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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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    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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As anon points you out, try to use $\overline{z\cdot w} = \overline{z}\cdot \overline{w}$ and simplify notation.

For instance, you could just write vectors in $\mathbb{C}^2$ like this: $(z_1, z_2)$, with $z_1, z_2 \in \mathbb{C}$.

Second, I would try to convince myself that that product/conjugacy rule works too for products of matrices and vectors. That is:

$ \overline{A\cdot v} = \overline{A}\cdot \overline{v} \ , $

for, say, $A$ a $2\times 2$ complex matrix and $v\in \mathbb{C}^2$. What would happen if $A$ had real coefficitients?

Finally, I would write the equality I already know, namely

$ A \cdot Y_0 = \lambda Y_0 \ . $

Staring at it should force inspiration to come. :-)