Eigenvalues

PensFan10 · #1 $Old$ November 6th, 2009, 07:29 AM

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

Defunkt · #2 $Old$ November 6th, 2009, 07:41 AM

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 --

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

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

PensFan10 · #3 $Old$ November 6th, 2009, 07:52 AM

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

PensFan10 · #4 $Old$ November 6th, 2009, 08:05 AM

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

Defunkt · #5 $Old$ November 6th, 2009, 08:27 AM

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.