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

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 $v_3= \begin{pmatrix} 3\\ 1 \\ 1\end{pmatrix}.$

Erdos
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