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?

You can check the correctness of your eigenvectors if you apply the matrix on them.
By the way: Since there is only one not-zero entry (2) in the first row of the matrix and this is the first entry, this is the only candidate for an eigenwert.
