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**jakncoke** So you solved the characteristic equation to get $\displaystyle \lambda = 3 $ and $\displaystyle \lambda = 1 $. I got the same eigen values.

Now to find the eigen vectors

$\displaystyle \begin{bmatrix}2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \lambda \begin{bmatrix} x \\ y \end{bmatrix} $ in which you get to solve a homogenous equation $\displaystyle \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $. Which yields infinetly many solutions of the form $\displaystyle \begin{bmatrix} -p \\ p \end{bmatrix}$ So one eigen vector for the eigen value $\displaystyle \lambda = 3 $ would be p = 1, $\displaystyle \begin{bmatrix} -1 \\ 1 \end{bmatrix} $

For $\displaystyle \lambda = 1 $ i got eigen vectors using the same processing above, infinitely many solutions of the form $\displaystyle \begin{bmatrix} p \\ p \end{bmatrix} $. So p = 1, $\displaystyle \begin{bmatrix} 1 \\ 1 \end{bmatrix} $ is a eigen vector for eigen value $\displaystyle \lamba = 1 $