= Solution
The given vectors satisfy
$$
y_{k+2}=2y_{k+1}-2y_k,
$$
and $y_k,y_{k+1}$ are <independent>, so their span is invariant under $A$. On that plane the <characteristic polynomial> is $\lambda^2-2\lambda+2$. Therefore
$$
\boxed{\lambda_\pm=1\pm i,\qquad
v_\pm=y_{k+1}-(1\mp i)y_k=(1\pm i,2\pm i,3\pm i)^T.}
$$
Indeed $Av_\pm=y_{k+2}-(1\mp i)y_{k+1}=(1\pm i)v_\pm$. No third <eigenvalue> is determined by these data.
The displayed $x_k=(1,1,1)$ has norm $\sqrt3$, although a large-$k$ normalized iterate should have norm one. This harmless source normalization is resolved by dividing all three displayed vectors by $\sqrt3$; the recurrence, <eigenvalues> and <eigenvector> directions are unchanged.
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