1. ## Matrix Eigenvectors

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?

2. Originally Posted by ct5845
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 exactly are you trying to do?
Do you want to find the Eigenvectors?

3. yer the eigenvectors.

The eigenvalues are correct, if someone could show me step by step how to work out one of the eigenvectors from one of the eigenvalues that would be a great help.

4. Here is an outline.

5. Originally Posted by ct5845
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?
These are to the limits of numerical precision the same equation, so you realy
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.

RonL

6. 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.

So

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.

7. Originally Posted by ct5845
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.

So

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.

If [0.53, 0.85]' is a unit e-vector then so is [-0.53, -0.85]' and vice versa.
They are both unit e-vectors for the given matrix correspomding to the same
e-value.

RonL