# Eigenvalue

• May 13th 2008, 02:47 AM
Sabita
Eigenvalue
I am sure there is an easier way than the one I am trying.
Prove that the matrix has one negative and onepositive eigenvalue:

1 a 1
a 1 a
1 a 1
where a=1.0001
• May 13th 2008, 05:23 AM
mr fantastic
Quote:

Originally Posted by Sabita
I am sure there is an easier way than the one I am trying.
Prove that the matrix has one negative and onepositive eigenvalue:

1 a 1
a 1 a
1 a 1
where a=1.0001

What is the way you tried?
• May 13th 2008, 12:01 PM
Moo
Hello,

Quote:

Originally Posted by Sabita
I am sure there is an easier way than the one I am trying.
Prove that the matrix has one negative and onepositive eigenvalue:

1 a 1
a 1 a
1 a 1
where a=1.0001

Try to transform this matrix by combining rows with each other. I think it's called "similar matrix", am not sure...
If you manage to find a diagonal matrix (which is the representation of the matrix in a certain basis), then the numbers in the diagonal will be the eigenvalues.

Let e1 be the first row, e2 the second and e3 the third.

e2 -> e2-a*e3
Matrix is now (...)

e3 -> e3-e1
Matrix is now (...)

Conclude (Sun)
• May 13th 2008, 12:32 PM
abender
I will partially guide you on your way to finding the eigenvalues of your matrix.
Note: I know you do not need to find the actual eigenvalues here, but you should know this without a doubt.

First give your matrix a name: A = $
\left( \begin{array}{ccc} 1 & a & 1 \\ a & 1 & a \\ 1 & a & 1 \end{array} \right)
$

Your first real step: determine the characteristic polynomial of A.

To do this, subtract lambda from each term along the main diagonal in A.
Like this: $
\left( \begin{array}{ccc} 1-\lambda & a & 1 \\ a & 1-\lambda & a \\ 1 & a & 1-\lambda \end{array} \right)
$
.

Compute the determinant of this modified matrix. You should get a polynomial in terms of lambda. This is the characteristic polynomial of A. Factor this and then set your factored result equal to zero and you have your characteristic equation. Solve for lambda. Have fun.

-Andy
• May 13th 2008, 09:34 PM
Sabita
Thanks for the replies.
I did try the lambda method, but I am getting an equation which I am unable to solve due to 1.0001 * 1.0001 in some terms. Hence I assumed there would be an easier way to solve it .
• May 13th 2008, 11:20 PM
mr fantastic
Quote:

Originally Posted by Sabita
Thanks for the replies.
I did try the lambda method, but I am getting an equation which I am unable to solve due to 1.0001 * 1.0001 in some terms. Hence I assumed there would be an easier way to solve it .

The determinant is $-\lambda (\lambda^2 - 3 \lambda - 2(a^2 - 1))$.

Solving $0 = -\lambda (\lambda^2 - 3 \lambda - 2(a^2 - 1))$ for $\lambda$ looks routine ........

Clearly there's one negative eigenvalue and one positive eigenvalue. And an eigenvalue equal to zero.
• May 14th 2008, 10:11 PM
Sabita
I got the equation . But how do I solve it?
Also what is the significance of 1.0001?.
• May 14th 2008, 11:31 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
The determinant is $-\lambda (\lambda^2 - 3 \lambda - 2(a^2 - 1))$.

Solving $0 = -\lambda (\lambda^2 - 3 \lambda - 2(a^2 - 1))$ for $\lambda$ looks routine ........

Clearly there's one negative eigenvalue and one positive eigenvalue. And an eigenvalue equal to zero.

Quote:

Originally Posted by Sabita
I got the equation . But how do I solve it?
Also what is the significance of 1.0001?.

Either $\lambda = 0$ or $\lambda^2 - 3 \lambda - 2(a^2 - 1) = 0$.

The latter is solved using the quadratic formula:

$\lambda = \frac{3 \pm \sqrt{8a^2 +1}}{2}$.

If a = 1.0001 then clearly $\sqrt{8a^2 + 1} > 9$ .......
• May 15th 2008, 12:43 AM
Isomorphism
Quote:

Originally Posted by mr fantastic
If a = 1.0001 then clearly $\sqrt{8a^2 + 1} > 9$ .......

Hi Sabita, (Hi)

mr fantastic means:

If a = 1.0001 then clearly $\sqrt{8a^2 + 1} > \sqrt{9} = 3$ .......
• May 15th 2008, 04:27 AM
Sabita
(Rofl)
Thanks mr fantastic, Isomorphism and Andy.

I got it!