Determine the eigenvalues and corresponding eigenvectors of $$ A= \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ Can you find a basis of $\mathbb{R}^3$ consisting of eigenvectors of $A$?
The characteristic polynomial is
$ \lambda^2 (1-\lambda) -1 + \lambda = -\lambda^3 + \lambda^2 + \lambda -1 $
The roots are
$\lambda_{1,2} = 1 $ and $\lambda_3 = -1$
now calculating
$ (A - \lambda_{i}E_3) \cdot (x,y,z)^T = 0$
results in
$\begin{pmatrix} t\\t\\z \end{pmatrix}$ with $t,z \in \mathbb{R}$ for $\lambda_{1,2}$
and
$\begin{pmatrix} t\\-t\\0 \end{pmatrix}$ with $t \in \mathbb{R}$ for $\lambda_{3}$
therefore a possible basis would be
$ b_1 = \begin{pmatrix} 1\\-1\\0 \end{pmatrix}$ $ b_2 = \begin{pmatrix} 1\\1\\0 \end{pmatrix}$ $ b_3 = \begin{pmatrix} 0\\0\\1 \end{pmatrix}$
Is this correct?