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Let $A$ be a $2\times 2$ real matrix without eigenvalues, and the roots of its characteristic polynomial be $\alpha+i \beta$ and $\alpha - i \beta$. Show that there exists a basis of $\mathbb{R^2}$ such that $A=\begin{pmatrix} \alpha & \beta \\ - \beta & \alpha \end{pmatrix}$.

I've tried to use the fact that those are its eigenvalues, when considering it a complex matrix, and try to construct a basis to $\mathbb{R^2}$ from the eigenvectors somehow, but I didn't achieve anything.

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    $\begingroup$ You're close. How are the (complex) eigenvectors for the two eigenvalues related? $\endgroup$ – Daniel Fischer Jul 4 '13 at 17:46
  • $\begingroup$ How is this true? $$\begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$$ is a counterexample? $\endgroup$ – Inquest Jul 4 '13 at 17:48
  • $\begingroup$ @Inquest $\alpha = \beta = 0 = -\beta$. However, for $\beta = 0$, $$\begin{pmatrix}\alpha & 1\\0 & \alpha \end{pmatrix}$$ would be a counterexample. $\endgroup$ – Daniel Fischer Jul 4 '13 at 17:50
  • $\begingroup$ As Daniel points out, you need the assumption that $\beta\ne 0$. $\endgroup$ – Cameron Buie Jul 4 '13 at 17:51
  • $\begingroup$ I forgot an hypothesis, sorry... editing $\endgroup$ – Aloizio Macedo Jul 4 '13 at 17:52
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Hint: Suppose that $x$ is a complex eigenvector of $A$ corresponding to the eigenvalue $\alpha+i\beta$ and note that $$A\overline x=\overline{Ax}=\overline{(\alpha+i\beta)x}=(\alpha-i\beta)\overline x.$$

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