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?

  • $\begingroup$ Do you really want to migrate this question? It is a numerical linear algebra problem, which is on-topic. $\endgroup$ Nov 13, 2013 at 5:42
  • $\begingroup$ I'm not sure; I think the reason I couldn't find a solution right now is because I don't understand the nature behind the problem, and I thought I should understand that first before determining the best way to solve it. $\endgroup$
    – syockit
    Nov 13, 2013 at 8:24
  • $\begingroup$ Just out of curiousity - I am assuming that the tensors A_ij are not diagonal? $\endgroup$
    – OscarB
    Nov 13, 2013 at 10:25
  • $\begingroup$ @OscarB Yes, they're not. $\endgroup$
    – syockit
    Nov 13, 2013 at 15:40
  • $\begingroup$ Another thing, since I work on multipole in relation to Helmholtz - sometimes, the region of the solution is known. This means that you could convert to spherical coordinates, and avoid the r-component since v_r might be known a priori - this would result in only 2n unknowns instead of 3n. Is this possible in your case? $\endgroup$
    – OscarB
    Nov 14, 2013 at 10:50

1 Answer 1


Either way, the number of generalized eigenvectors of your system should be $3n$; the eigenvectors of your linear system may not form a complete basis. Either way you pose the problem (expanding the matrix of tensors, or not expanding the matrix of tensors), you should have the same number of eigenvalues. Normalizing the eigenvectors (which is what your constraint does, in essence) does not change the number of eigenvalues of a matrix. (You can show that if $v$ is an eigenvector of matrix $A$, then $v / \|v\|$ is also an eigenvector of matrix $A$; by definition, $v \neq 0$.)


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