2
$\begingroup$

Suppose I have a symmetric matrix $A$ and several diagonal matrices $D_1,D_2,\dots$. Are there any matrix transformations such as $P^\top A P$ so that

$$P^\top A P, P^\top D_1 P, P^\top D_2 P, \dots$$

are either all tridiagonal, or all have minimimal bandwidth in some sense?

If, for example, I only had $D_1$ then solving the generalized eigenvalue problem for the matrix pencil $A, D_1$ would give me a basis that simultaneously diagonalizes both $A$ and $D_1$. Obviously, one basis will not simultaneously diagonalize more matrices in general, but a set of banded matrices would also be pretty nice.

I am aware of Tisseur and Garvey's papers on simultaneous tridiagonalization of two matrices.

Improve this question
$\endgroup$
4
  • $\begingroup$ Sorry, I should not have chosen the symbol $P$ in that case. $P$ is not a permutation matrix in my example, and $A$ is assumed to be full. As an example, suppose I have the generalized eigenvalue problems [ A V = V D_1 \Lambda_1 ] [ A U = U D_2 \Lambda_2 ] Now I have a matrice $V$ and $U$ such that the following matrices are diagonal: $V^\top A V$, $U^\top A U$, $V^\top D_1 V$, $U^\top D_2 U$ It seems, in general, impossible to find a single matrix that diagonalized $A$, $D_1$ and $D_2$, however, if it produced a banded matrix, that would still be useful for my purposes. $\endgroup$
    – Greg von Winckel
    Apr 7, 2012 at 13:26
  • $\begingroup$ In general if $A$ is symmetric and $D$ is diagonal, it is not possible to simultaneously diagonalize them because this would imply that they commute. $\endgroup$
    – Chris Godsil
    Apr 7, 2012 at 13:29
  • $\begingroup$ Yes. That is exactly what I said. That is precisely why I was asking if there was a way to reduce them both to banded form. $\endgroup$
    – Greg von Winckel
    Apr 7, 2012 at 20:29
  • $\begingroup$ Additionally, it is possible to simultaneously diagonalize two matrices, but not with a similarity transformation. The matrix of eigenvectors to a symmetric definite matrix pencil $Ax=\lambda B x$ does diagonalize both $A$ and $B$, however, it is not an orthogonal matrix (see my first comment above). $\endgroup$
    – Greg von Winckel
    Apr 7, 2012 at 20:33

1 Answer 1

Reset to default
1
$\begingroup$

If you use the typical reorderings (like reverse Cuthill-McKee ordering), $P^T A P$ will have a smaller bandwidth (in general, not tridiagonal though). Since $P$ is a permutation matrix, all $P^T D_i P$ will remain diagonal too.

Bart

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.