Answer to Question #223169 in Linear Algebra for sunday

Question

Question1

 

Given   matrix A =        2      2     1
                          1      3     1
                          1      2     2

 
 

1.1 find the eigenvalues of   A

 
1.2.   Determine an eigenvector for the eigenvalue  =  5

Answer 1

$A=\begin{bmatrix} 2 & 2&1 \\ 1 & 3&1\\ 1&2&2 \end{bmatrix}$

1.1. $\det (A-\lambda I)=\begin{vmatrix} 2-\lambda & 2&1 \\ 1 & 3-\lambda &1\\ 1&2&2-\lambda \end{vmatrix}=-\lambda^3+7\lambda^2-11\lambda+5=-(\lambda-5)(\lambda-1)^2=0$

Eigenvalues: $\lambda _1=5$ and $\lambda_2=1$

1.2.

$\lambda_1=5$ $: \ \ (A-5I)\mathbf{x}= \begin{bmatrix} -3 & 2&1 \\ 1 & -2&1\\ 1&2&-3 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 0\\0\\0 \end{bmatrix}$ gives the eigenvector $\mathbf{x}_1=\begin{bmatrix}1\\1\\1\end{bmatrix}$ .

Answer: 1.1. $\lambda_1=5$ and $\lambda_2=1$ ; 1.2. $\mathbf{x}_1=\begin{bmatrix}1\\1\\1\end{bmatrix}$