Consider 3x3 matrix with unknowns $v_0$,$v_1$,$v_2$:
$$A \vec{v} =A\left[\begin{array}{l}v_0\\v_1\\v_2\end{array}\right] = \left[\begin{array}{l}v_0 v_0\\v_0 v_1\\v_1v_2\end{array}\right]$$
As you can see, this is almost eigenvalue problem $A \vec{v} = v_0 \vec{v}$, except last equation. Is it possible to find some matrix $A'$, so that $A' \vec{u} = v_0 \vec{u}$, i.e. convert initial problem to the eigenvalue problem? I am of course interested in generalized solition, which could be somehow applied to n*n matrix, where few last lines are $v_i v_j$ instead of $v_0 v_j$...