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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. –  Qiaochu Yuan Jan 6 '11 at 13:28
2  
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. –  Alex B. Jan 6 '11 at 13:32
    
@Alex: the absolute value 1 condition is important for orthonormality, which doesn't make sense over an arbitrary field. –  Qiaochu Yuan Jan 6 '11 at 13:55
    
@Qiaochu You are right. Read that as "real or complex, depending on..." –  Alex B. Jan 6 '11 at 14:38

2 Answers 2

0 and 2 are the correct eigenvalues to your matrix; (1, -1) is one eigenvector, (1, 1) the other. Your solution is correct.

Span is the set of all linear combinations, so if you consider a vector space over $\mathbb{R}$, it absolutely doesn't matter what scalar in $\mathbb{R}$ you multiply your vectors with inside Span. This does not affect the set Span at all. So the solution is the same.

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There is no such thing as the eigenvector of a matrix, or the orthonormal basis of eigenvectors. There are usually many choices.

Remember that an eigenvector $\mathbf{v}$ of eigenvalue $\lambda$ is a nonzero vector $\mathbf{v}$ such that $T\mathbf{v}=\lambda\mathbf{v}$. That means that if you take any nonzero multiple of $\mathbf{v}$, say $\alpha\mathbf{v}$, then we will have $$T(\alpha\mathbf{v}) = \alpha T\mathbf{v} = \alpha(\lambda\mathbf{v}) = \alpha\lambda\mathbf{v}=\lambda(\alpha\mathbf{v}),$$ so $\alpha\mathbf{v}$ is also an eigenvector corresponding to $\lambda$. More generally, if $\mathbf{v}_1,\ldots,\mathbf{v}_k$ are all eigenvectors of $\lambda$, then any nonzero linear combination $\alpha_1\mathbf{v}_1+\cdots+\alpha_k\mathbf{v}_k\neq \mathbf{0}$ is also an eigenvector corresponding to $\lambda$.

So, of course, since $\left(\begin{array}{r}-1\\1\end{array}\right)$ is an eigenvector (corresponding to $x=0$), then so is $\alpha\left(\begin{array}{r}-1\\1\end{array}\right)$ for any $\alpha\neq 0$, in particular, for $\alpha=-1$ as you take.

Now, a set of vectors $\mathbf{w}_1,\ldots,\mathbf{w}_k$ is orthogonal if and only if $\langle \mathbf{w}_i,\mathbf{w}_j\rangle = 0$ if $i\neq j$. If you have an orthogonal set, and you replace, say, $\mathbf{w}_i$ by $\alpha\mathbf{w}_i$ with $\alpha$ any scalar, then the result is still an orthogonal set: because $\langle\mathbf{w}_k,\mathbf{w}_j\rangle=0$ if $k\neq j$ and neither is equal to $i$, and for $j\neq i$, we have $$\langle \alpha\mathbf{w}_i,\mathbf{w}_j\rangle = \alpha\langle\mathbf{w}_i,\mathbf{w}_j\rangle = \alpha 0 = 0$$ by the properties of the inner product. As a consequence, if you take an orthogonal set, and you take any scalars $\alpha_1,\ldots,\alpha_k$, then $\alpha_1\mathbf{w}_1,\ldots,\alpha_k\mathbf{w}_k$ is also an orthogonal set.

A vector $\mathbf{n}$ is normal if $||\mathbf{n}||=1$. If $\alpha$ is any scalar, then $||\alpha\mathbf{n}|| = |\alpha|\,||\mathbf{n}|| = |\alpha|$. So if you multiply any normal vector $\mathbf{n}$ by a scalar $\alpha$ of absolute value $1$ (or of complex norm $1$), then the vector $\alpha\mathbf{n}$ is also a normal vector.

A set of vectors is orthonormal if it is both orthogonal, and every vector is normal. By the above, if you have a set of orthonormal vectors, and you multiply each vector by a scalar of absolute value $1$, then the resulting set is also orthonormal.

In summary: you have an orthonormal set of two eigenvectors. You multiply one of them by $-1$; this does not affect the fact that the two are eigenvectors. The set was orthogonal, so multiplying one of them by a scalar does not affect the fact that the set is orthogonal. And the vectors were normal, and you multiplied one by a scalar of absolute value $1$, so the resulting vectors are still normal. So you still have an orthonormal set of two eigenvectors. I leave it to you to verify that if you have a linearly independent set, and you multiply each vector by a nonzero scalar, the result is still linearly independent.

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