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I am a little apprehensive to ask this question because I have a feeling it's a "duh" question but I guess that's the beauty of sites like this (anonymity):

I need to find an orthonormal eigenbasis for the $2 \times 2$ matrix $\left(\begin{array}{cc}1&1\\ 1&1\end{array}\right)$. I calculated that the eigenvalues were $x=0$ and $x=2$ and the corresponding eigenvectors were $E(0) = \mathrm{span}\left(\begin{array}{r}-1\\1\end{array}\right)$ and $E(2) = \mathrm{span}\left(\begin{array}{c}1\\1\end{array}\right)$. Therefore, an orthonormal eigenbasis would be: $$\frac{1}{\sqrt{2}}\left(\begin{array}{r}-1\\1\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right).$$

Here my question: Could the eigenvalues for $E(0)$ been $\mathrm{span}\left(\begin{array}{r}1\\-1\end{array}\right)$?? This would make the final answer $\frac{1}{\sqrt{2}}\left(\begin{array}{r}1\\-1\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right)$. Is one answer more correct than the other (or are they both wrong)?

Thanks!

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    I don't quite understand the question as written, but if it's what I think it is, then yes, orthonormal eigenbases are not unique; you can multiply any eigenvector by a complex number of absolute value 1.2011-01-06
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    Or by a real number or whatever the field is that you are working over. One more thing: professional mathematicians have "duh" moments all the time. It's part of the job description, so to speak. There is absolutely no need to hide behind anonymity. On the contrary, it is important for a mathematician to learn to live with those "duh" moments. Otherwise he will never be able to freely talk with his colleagues, which would greatly hinder his progress.2011-01-06
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    @Alex: the absolute value 1 condition is important for orthonormality, which doesn't make sense over an arbitrary field.2011-01-06
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    @Qiaochu You are right. Read that as "real or complex, depending on..."2011-01-06

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