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Consider the matrix $ A=\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & g \end{pmatrix}, $ where $g$ is a real parameter. If I set $g=0$ and calculate the normalized eigenvectors of $A|_{g=0}$ with Mathematica, I find that they are $ v_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\ v_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix},\ v_3 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}. $

If instead I calculate the eigenvectors of $A$ leaving $g$ as an unknown and then take their limit as $g\to 0$, I find $ u_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\ u_2 = \frac{1}{\sqrt{6}}\begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix},\ u_3 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}. $

My question is, why are these two sets of eigenvectors different?

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    In physics, eigenvectors like $v_1$ and $v_2$ resp. $u_1$ and $u_2$ are called _degenerate_ (perhaps it's the same in mathematics), because they correspondent to the same eigenvalue and therefore span an entire 2-dimensional subspace in which _all_ vectors are eigenvectors to the eigenvalue $-1$.2011-05-25

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Both results are correct. $u_1$ and $u_2$ correspond to the same eigenvalue $-1$, and $\left( \matrix{-1\cr 0\cr 1\cr} \right) = \frac{1}{2} \left(\matrix{-1 \cr 1 \cr 0\cr} \right) + \frac{1}{2} \left(\matrix{-1 \cr -1 \cr 2\cr} \right)$, so $u_1$ and both versions of $u_2$ span the same vector space. Any nonzero vector in this space is an eigenvector for eigenvalue $-1$.

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    Aah, I knew it had to be something simple like that, but when I couldn't rotate one basis onto the other to explain it, I forgot to think about other legitimate ways of comparing bases :p ... Thanks!2011-05-25
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Both $(-1,1,0)^T$ and $(-1,0,1)^T$ are eigenvectors of $-1$; so is $(-1,-1,2)^T$, as $(-1,-1,2) = -(-1,1,0) + 2(-1,0,1).$ You are just taking a different basis for the eigenspace corresponding to $-1$.

It's likely just an artifact of how Mathematica finds a basis for the eigenspace; the eigenvalues of the matrix are $-1$, $\frac{1+g+\sqrt{g^2-2g+9}}{2},$ and $\frac{1+g-\sqrt{g^2-2g+9}}{2}$ so that there is, up to sign, only one normal eigenvector for each eigenvalue when $g\neq 0$ (note the quadratic in the square root is always positive, so those two eigenvalues never coincide, and neither is equal to $-1$ unless $g=0$). But at the limit you end up with a matrix that has a repeated eigenvalue (corresponding to $\lambda=-1$) and in that case you have many different ways of obtaining a basis.