- #1

Mindscrape

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- 1

[tex]\mathbf{x'} = \left(

\begin{array}{ccc}

3 & 0 & -1\\

0 & -3 & -1\\

0 & 2 & -1

\end{array}

\right) \mathbf{x} [/tex]

solving for the eigenvalues by taking the determinate and using the "basketweave" yields

[tex] (3 - \lambda)(-3-\lambda)(-1-\lambda) + 2(3-\lambda) = 0 [/tex]

and further simplification shows that

[tex] -\lambda^3 - \lambda^2 +7\lambda +15 = 0 [/tex]

guessing roots, I found that 3 is one and divided the polynomial by the root

[tex](\lambda - 3)(-\lambda^2 - 4\lambda -5)=0[/tex]

so the eigenvalues are (solving for both eqns in the brackets)

[tex]\lambda = 3[/tex]

[tex]\lambda = 2 + i[/tex]

[tex]\lambda = 2 - i[/tex]

The thing is that when I put in the real eigenvalue, it doesn't seem to have an eigenvector. Is this right? I think it would be because I don't have any idea how to get the solution for a mixture of real and complex eigenvalues. The solutions I know how to solve are when the eigenvectors are of the form v1,v2=p±iq. Though I imagine that if the real eigenvalue made an eigenvector I could just take the general solution of that and add it to the general solution of the complex eigenvectors.

Basically what I want to know is, what do I do know?