$\dot{x}=Ax=\begin{bmatrix}-1 && 3 && 0 \\ -3 && -1/2 && 0 \\ 1 && 3 &&1 \end{bmatrix} (x_1x_2 x_3)'$

From doing the computation to get the eigenvectors, I got the following equation.

$-\lambda^3-\frac{1}{2}\lambda^2-8\lambda+9.5$ or $(-\lambda^2-\frac{3}{2}\lambda-9.5)(\lambda-1)$

with this characteristic equation I was able to find the one real eigenvalue (1) and the corresponding eigenvector for it,

$\psi_1= \begin{bmatrix}0 \\ 0 \\ x_3 \end{bmatrix}$

Now I know the other two eigenvalues are complex but I don't know how to obtain the eigenvectors.

The eigenvalues I obtained from doing the quadratic equation to $(-\lambda^2-\frac{3}{2}\lambda-9.5)$ were the following,

$\lambda_{2,3}=-.75\pm2.99i $

I know the traditional format to obtaining eigenvectors is plugging the eigenvalue back into the diagonal and just brute force solving the matrix and getting into reduced form...My problem is that every time I plug one of these complex eigenvalues back into my original equation, I get very ugly values at the end.

Could someone please show me the steps you are taking into finding one of these complex eigenvectors?


1 Answer 1


They are indeed not so nice. Here is a set of eigenvectors of $A$, $$ \left[\begin{array}{c} -36\frac{\text{i}\sqrt{143} - 7}{\left(\text{i}\sqrt{143} - 37\right)\left(\text{i}\sqrt{143} +1\right)} \\ -3\frac{\text{i}\sqrt{143} - 7}{\text{i}\sqrt{143} - 37} \\ 1 \end{array}\right] , \left[\begin{array}{c} 36\frac{\text{i}\sqrt{143} + 7}{\left(\text{i}\sqrt{143} + 37\right)\left(\text{i}\sqrt{143} -1\right)} \\ -3\frac{\text{i}\sqrt{143} + 7}{\text{i}\sqrt{143} + 37} \\ 1 \end{array}\right] , \left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] , $$ corresponding to the eigenvalues $\left\lbrace -\frac{3}{4}+\text{i}\frac{\sqrt{143}}{4}, -\frac{3}{4}-\text{i}\frac{\sqrt{143}}{4}, 1 \right\rbrace$, respectively.


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