totally naive approach:

the formula for T (provided we CAN find one) is going to be something like:

T(x,y) = (ax+by,cx+dy) <---general linear function from R2 to R2.

so what we really want to do is find a,b,c and d.

now T(4,7) = (3,4). to me this means:

4a+7b = 3
4c+7d = 4

that by itself isn't enough to say what a,b,c and d are, but it's a start.

we also know T(3,5) = (4,9) so:

3a+5b = 4
3c+5d = 9

looking at just a&b, we have the two equations:

4a+7b = 3
3a+5b = 4

so:

12a+21b = 9
-12a-20b = -16

thus b = -7 (and then from 4a+7b = 3, we get: 4a - 49 = 3, so 4a = 52, thus a = 13).

similarly:

12c+21d = 12
=12c-20d = -36

so d = -24 (and then from 4c+7d = 4, we get: 4c - 168 = 4, so 4c = 172, so c = 43).

so our formula for T is:

T(x,y) = (13x-7y,43x-24y)

yes, you CAN consider a basis B, and then find the matrix of T relative to that basis. but since we're in R2, and using coordinates (x,y) to denote elements of R2, we can by-pass that altogether, and just deal with "coordinate functions":

T1(x,y) = ax+by (the first coordinate of T(x,y))
T2(x,y) = cx+dy (the second coordinate of T(x,y)).

again, for emphasis: linear functions are linear in EACH coordinate.

*******

NOTE: there is nothing wrong with any of the previous answers. realize that if you are going to find a FORMULA for a linear transformation T in terms of COORDINATES, one really should say "which basis the coordinates are IN". writing (x,y) usually tacitly assumes that the basis is B = {(1,0),(0,1)}, but the matrix for T will have "different numbers" (i.e., a different formula), if you use some OTHER basis (in the domain, range, or both).

Excellent. I actually tired this approach, but I made a mistake in my arithmetic and didn't get he right answer. And it really confused me, as to why it didn't work.
• Oct 27th 2012, 02:23 PM
M.R
Re: Linear transformation formula
One final question. Say I have a vector $v = v_{1} + v_{2}$ in a non-standard basis, like the above. So if I express x in terms of $v$, that is $x = x_{1}v_{1}+x_{2}v_{2}$. Now if I apply transformation to $x$ (ie, $T(x)$), would the answer of the transformation be in the same non-standard basis (of $v$)? Our would we assume that after the transformation, the answer is in the standard basis (ie. ${(1,0), (0,1)}$?
• Oct 27th 2012, 06:04 PM
Deveno
Re: Linear transformation formula
when you are specifying a matrix, [T], for a linear transformation T, you actually need TWO bases: one for the domain of T, and one for the co-domain (the vector space the image lies in).

if T is a linear endomorphism (that is, the domain, and the co-domain are the same) one can use the same basis for BOTH, but you don't HAVE to.

let me illustrate with a simple example:

suppose T:R2-->R2 is the identity function: T(x,y) = (x,y).

FROM the standard basis TO the standard basis, the matrix for T is (as you would suspect):

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

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.

if we want the matrix for T with "inputs" as B-coordinates, and "outputs" in the standard basis, this matrix has to map (1,0) to (1,1), and (0,1) to (1,1) so we have:

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

if we want the matrix for T with "inputs" as standard coordinates, and "outputs" in B-coordinates, note that:

(1,0) = 1b1 + 0b2 = [1,0]B, so the first column of [T] will be as before. BUT:

(0,1) = (-1)b1 + (1)b2 = [-1,1]B, so the second column of [T] will be (-1,1) (which is what (0,1) turns into in "B-coordinates").

so NOW the matrix for T is:

$[T] = \begin{bmatrix}1&-1\\0&1 \end{bmatrix}$

you may verify yourself that this matrix is the inverse of the matrix from the B-coordinates to the standard coordinates.

if we want [T] to go from B-coordinates to B-coordinates, we can use the same matrix as from the standard basis to the standard basis. but this is because T is the identity function, this doesn't work for EVERY linear transformation.

in general, if P is the matrix that takes the B-coordinates to the standard basis (a change of basis matrix), and [T] is the matrix of of a linear transformation T (with respect to the standard basis), then the matrix for T from B-coordinates to B-coordinates will be:

[T]B = P-1[T]P.

any vector in R2 can be REPRESENTED as a pair of numbers. what these numbers MEAN depends on what AXES we've chosen (the x-axis and the y-axis are the "standard" ones). if we choose different axes to measure along, we're going to get different coordinates in different coordinate systems, but the vector itself remains the same (we just give it "a different address").

what makes this confusing is that the standard basis is "invisible", the standard coordinates of a vector (x,y) are just x and y (in the standard basis, the coordinates of a vector are "themselves"). or to put it another way: (4,3) doesn't really tell us which vector we have...one needs to ask 4 which way, and 3 which other way?
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