# Thread: Find the general relationship between coordinates

1. ## Find the general relationship between coordinates

Hi, this is the question from My book, i Will show
My progress aswell

Problem: A vector u has got the coordinates
(1,2) in Two different basis (e1,e2) and (f1,f2).
We know That f1-f2 = e1+e2. Find the
General relationship between the coordinates
In both of the basis

My progress

First of all.

I use Two formulas

(1) y=t(inverse)*x. T = change of basis Matrix
(2) x=t*y

So i try to set up two different change of basis
Matrices one for the basis of e and with f
To be exact i mean e1+2e2 and F1+2f2

So i use these values to create Two different matrices

T(f)= 1 1
-1 2

And t(e) = 1 1
1 2

I have written the basis values from their
Rows to columns and placed them in the
First columns

I need the general relationship
So i assume That i need the inverse
Of the f:s basis and because formula
With the y has got t inverse
Y represents f basis and i assume
That i need to invert the matrix
Of t(f)

Through gaussian elimination
Or determinants i get inverse of t(f)

1/3. * 2 -1
1 1

How can i proceed to get
The general relationship between
The coordinates in the Two basis?

Can someone Solve this for me

(y1,y2) = (x1, -4*x1 + 3*x2)

2. ## Re: Find the general relationship between coordinates

i don't understand where you are getting those two matrices. suppose T is the matrix that turns the f-basis coordinates into e-basis coordinates. so T =

[a b]
[c d], and we know that:

a+2b = 1
c+2d = 2, since T([1,2]f) = [1,2]e.

we also know that:

T([1,-1]f) = [1,1]e (because f1-f2 = e1+e2).

thus a+b = 1 and c-d = -1.

from the two equations involving a and b, we have b = 0, so a = 1. from the two equations involving c and d, we have:

3d = 1, so d = 1/3, so c = 4/3. that gives us T, but what we really WANT is T-1, which is found to be:

$T^{-1} = \begin{bmatrix}1&0\\-4&3 \end{bmatrix}$.

this, of course, means that:

T-1(x1,x2) = (x1,-4x1+3x2).

it is easy to check that (1,2) is invariant under T-1:

T-1(1,2) = (1,(-4)(1) + (3)(2)) = (1,-4+6) = (1,2) and that:

T-1(1,1) = (1,(-4)(1) + (3)(1)) = (1,-4+3) = (1,-1), as required.