Limit of eigenvectors versus eigenvectors of limit 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?
 A: 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 $-1$.
A: 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.
