<s>[INST] Calculate the eigenvectors We calculate the eigenvectors for the matrix 
$$
\begin{equation*}
  \mathbf{A} = \left(
    \begin{array}{ccc}
      2 & 0 & 0 \\
      1 & 1 & 1 \\
      1 & -1 & 3 \\
    \end{array}
  \right)
\end{equation*}
$$
First, I calculted the eigenvalue polynomial $\det(\mathbf{A}-\lambda \mathbf{I})=0$, and got triple overlapping eigenvalues $\lambda=2$
$$\begin{equation*}
  \mathbf{A} - \lambda \mathbf{I} = \left(
    \begin{array}{ccc}
      0 & 0 & 0 \\
      1 & -1 & 1 \\
      1 & -1 & 1 \\
    \end{array}
  \right)
\end{equation*}$$
So, I got only two eigenvectors $x_1=\left(
\begin{array}{ccc}
      1 \\
      1 \\
      0 \\
    \end{array} \right)$
and $x_2=\left(
\begin{array}{ccc}
      0 \\
      1 \\
      1 \\
    \end{array} \right)$. So the matrix $\mathbf{A}$ is a degenerate matrix.
Did I calculate it correctly?
 [/INST] Your results are correct, but that does not mean that $A$ is a degenerate matrix, just that $A$ is not diagonalizable.
Just one little thing:
$x_1=\left(
\begin{array}{ccc}
      1 \\
      1 \\
      0 \\
    \end{array} \right)$
and $x_2=\left(
\begin{array}{ccc}
      0 \\
      1 \\
      1 \\
    \end{array} \right)$ are not the only eigenvectors but every vector in the whole subspace $\operatorname{span}\left(\left(
\begin{array}{ccc}
      1 \\
      1 \\
      0 \\
    \end{array} \right),\left(
\begin{array}{ccc}
      0 \\
      1 \\
      1 \\
    \end{array} \right)\right)$
</s>