How can I find out the eigenvectors for this matrix:

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

I found the eigenvalues: $\lambda_{1}=-3$, $\lambda_{2}=2-i$, $\lambda_{3}=2+i$. The eigenvector for $\lambda_{1}=-3$ is $$\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}.$$ For $\lambda_{3}=i+2$ I write $$A(u+iv)=(i+2)(u+iw),$$ but I don't know how to do$\ldots$ Thanks!


I do not recommend separating the eigenvector into real and imaginary parts. In my opinion it only makes it more troublesome, complex numbers are perfectly good scalars to work with. Note the following $$\begin{align} Av=\lambda _2v&\iff (A-\lambda _2I)v=0_{3\times 1}\\ &\iff \begin{pmatrix} -5+i & 0 & 0\\ 0 & 1+i & -2\\ 0 & 1 & -1+i\end{pmatrix}\begin{pmatrix} v_1\\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}0\\0\\0 \end{pmatrix}\\ &\iff \begin{cases}(-5+i)v_1&=0\\ (1+i)v_2-2v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\ &\iff \begin{cases}v_1&=0\\ 2v_2-2(1-i)v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\ &\iff \begin{cases}v_1&=0\\ 2v_2+2(-1+i)v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\ &\iff \begin{cases}v_1&=0\\ 2v_2-2(1-i)v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\ &\iff \begin{cases}v_1&=0\\ v_2+(-1+i)v_3&=0\end{cases}\\ &\iff (v_1, v_2, v_3)=(0,(1-i)v_3, v_3). \end{align}$$

You can thus take the eigenpair $\left(2-i, \begin{pmatrix} 0\\ 1-i\\ 1\end{pmatrix}\right)$.

It is very important to note that given a square real matrix $M$ and $(\lambda, u)$ one of its eigenpairs, the following holds: $$Mu=\lambda u\iff \overline{Mu}=\overline{\lambda u}\iff M\overline u=\overline \lambda \overline u.$$

What does this tell you?

  • $\begingroup$ Cohen had said, 'There's men here who can push a wheelbarrow for thirty miles on a bowl of millet with a bit of scum in it. What does that tell you? It tells me someone's porking all the beef.' $\endgroup$ – Will Jagy May 26 '14 at 20:08
  • $\begingroup$ @WillJagy I hoping 'porking' doesn't mean what the link says it means. $\endgroup$ – Git Gud May 26 '14 at 20:14
  • $\begingroup$ In the U.S. we would say "hogging" for this sentiment, meaning that the rich eat beef and the poor eat cheap food. I got this from a website. I do have a copy of "Interesting Times," maybe I will be able to find the genuine quote. Cohen the Barbarian is often given to these blunt and cynical observations. $\endgroup$ – Will Jagy May 26 '14 at 20:17
  • $\begingroup$ strange discussion... $\endgroup$ – Iuli May 26 '14 at 20:17
  • 1
    $\begingroup$ @Iuli Welcome to Will's domain xD $\endgroup$ – Git Gud May 26 '14 at 20:23

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