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

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- May 13th 2008, 01:47 AMSabitaEigenvalue
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, 04:23 AMmr fantastic
- May 13th 2008, 11:01 AMMoo
Hello,

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, 11:32 AMabender
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 = $\displaystyle

\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: $\displaystyle

\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, 08:34 PMSabita
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, 10:20 PMmr fantastic
The determinant is $\displaystyle -\lambda (\lambda^2 - 3 \lambda - 2(a^2 - 1))$.

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

Clearly there's one negative eigenvalue and one positive eigenvalue. And an eigenvalue equal to zero. - May 14th 2008, 09:11 PMSabita
I got the equation . But how do I solve it?

Also what is the significance of 1.0001?. - May 14th 2008, 10:31 PMmr fantastic
Either $\displaystyle \lambda = 0$ or $\displaystyle \lambda^2 - 3 \lambda - 2(a^2 - 1) = 0$.

The latter is solved using the quadratic formula:

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

If a = 1.0001 then clearly $\displaystyle \sqrt{8a^2 + 1} > 9$ ....... - May 14th 2008, 11:43 PMIsomorphism
- May 15th 2008, 03:27 AMSabita
(Rofl)

Thanks mr fantastic, Isomorphism and Andy.

I got it!