$A = \begin{bmatrix}0&1&0\\0&0&1\\-2&-4&-3 \end{bmatrix}$

The eigenvalues are: $\lambda =-1,\ \lambda = 1+i\ $ and $\ \lambda =1-i$

but I'm having some trouble determining the eigenvectors.

For $\lambda =1$ we get: $[A-\lambda I]=\begin{bmatrix}1&1&0\\0&1&1\\-2&-4&-2\end{bmatrix}\underline{x}=0$

The corresponding eigenvector is $\begin{bmatrix}1\\-1\\1 \end{bmatrix}$

For $\lambda=1+i$ we get: $[A-\lambda I]=\begin{bmatrix}-1-i&1&0\\0&-1-i&1\\-2&-4&-4-i\end{bmatrix}\underline{x}=0$

I am not able to correctly row reduce this matrix in order to find the eigenvector. The same geos for the third eigenvector. But if someone can show me how to do this one then I can figure out the last by myself.

  • $\begingroup$ Basically the same procedure. $\endgroup$ – xbh Sep 10 '18 at 8:38
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    $\begingroup$ You are wrong about the eigenvalues of $A$. They are $-1$ and $-1\pm i$. $\endgroup$ – José Carlos Santos Sep 10 '18 at 8:43
  • $\begingroup$ Thanks for pointing it out. $\endgroup$ – user463102 Sep 10 '18 at 9:52
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    $\begingroup$ You only need to do this for one of the complex eigenvalues. You can find an eigenvector of the other complex eigenvalue “for free:” For a real matrix, complex eigenvalues come in conjugate pairs, and for such an eigenvalue $\lambda$, you have $Av=\lambda v \implies A\overline v = \overline\lambda \overline v$. $\endgroup$ – amd Sep 10 '18 at 23:47
  • $\begingroup$ So the second complex eigenvector is just the complex conjugate of the first complex eigenvector?? $\endgroup$ – user463102 Sep 11 '18 at 11:12

Assuming that you want to row reduce \begin{align} \begin{bmatrix}-1-i&1&0\\0&-1-i&1\\-2&-4&-4-i\end{bmatrix} \end{align} you can multiply the first row by $$ \frac{2}{-1-i} = \frac{2}{-1-i} \cdot \frac{-1+i}{-1+i} = -1+i $$ to get \begin{align} \begin{bmatrix}2&-1+i&0\\0&-1-i&1\\-2&-4&-4-i\end{bmatrix}. \end{align} Now add it to the third row and proceed. Can you continue on your own?

  • $\begingroup$ Yes, thank you. $\endgroup$ – user463102 Sep 10 '18 at 10:00

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