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Find the general solution:

$$x'= \pmatrix{-2&-1\\2&0}x$$

I got the eigenvalues to be $\lambda = -1 \pm i$.

Now finding the eigenvectors:

$(\lambda = -1 + i )= v_1 =\pmatrix{-2 -(-1 + i) &-1\\2&-(-1 + i)} =\pmatrix{2 &1-i\\2&1 -i}= \pmatrix{2 &1-i\\0&0} $

Thus the eigenvector is $v_1 = \pmatrix{1\\ -\frac{2}{1-i}}$

but the eigenvector in the solution set is $v_1 = \pmatrix{1\\ -1-i}$

how did they get that?

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$$ -\frac{2}{1-i} = -\frac{2}{1-i} \left(\frac{1+i}{1+i}\right) = - 1 - i $$

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  • $\begingroup$ But then wouldn't the 1 in row 1 column 1 be (1+ i) then ? $\endgroup$ – Q.matin Dec 6 '12 at 23:24
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    $\begingroup$ @Q.matin Eric isn't saying multiply by $(1+i)$, he is pointing out that $-2/(1-i)$ is equal to $-1-i$. $\endgroup$ – Matt Dec 6 '12 at 23:35
  • $\begingroup$ @Matt damn you complex numbers. Thank you! $\endgroup$ – Q.matin Dec 6 '12 at 23:38

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