0

This is a question that I've been mulling on for a bit. Given some linear combination of matrices:

\begin{equation*} w_{0}A_{0} + w_{1}A_{1} + \ldots + w_{n}A_{n} \end{equation*}

Assuming that all of these matrices are invertible and known (i.e., each $A_{i}^{-1}$ is known beforehand for all $i$.), and additionally assuming that the resulting matrix inverse actually exist (e.g., under floating point arithmetic), is there some way to efficiently compute the inverse of this matrix?

Or in other words, can I somehow compute the matrix:

\begin{equation*} (w_{0}A_{0} + w_{1}A_{1} + \ldots + w_{n}A_{n})^{-1} \end{equation*}

Using the matrices $A_{i}^{-1}$? Alternatively, is there a way to approximate this matrix under the same assumptions?

In my case, these matrices are all 4x4 affine transformation matrices (i.e., some combination of a scaling, rotation and translation transform) which may simplify things in practice, but I'm also curious if this is possible for general matrices.

I think this is related to this question Inverse of the sum of matrices, however, it is not exactly clear to me how this extends to multiple (weighted) matrices, if it actually does so.

Xaldew
  • 103
  • I'm afraid that nothing can be said for more than two matrices. 2) The fact it is weighted sum is unimportant because you can incorporate the weights into the matrices.
  • – Jean Marie Feb 20 '23 at 12:36
  • I don't see why the linked question doesn't answer your question immediately. You just iterate the formula: $(A_1+A_2+A_3+\dots+A_k)^{-1} = (A+B)^{-1} = C_r^{-1}-g_rC_r^{-1}B_rC_r^{-1}$, where $A=A_1$ and $B=A_2+A_3+\dots+A_k$. In terms of efficiency, however, I do not know. – SomeCallMeTim Feb 20 '23 at 12:42
  • Just adding this here since I made an edit: The weights are not known "before-hand", so only the non-weighted matrix inverses would be known to make the computations efficient (if it would be possible) – Xaldew Feb 20 '23 at 13:31
  • @SomeCallMeTim That seems feasible at least, but yeah, probably a bit inefficient. Still, I didn't think of such a simple re-arrangement, I'll see if that could be workable – Xaldew Feb 20 '23 at 13:49
  • @Xaldew Just a small comment: even though the weights are not given before-hand, they are still no problem. $(wA)^{-1} = w^{-1}A^{-1}$ (when the inverse exists). – SomeCallMeTim Feb 20 '23 at 18:22