
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 . Now, note that 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 .
So we know that 2 is an eigenvalue of geometric multiplicity and 0 is an eigenvector with geometric multiplicity 1. Since algebraic multiplicity geometric multiplicity for each eigenvalue, we get that the algebraic and geometric multiplicities for each eigenvector are 1 and 1.
Therefore:
With 0,2 as eigenvalues
and the corresponding eigenvectors.
However, if you still want to know what you did wrong 

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 . In this case, .
From this, could all be eigenvectors as well.