The problem is how to solve $$AV=V\Lambda, \quad\quad (1)$$ where $A$ is an $n\times n$ matrix, $V$, the "generalized eigen-vector-matrix", is $n\times q$, and, $\Lambda$, the "generalized eigen-value-matrix", is $q\times q$. Given $A$ and $q$ the objective is to find $V$ and $\Lambda$. The requirement to filter out trivial solutions seems to be that $V$ has rank $q$.
One solution is when $\Lambda$ is a diagonal matrix with some of the ordinary eigenvalues of $A$ on the diagonal; and $V$ a corresponding collection of ordinary eigenvectors of $A$. The focus of this question is to find other solutions.
In general $V$ is not square. The special case in which $A$ and $\Lambda$ are symmetric is of particular interest.
Equation (1) can be written in a block-matrix notation. It becomes a $q\times q$ block matrix $B$, where each block, $B_{ij}$, is $n\times n$. The diagonal blocks are $B_{ii}=A-\lambda_{ii}I$ and the off-diagonals are $B_{ij}=-\lambda_{ji} I$. The matrix $B$ is $qn \times qn$, but the blocks commute and hence the determinant simplifies to the formula for the determinant of a $q \times q$ matrix where each entry is replaced by the corresponding block of $B$. For example if $q=2$ then $$\det(B)=\det((A-\lambda_{11}I)(A-\lambda_{22}I)-\lambda_{21}\lambda_{12}I).$$ Setting $\det(B)=0$ produces one equation for the $q^2$ elements of $\Lambda$.