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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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3  
It is going to help. –  Mariano Suárez-Alvarez Jul 25 '11 at 5:14
    
Just do it ! :D –  Olivier Bégassat Jul 25 '11 at 5:18
    
Try typing out the algebra you did in your question, and we'll see where you went wrong... –  J. M. Jul 25 '11 at 5:22
7  
@mary: no problem is trivial until it becomes trivial for you. –  Mariano Suárez-Alvarez Jul 25 '11 at 5:44
2  
@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. –  Jyrki Lahtonen Jul 25 '11 at 6:02

1 Answer 1

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. :-)

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