Find an eigenvector of A=[2112]A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}A=[2112] corresponding to eigenvalue λ=3\lambda = 3λ=3.
[1−1]\begin{bmatrix} 1 \\ -1 \end{bmatrix}[1−1]
[11]\begin{bmatrix} 1 \\ 1 \end{bmatrix}[11]
[21]\begin{bmatrix} 2 \\ 1 \end{bmatrix}[21]
[−11]\begin{bmatrix} -1 \\ 1 \end{bmatrix}[−11]