# Math Help - Eigenvalues

1. ## 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

2. 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 $rank(A) = 1$. Now, note that $trace(A) = 1+1 = 2$ 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 $(1 \ 1)^T$.

So we know that 2 is an eigenvalue of geometric multiplicity $\geq 1$ and 0 is an eigenvector with geometric multiplicity 1. Since algebraic multiplicity $\geq$ geometric multiplicity for each eigenvalue, we get that the algebraic and geometric multiplicities for each eigenvector are 1 and 1.

Therefore: $\Delta_A(x) = x(x-2)$
With 0,2 as eigenvalues
and $(1\ -1)^T, (1 \ 1)^T$ the corresponding eigenvectors.

However, if you still want to know what you did wrong --

$det(\lambda I-A) = (\lambda-1)(\lambda-1) - (-1)(-1) = (\lambda-1)^2 -1 \neq (\lambda-1)^2+1$

$=\Rightarrow det(\lambda I-A) = \lambda^2 -2\lambda +1 -1 = \lambda^2 -2\lambda = \lambda(\lambda-2)$

3. ## Thanks

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

4. 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

5. That is fine. The eigenvector can be any non-zero scalar multiple of $(1 \ -1)^T$. In this case, $(-1)(1 \ -1)^T = (-1 \ 1)^T$.

From this, $(2\ -2)^T, \ (-8\ 8)^T,...$ could all be eigenvectors as well.