Thread: 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(Lamda-A) And got

[lamda - 1 -1]
[-1 lamda - 1]

The det of that gives (lamda - 1)(lamda-1) + 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 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 non-zero scalar multiple of . In this case, .

From this, could all be eigenvectors as well.