The matrix;

Question

[ 1 -2 0]
[-2 -1 1]
[ 0 0 -1]
I have found the eigenvalues to be -1 and +/- 5^1/2
but am having problems putting the root 5 values a eigenvectors.
I know I sub it back into the matrix
{ x-1 -2 0 }
{ -2 x+1 1 }
{ 0 0 x+1}
but then it gets messy. Can someone finish this off?? Please and Thanks.

You wrote the last matrix wrong, it should be:

{ x-1 2 0 }
{ 2 x+1 -1 }
{ 0 0 x+1}

The case of x = +/- 5^1/2
is simpler than the case of x = -1, because now the third component of the eigenvector must be zero as you can see from the last equation. If we call the first component of the eigenvector a, the second b and the third c, then:

(x-1)a + 2b + 0 c = 0 --->

a = -1/2[sqrt(5) + 1]b

This fixes the eigenvector, because the normalization does not matter. You can choose b = 1, then a = -1/2[sqrt(5) + 1]. The third component is zero. We didn't need the other equation:

2a (x+1)b -c = 0

this is automatically satisfied, provided you determined the aigenvalues correctly.

How do you get from (x-1)a + 2b + 0c = 0
to a = -1/2[sqrt(5) + 1]b.
And just to clarify the final eigenvector for +(sqrt 5) is;
(-1/2[sqrt5 + 1), 1, 0)
or in terms of arbitrary t;
(-1/2(sqrt5 +1)t, t, 0)
and then for -(sqrt5);
(1/2(sqrt5 -1)t, t, 0)
Thanks for the help. Hope you can clear up this last bit for me.
Oh yeah, and you say it is easier than the -1 value but I find that to be much easier as I just solve the equations;
-2x-2y=0
-2x+z=0 ====> (t, -t, 2t)