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Unregistered
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Can anyone help me out please? I have serious problems with transformations in p6.

For example (off the top of my head)

the transformation marks z to w by:

w = (z+3i)/(z-2)

As z is a circle centre 3+i radius 4, find the locus of w.

Any quick tips?

Also does anyone have a clue what the heck an eigenvector or eigen value is....and what it means?

Thanks

Rich
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Bigcnee
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For the transformation you're probably best multiplying through by the denominator, then substituting for x+iy and u+iv and seeing what happens. Failing that try to rationalise the denominator. I think transformations are the hardest part of P6.

In linear algebra, the eigenvectors of a linear operator are non-zero vectors which, when operated on by the operator, result in a scalar multiple of themselves. The scalar is then called the eigenvalue associated with the eigenvector. The set of all the eigenvalues is called the matrix.
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Expression
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Originally posted by Unregistered
Also does anyone have a clue what the heck an eigenvector or eigen value is....and what it means?
Here comes the science bit, concentrate.

A fixed direction or Eigenvector of a Matrix M, is defined to be a "non zero vector e such that Me=Lamdae for some number Lamda"

The number Lamda, is the stretch factor parallel to the eigenvector, and is called the eigenvalue of e.

Finding Eigenvalues - Example 1

Let M =

(1 -1 )
(2 4 )

and suppose that
e =
x
y is an eigenvector of M, with eigenvalue LAMDA (L)

Then
Me=Le

=>

(1 -1) (x) = L (x)
(2 4) (y) (y)

=>

(1 -1) (x) = (L 0) (x)
(2 4 ) (y) (0 L) (y)

==>

(1 -1) (x) - (L 0) (x) = (0)
(2 4 ) (y) (0 L) (y) (0)

Factorising ==>

[ (1 -1) - (L 0) ] (x) = (0)
[ (2 4 ) (0 L) ] (y) (0)

Peforming the Subtraction

(1-L -1) (x) = (0)
(2 4-L) (y) (0)

If there is a non zero solution of this equation then the determinant of the matrix is zero:

| 1-L -1 | = 0
| 2 4-L |

==> (1-L)(4-L) - 2(-1) = 0
==> 4 - L- 4L + L^2 +2 = 0
==> L^2 - 5L +6 =0
==> (L-3) (L-2) = 0
==> L = 3 or L=2

These roots are the Eigenvalues.

The equation | M - L I | = 0 is called the Characteristic Equation of the matrix M. Its roots are Eigenvalues of M.

Finding Eigenvectors

Once the Eigenvalues (L) have been found, the eigenvectors can then be found by considering the equation

( a b ) (x) = L (x)
( c d ) (y) (y)

=>
( a b ) (x) = (L 0) (x)
( c d ) (y) (0 L) (y)

=>
( a b ) (x) - (L 0) (x) = (0)
( c d ) (y) (0 L) (y) (0)

=>
( a-L b ) (x) = (0)
( c d-L ) (y) (0)

Example 2
The Eigenvalues of the matrix M=
(3 5 ) are L= -2 and L=4
(1 -1)

Find the Eigenvectors.
Describe the geometric effect of this transformation as fully as possible.

(3-L 5 ) (x) = (0)
( 1 -1-L) (y) (0)

When L=4
(3-4 5 ) (x) = (0)
( 1 -1-4) (y) (0)

==>

(-1 5 ) (x) = (0)
( 1 -5) (y) (0)

==>

-x + 5y = 0
x - 5y = 0 ==> x = 5y

So for L=4 a possible Eigenvector is

(5)
(1)

For L= -2

(3- -2 5 ) (x) = (0)
( 1 -1- -2) (y) (0)

==>

(5 5 ) (x) = (0)
( 1 1) (y) (0)

==>

5x + 5y = 0
x + y = 0 ==> x = -y

So for L = -2 a possible Eigenvector is

(-1)
(1)


So M is a 2-way stretch of factors 4 and -2 (the Eigenvalues) in the directions
(5)
(1)

and

(-1)
(1)

respectively.

Summary

To find the Eigenvalues, L, and Eigenvectors e of a Matrix, M=
(a b)
(c d )

the equation Me = Le needs to be solved.

i) Find the Eigenvalues by solving the Characteristic Equation from |M - L I| = 0

ii) Substitute the Eigenvalues into:
(a-L b) (x) = (0)
( c d-L) (y) (0)

to find the Eigenvectors.
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theboyberman
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Thanks so much.
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Expression
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No problem, just hope all that is relavent. I pulled it out of my AQA Pure 6 notes which I sat last June as my very first Further Pure unit and got a B on !

Sorry it comes up a bit long, and slightly out of line on screen, but if thats some use then so be it.

Good Luck to everyone doing Edexcel P6.
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theboyberman
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hi

Just to confirm another thing.

Does the order in which you put eigen veectors also not matter in the following question:

Find an orthogonal matrix P such that P(to the T)AP diagonalises the symmetric matrix A, where A = bla bla (3x3 matrix)

Does its not matter which order you put the eigen vectors in.

However when you evaluate P(to the T)AP will be a different. It should be all zeros apart from the leading diagonal which will the be eigenvalues, will it matter if they are in different places.

Btw P(to the T) - i mean the transpose of P.

Many thanks,

Richard
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Expression
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Sorry mate, thats where theres a gap in the content we cover.

Andy.
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theboyberman
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OK. Thanks anyway
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