Hi just wondering if anybody could help me with a problem i've can't see to solve!
I can work out the Eigenvalues fine
Matrix A = [ 2 -1
-1 1 ]
Ax = Lx L = lamda
| [A-LI] | = 0
(2-L)(1-L) - 1 = 0
L^2 - 3L + 1 = 0
L(1) = 0.38
L(2) = 2.62
Now this is where i run into trouble. i try solve the simulataenous equations from this, these are.
1. 1.62x1 - x2 = 0
2. 0.62x2 - x1 = 0
What do i do now, i've tried solving these but all i end up with is Cx1 = 0, which doesn't help anything, any help on what to do?
only have one equation corresonding to lambda=0.38. Which is as it should
be since the property of being an eigen vector tells you nothing about the
magnitude of the vector.
You have x2=1.62 x1, or the eigen vector is k[1, 1.62]'. Now you might want
this to be a unit vector when you have something like:
x ~= [0.5253, 0.8509]'
Now try it with the other e-value.
Hey, thanks thats answered half my questions but unfortunatly isn't right. ( is a past exam paper have the answer but didn't know how to get there )
The unit vector part is the part i was missing thanks.
A = [ 2 -1, -1 1 ] Eigenvectors = [ -0.53, -0.85 ], [ -0.85,0.53 ]
Eigenvalues = 0.38, 2.62
So solving as simulataneous equations ( for L = 0.38 )
(1) ( 2 - Lamda )x(1) + (-1)x(2) = 0 => 1.62x(1) - x(2) = 0
(2) (-1)x(1) + (1-Lamda)x(1) = 0 => -x(1) + 0.62x(2) = 0
now obviously from this we get the answer in the above post
1.62x(1) = x(2)
and 0.62x(2) = x(1)
So getting the unit vector = [0.53, 0.85], any idea why the signs would be different to the 'answer', I personally can't see why unless for some reason you take 1.62x(1) from a side in (1) so
(1) -x(2) = -1.62x(1)
but i'm not sure why on earth you would do this? Again thanks for any help given.