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Suppose the matrix A with real entries has complex eigenvalues $\lambda = \alpha + i\beta\,$ and $\overline{\lambda} = \alpha - i\beta\,$. Suppose that $Y_0 = (x_1 + iy_1, x_2 + iy_2)$ is an eigenvector for the eigenvalue $\lambda$. Show that $\overline{Y_0} = (x_1 - iy_1, x_2 - iy_2)$ is an eigenvector for the eigenvalue $\overline{\lambda}$. In other words, the complex conjugate of an eigenvector for $\lambda$ is an eigenvector for $\overline{\lambda}$.

This is what I have:

$Av = \lambda v$ but lost with notations. Can somebody explain how to do this?

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    It *is* going to help.2011-07-25
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    Just do it ! :D2011-07-25
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    Try typing out the algebra you did in your question, and we'll see where you went wrong...2011-07-25
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    A(x0, y0) = (x1 + iY1)(x0, y0) is the start2011-07-25
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    ...and what happens if you conjugate both sides?2011-07-25
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    we get A bar (xo, yo) = (x1 - iy1)(xo,yo) ?2011-07-25
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    You did remember to conjugate both the eigenvalue and eigenvector?2011-07-25
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    Yes; is this problem supposed to be trivial?2011-07-25
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    @mary: no problem is trivial until it becomes trivial *for you*.2011-07-25
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    I am following JM's hints, but not sure how to proceed next2011-07-25
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    @mary: You will get there, but you seem to have forgotten $\lambda$ from your equation, and you also forgot to use the same notation for the vectors $Y_0$ and $\overline{Y_0}$. In other words you could start by writing out the eigenvalue equation $AY_0=\lambda Y_0$. Meanwhile, I TeXified your question. If it is not ok, you can roll it back.2011-07-25
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    Perhaps a lot of unnecessary writing and typing could be avoided if we simply used $\overline{z\cdot w}=\overline{z}\cdot\overline{w}$...2011-07-25

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