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I have this matrix

\begin{pmatrix} -3 & 1 \\ -2 & -1 \end{pmatrix}

and I found the eigenvalues, which are $\lambda_{1} = -2+i \ $ and $\ \lambda_{2}= -2-i$,

and now I have to find the eigenvectors, but they are with complex values and I don't know the steps for that.

Edit: So for the first eigenvalue $\lambda_{1}$= -2 + i, I tried:

\begin{pmatrix} -3-(-2+i) & 1 \\ -2 & -1-(-2+i) \end{pmatrix}

and then I've got:

(-1-i)x + y = 0
 -2x + (1-i)y = 0

What should I do next?

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    $\begingroup$ Welcome to Mathematics Stack Exchange. How would you find the eigenvectors if the eigenvalues were real? $\endgroup$ – J. W. Tanner Jun 23 at 10:33
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    $\begingroup$ just do the same operations you would do for real eigenvalues, remember you can multiply, add, substract and divide complex numbers like you would do with real numbers. so the algorithm is basically the same $\endgroup$ – mathma Jun 23 at 10:33
  • $\begingroup$ Next notice your two lines are compatible, line2 = line1 $\times(1-i)$. So you have two variables and one equation. Means you can choose $x=1$ for instance and get eigenvector $(1,1+i)$. $\endgroup$ – zwim Jun 23 at 14:15
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Here's a solution of an analogous problem as a starting point for you:

Here's a matrix with eigenvalues $\pm i$: $$ A = \pmatrix{0 & 1 \\ -1 & 0} $$ I'll find the eigenvector for $+i$. I want to find a vector $v$ with $$ Av = i ~v, $$ right? I convert that to $$ Av = iI ~v\\ (A-iI)v = 0 $$ So now I want to find the kernel of $$ \pmatrix{-i & 1 \\ -1 & -i}, $$ so I need to solve $$ \pmatrix{-i & 1 \\ -1 & -i} \cdot \pmatrix{x \\ y} = \pmatrix{0 \\ 0} $$ which becomes $$ -ix + y = 0 \\ -x - iy = 0 $$ we see that the second equation is just $-i$ times the first one, so we can simply solve the first equation. Picking $x = 1$ (because it's easy!), we get $y = i$, so one eigenvector is $$ \pmatrix{1\\i}. $$ Any (complex) nonzero multiple of this is also an eigenvector, so $$ \pmatrix{i\\-1}, \pmatrix{2\\2i}, \pmatrix{1+i\\-1+i} $$ are all eigenvectors as well. (If we'd chosen $x = i, x = 2, x = 1 + i$, they're what we'd have found when we solved for $y$.)

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  • $\begingroup$ What is the need for the $(A-iI)$ transform ? You can just go for $\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=i\begin{pmatrix}x\\y\end{pmatrix}$ and get $\begin{cases}y=ix\\-x=iy\end{cases}$ all the same. $\endgroup$ – zwim Jun 23 at 10:49
  • $\begingroup$ If, this matrix were represented by some linear transformation on real space, then how could you say that eigenvectors can exist!! $\endgroup$ – A learner Jun 23 at 10:49
  • $\begingroup$ Even eigenvalue doesn't exist!!! $\endgroup$ – A learner Jun 23 at 10:51
  • $\begingroup$ So I tried to find the eigenvector with your solution, but then I've got lost again, can you please help me? $\endgroup$ – Martina Jun 23 at 11:00
  • $\begingroup$ @Subhajit The matrix $A$ represents a rotation in $\mathbb{R}^2$ by $90^°$ clockwise, and thus has no real eigenvalues, anyway we are talking about complex eigenvalues $\endgroup$ – Peter Melech Jun 23 at 11:00

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