1. ## Finding eigenvalues

I've worked out my characteristic equation to be $\lambda^{3} - 4\lambda^{2} - 4\lambda + 16 = 0$

I have been letting $p(\lambda) = \lambda^{3} - 4\lambda^{2} - 4\lambda + 16$ and filling in values for $\lambda$ which are the factors of 16 i.e. $\lambda = 1,-1,2,-2,4,-4,8,-8,16,-16$.

However this method cannot tell me if an the same eigenvalue occurs twice... which i've a feeling in this case occurs...

Any thoughts?

2. Hi!

Originally Posted by Tekken
I've worked out my characteristic equation to be $\lambda^{3} - 4\lambda^{2} - 4\lambda + 16 = 0$

I have been letting $p(\lambda) = \lambda^{3} - 4\lambda^{2} - 4\lambda + 16$ and filling in values for $\lambda$ which are the factors of 16 i.e. $\lambda = 1,-1,2,-2,4,-4,8,-8,16,-16$.

However this method cannot tell me if an the same eigenvalue occurs twice...
I agree!

Do you know that a polynomial long division is? After finding a zero/eigenvalue polynomial long division leads to a quadratic equation including the two remaining zeros.

Originally Posted by Tekken
Any thoughts?
I think the eigenvalues are
+4
+2
-2

so there are 3 different zeros - in this case no zero occurs twice, because the polynom

$p(\lambda)= \lambda^{3} - 4\lambda^{2} - 4\lambda + 16 = 0$

has the degree 3, so there are 3 solutions in $\mathbb{R}$ max

3. Originally Posted by Rapha
Hi!

I agree!

Do you know that a polynomial long division is? After finding a zero/eigenvalue polynomial long division leads to a quadratic equation including the two remaining zeros.

I think the eigenvalues are
+4
+2
-2

so there are 3 different zeros - in this case no zero occurs twice, because the polynom

$p(\lambda)= \lambda^{3} - 4\lambda^{2} - 4\lambda + 16 = 0$

has the degree 3, so there are 3 solutions in $\mathbb{R}$ max
Yes i know polynomial long division. So what your saying is if i found the first eigenvalue to be $\lambda = 4$ i could divide the characteristic equation by $\lambda - 4$ which will leave me a quadratic equation to which the solutions would be $\lambda = 2$ & $\lambda = -2$

Correct?

4. Originally Posted by Tekken
Yes i know polynomial long division. So what your saying is if i found the first eigenvalue to be $\lambda = 4$ i could divide the characteristic equation by $\lambda - 4$ which will leave me a quadratic equation to which the solutions would be $\lambda = 2$ & $\lambda = -2$

Correct?
That's correct.
Your remaining quadratic equation should be $\lambda^2-4$

5. Originally Posted by Rapha
That's correct.
Your remaining quadratic equation should be $\lambda^2-4$
Many thanks for the help, you've just given me a bit more confidence coming up to exam time

6. Originally Posted by Tekken
Many thanks for the help, you've just given me a bit more confidence coming up to exam time
You're welcome.