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- October 8th 2006, 07:51 PM #1

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- October 8th 2006, 10:12 PM #2

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- October 9th 2006, 04:29 AM #3

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- October 9th 2006, 05:08 AM #4

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- October 9th 2006, 05:42 AM #5

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The charateristic equation is:

det(A-lambda I) = 0

(as a polynomial in lambda). Now as 3, 2, 1 are the roots of this equation

we know that if we expand the above expression we will have:

(lambda-3)(lambda-2)(lambda-1)=0

for the charateristic equation, and as A should satisfy this we should find

that:

(A-3)(A-2)(A-1)=0

where there is an impicit multiplication of the 3, 2, 1 and 0 by I.

This expands to:

(A-3)(A-2)(A-1) = (A-3)(A^2 - 3A +2) = A^3 - 6A^2 + 11A -6

The powers of A are simply A^2=AA, A^3=A(AA), and the 6 is short hand

for 6 I.

RonKL

- October 9th 2006, 06:08 AM #6

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- October 9th 2006, 08:10 AM #7

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I'm a bit confused by what you need to show. With a particular matrix you

can compute the powers and substitute them into the equation and show

that the matrix satisfies it.

For a general matrix you could show this is in general true.

If D is the diagonal matrix then as each element on the diagonal is an eigen

value of A, we have tivialy that D satisfies the charateristic equation.

Also as P D P^-1 = A, and P D^2 P^-1 = P D P^-1 P D P^-1 = A^2,

and so on for higher powers, so P D^n P^-1 = A^n and we have:

if

sum_{r=0,n} a_r x^r = 0

is a polynomial satisfied by D, then it is also satisfied by A, since:

P sum_{r=0,n} a_r D^r P^-1 = sum_{r=0,n} a_r P D^r P^-1

..................=sum_{r=0,n} a_r A^r = 0.

RonL

- October 10th 2006, 04:31 PM #8

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Ok, I think I get it. Thankyou.

However I have a question for which I am completely stumped.

First, it asked you to find the eigenvectors and eigenvalues for matrix A.

Code:Matrix A: |1 3| |2 6| Here's what I did: |1 3| |2 6|x = Yx |1 3| |2 6|x - Yx = 0 |1-Y 3 | | 2 6 - Y|x = 0 .: (1 - Y)(6 - Y) - 6 = 0 6 + Y^2 - 6Y - 1Y - 6 = 0 Y^2 - 7Y + 0 = 0 (Y + 0)(Y - 7) = 0 When Y = 0 |1 3| |x1| |2 6| |x2| = 0 Since |1 3| | 3| |2 6| |-1| = 0 | 3| |-1| is an eigenvector corresponding to the eigenvalue 0. When Y = 7 |1-7 3| |2 6-7|x = 0 |-6 3| |x1| | 2 -1| |x2| = 0 Since |-6 3| |1| | 2 -1| |2| = 0 |1| |2| is an eigenvector corresponding to the eigenvalue 7. Good, so I got the eigenvalues and vectors. The question then asked to find P for which P^-1AP is a diagonal matrix. P = |3 1| |-1 2| P-1 = |3 1| |-1 2| = | 3 1|1 0| |-1 2|0 1| = |1 0|2 -1| |0 1|1 3| = |2 -1| |1 3| .: P^-1AP = |2 -1| |1 3| | 3 1| |1 3| |2 6| |-1 2| = |0 0| |3 1| |7 21| |-1 2|

- October 10th 2006, 08:10 PM #9

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- October 11th 2006, 06:42 AM #10

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- October 11th 2006, 07:52 AM #11

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some sort of brain fade has occured, I must have inverted the wrong matrix

(this must be the case as I multiplied what I inverted by this "inverse" and

got the identity matrix)

Checking this, the real inverse is

P^-1= [2/7 -1/7]

.........[1/7...3/7]

obtained by Gaussian elimination with multiple right hand sides.

There are many ways of inverting a 2x2 matrix, there is Gaussian elimination,

Cramers rule, inspection, ..

Code:Demo of Gaussian elimination ---------------------------- Initial set-up: 3 1 | 1 0 -1 2 | 0 1 multiply the second row by 3 and add the first: 3 1 | 1 0 0 7 | 1 3 multiply the first row by 7 and subtract the second: 21 0 | 6 -3 0 7 | 1 3 Divide the first row through by 21, and the second by 7: 1 0 | 2/7 -1/7 0 1 | 1/7 3/7 So the inverse the the matrix: [2/7 -1/7] [1/7 3/7] Check: > p=[3,1;-1,2] 3 1 -1 2 >ip=[2/7,-1/7;1/7,3/7] 0.285714 -0.142857 0.142857 0.428571 > > >ip.p 1 0 0 1 >

- October 11th 2006, 07:06 PM #12

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