# Thread: Eigenvectors and Lorenz Equations

1. ## Eigenvectors and Lorenz Equations

Hey MHF,

So I am working through the Lorenz Equations and have run into a road block that has stumped me all day, I have turned to my beloved internet for help again.

So I have managed to work through the system of equ.

and have used θ=10 and b=8/3... I know original... which gives me this matrix:::

Now I get stumped on producing the right the eigenvectors, which is a laborious process most of the time but it was beyond me today...
anyways, Eigenvalues are
λ= -8/3
λ= (-11-(81+40r)^(1/2))/2 and
λ= (-11+(81+40r)^(1/2))/2
Duhhh... this is easy but when I use λ= -8/3
I do not get the ε=(0,0,1) which my solutions say I should get....
22/3ε1=10ε2
I am so very lost....Could someone please point me in the right direction.

2. ## Re: Eigenvectors and Lorenz Equations

Originally Posted by LeraysJeans
Duhhh... this is easy but when I use λ= -8/3
I do not get the ε=(0,0,1) which my solutions say I should get....
We have:

$\displaystyle E=\ker (A+(8/3)I)\equiv\begin{Bmatrix}(-22/3)x_1+10x_2=0\\ rx_1+(5/3)x_2=0\\0=0\end{matrix}$

So $\displaystyle \epsilon=(0,0,1)^t$ is a non null solution of the system, as a consequence $\displaystyle \epsilon$ is an eigenvector associated to $\displaystyle \lambda=-8/3$ . Notice that $\displaystyle \{\epsilon\}$ is a basis of $\displaystyle E$ (dimension 1) if and only if $\displaystyle D=\det \begin{bmatrix}{-22/3}&{10}\\{r}&{5/3}\end{bmatrix}\neq 0$ , if $\displaystyle D=0$ we have a basis $\displaystyle \{\epsilon\;,\;\epsilon_1\}$ (dimension 2).

3. ## Re: Eigenvectors and Lorenz Equations

Just a random, stupid thought here, but aren't the Lorenz equations nonlinear? Are you working with a linearization? Or are you trying to use a technique that is really only appropriate for linear systems?

4. ## Re: Eigenvectors and Lorenz Equations

Originally Posted by FernandoRevilla

$\displaystyle D=\det \begin{bmatrix}{-22/3}&{10}\\{r}&{5/3}\end{bmatrix}\neq 0$ , if $\displaystyle D=0$ we have a basis $\displaystyle \{\epsilon\;,\;\epsilon_1\}$ (dimension 2).
FernandoRevilla! Thank you! The determinate of the matrix was the icing on the cake.

Originally Posted by Ackbeet
Just a random, stupid thought here, but aren't the Lorenz equations nonlinear? Are you working with a linearization? Or are you trying to use a technique that is really only appropriate for linear systems?
Thanks for the concern Ackbeet, I forgot to include the fact that I input Lorenz's equations into a Jacobian Matrix. Which looks at locally linear systems around point (x,y,z). My bad. I'm always skipping steps...

5. ## Re: Eigenvectors and Lorenz Equations

Originally Posted by LeraysJeans
Thanks for the concern Ackbeet, I forgot to include the fact that I input Lorenz's equations into a Jacobian Matrix. Which looks at locally linear systems around point (x,y,z). My bad. I'm always skipping steps...
Jolly good. Thanks for the info.