Find eigenvalues and eigenvectors of  the matrix $\begin{pmatrix} 1 & 6 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix} $ 

(i) To find the eigenvalues we should set
\[\text{det}(A-\lambda I)=0.\]
Hence
\[\begin{vmatrix} 1-\lambda & 6 & 0 \\ 0 & 2-\lambda & 1 \\ 0 & 1 & 2-\lambda \end{vmatrix}=(1-\lambda) \begin{vmatrix} 2-\lambda & 1 \\ 1& 2-\lambda \end{vmatrix} \]
\[=(1-\lambda)((2-\lambda)(2-\lambda)-1)=(1-\lambda)(\lambda^2-4\lambda+3)\]
\[=(1-\lambda)(1-\lambda)(\lambda-3)=0\]
\[\Rightarrow \lambda_1=\lambda_2=1, \lambda_3=3.\]
(ii) Next we find the eigenvectors for $\lambda_1=\lambda_2=1$. 
\[(A-I)v_1=0      \Rightarrow\]
\[\begin{pmatrix} 0 & 6 & 0 & 0 \\ 0 & 1 & 1 &0 \\ 0 & 1 & 1 & 0 \end{pmatrix}     \]
$-R_2+R_3 \rightarrow R_3$ gives 
\[\begin{pmatrix} 0 & 6 & 0 & 0 \\ 0 & 1 & 1 &0 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \]
Hence 
\[6y=0    \text{and}   y+z=0  \Rightarrow y=z=0.\]
$x$ is a free variable, and we can take $x=1$. So there is only one linearly independent eigenvector 
\[v= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}. \]

(iii) We can similarly compute the eigenvalues for $\lambda_3=3.$. \[(A-I)v_3=0 \Rightarrow\] \[\begin{pmatrix} -2 & 6 & 0 & 0 \\ 0 & -1 & 1 &0 \\ 0 & 1 & -1 & 0 \end{pmatrix} \] $R_2+R_3 \rightarrow R_3$ gives \[\begin{pmatrix} -2 & 6 & 0 & 0 \\ 0 & -1 & 1 &0 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \] Hence \[x=3y    \text{and}    y=z \Rightarrow y=z=0.\] $z$ is a free variable, and we can take $z=1$. So there is only one linearly independent eigenvector \[v_3= \begin{pmatrix} 3\\ 1 \\ 1\end{pmatrix}. \]