
Eigenvalues
I need to find the characteristic polynomial, eigenvalues, and eigenvectors or the following matrix.
[1 1]
[1 1]
So far, this is what I have done:
det(LamdaA) And got
[lamda  1 1]
[1 lamda  1]
The det of that gives (lamda  1)(lamda1) + 1 which is lamda squared  2 lamda +2
I used the quad formula on that and got 1 plus/minus i
I dont know where to go from here. I think Now i do this:
A(1+i)I =
[i 1]
[1 i]
i have no clue where to go from there
Please help. THanks

Hi!
There really is no need to calculate the characteristic polynomial that way.
First, you can easily observe that 0 is an eigenvalue of geometric multiplicty 1 (since $\displaystyle rank(A) = 1$. Now, note that $\displaystyle trace(A) = 1+1 = 2$ and A only has two eigenvalues, so the second one must be 2. Another way to know that 2 is an eigenvalue is to note that the sum of all rows is 2, therefore, according to a theorem which you should have learned, 2 is an eigenvalue with corresponding eigenvector $\displaystyle (1 \ 1)^T$.
So we know that 2 is an eigenvalue of geometric multiplicity $\displaystyle \geq 1$ and 0 is an eigenvector with geometric multiplicity 1. Since algebraic multiplicity $\displaystyle \geq$ geometric multiplicity for each eigenvalue, we get that the algebraic and geometric multiplicities for each eigenvector are 1 and 1.
Therefore: $\displaystyle \Delta_A(x) = x(x2)$
With 0,2 as eigenvalues
and $\displaystyle (1\ 1)^T, (1 \ 1)^T$ the corresponding eigenvectors.
However, if you still want to know what you did wrong 
$\displaystyle det(\lambda IA) = (\lambda1)(\lambda1)  (1)(1) = (\lambda1)^2 1 \neq (\lambda1)^2+1$
$\displaystyle =\Rightarrow det(\lambda IA) = \lambda^2 2\lambda +1 1 = \lambda^2 2\lambda = \lambda(\lambda2)$

Thanks
Thanks a bunch. It would be nice if i could add :P

So I was trying to do it the way i was. I got the (1,1), but instead of (1,1) I got (1,1).
0lamda  A =
1 1
1 1
then rref that and got
1 1
0 0
x1= r
x2 = r
1, 1

That is fine. The eigenvector can be any nonzero scalar multiple of $\displaystyle (1 \ 1)^T$. In this case, $\displaystyle (1)(1 \ 1)^T = (1 \ 1)^T$.
From this, $\displaystyle (2\ 2)^T, \ (8\ 8)^T,...$ could all be eigenvectors as well.