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Consider the matrix $$A = \begin{bmatrix}2&-10\\1&4\end{bmatrix}$$ The eigenvalues of this matrix are complex: $\lambda_{1,2} = 3 \pm 3i$. Now I calculated the eigenvectors to be $ \begin{bmatrix}10\\-1\mp3i\end{bmatrix}$respectively for each eigenvalue. So my first though is: The complex entries are conjugates. My question: Is that always so (I suspect YES, and that would make life a lot easier) and if so, is there a proof? (This is part of my learning process)

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  • $\begingroup$ Why a down vote? Is it too easy? $\endgroup$ – imranfat Aug 10 '16 at 16:02
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If the matrix $A$ is real then the eigenvalues will occur in conjugate pairs.

If $Av = \lambda v$, then $A \bar{v} = \bar{\lambda} \bar{v}$.

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  • $\begingroup$ Yes, I know that the eigenvalues appear in conjugate pairs, but I have trouble in understanding why the entries of the eigenvectors appear in conjugates as well... $\endgroup$ – imranfat Aug 10 '16 at 16:02
  • $\begingroup$ Why is what? Just take conjugates of both sides and note that $A$ is real so $\bar{A} = A$. Then if $(\lambda,v)$ is an eigenvalue, eigenvector pair, so is $(\bar{\lambda}, \bar{v})$. $\endgroup$ – copper.hat Aug 10 '16 at 16:05
  • $\begingroup$ Ok, It makes sense but it is not clicking. I need to think this over...my brain is slow as usual $\endgroup$ – imranfat Aug 10 '16 at 16:07

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