Linearly independent eigenvectors The matrix A=$\begin{pmatrix}0 & 1 &1 \\1& 0&1 \\ 1 & 1 & 0\end{pmatrix}$
has eigenvalues $\lambda=2$ with algebriac multiplicity $1$ and $\lambda=-1$ with multiplicity $2$
For $\lambda=-1$
$A+I=0 \implies$$\begin{pmatrix}1 & 1 &1 \\1& 1&1 \\ 1 & 1 & 1\end{pmatrix}\begin{pmatrix}x_1  \\x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix}0  \\0 \\ 0 \end{pmatrix}$
so $x_1+x_2+x_3=0$
$3$ possible cases of eigenvectors are:
$\begin{pmatrix}-2  \\1 \\ 1 \end{pmatrix}$, $\begin{pmatrix}1  \\-2 \\ 1 \end{pmatrix}$, $\begin{pmatrix}1 \\-1 \\ 0 \end{pmatrix}$
which are all linearly independent. 
So we can have more than $2$ distinct (non-linearly dependent) eigenvectors, even if the algebraic multiplicity is only $2$?

Compute
$$\begin{pmatrix}-2  \\1 \\ 1 \end{pmatrix} - \begin{pmatrix}1  \\-2 \\ 1 \end{pmatrix} + 3\begin{pmatrix}1 \\-1 \\ 0 \end{pmatrix}$$
