I'm trying to solve a multipole system. It involves a matrix of 3x3 tensors $A_{ij}$ and a vector of 3-tuples $\mathbf v_i$.
$$\left(\begin{matrix} A_{11} & A_{12} & \cdots & A_{1n}\\ A_{21} & A_{22} & \cdots & A_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ A_{n1} & A_{n2} & \cdots & A_{nn} \end{matrix}\right) \left(\begin{matrix} \mathbf v_1 \\ \mathbf v_2 \\ \vdots \\ \mathbf v_n \end{matrix}\right) = \lambda \left(\begin{matrix} \mathbf v_1 \\ \mathbf v_2 \\ \vdots \\ \mathbf v_n \end{matrix}\right) $$ It doesn't seem that I can trivially solve it using available linear algebra packages.
My idea is that I expand the matrix of tensors into a $3n\times3n$ matrix, and the vector of vectors into a $3n$ vector, $$\mathbf v_i = \left(\begin{matrix}v_{ix}\\v_{iy}\\v_{iz}\end{matrix}\right)$$ But it seems like I'd get up to a maximum of $3n$ eigenvalues instead of just $n$ if I do it this way. Is there a better way?
Edit1: since $v_{ix}$, $v_{iy}$, and $v_{iz}$ are components of a vector, they should fulfill the following constraint: $$ v_{ix}^2 + v_{iy}^2 + v_{iz}^2 = c_i^2 $$ So I suppose this reduces the number of eigenvalues from $3n$ to $2n$, right? But is this even a linear algebra problem anymore?