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Nov 16, 2014 at 21:44 comment added Reimundo Heluani Adding to Alex Degtyarev's comment: for any bounded operator on a Hilbert space (hence dealing with convergence of sums as you wanted to avoid) there exists a basis such that the corresponding matrix has your property on rows. This was proved by Toeplitz in 1910.
Nov 16, 2014 at 0:32 comment added Alex Degtyarev @RickyDemer Doesn't matter: just left vs. right modules.
Nov 16, 2014 at 0:21 comment added user5810 @AlexDegtyarev: $\:$ That be for "a finite number of non-zero values on each" column. $\hspace{1.34 in}$
Nov 15, 2014 at 0:07 comment added Jack M Where $A$ is such a matrix, I'm interested in deducing as much information as I can about $A^n$, ideally getting a closed form for it. In particular, I'd love to know if there are analogues to diagonalization and similar methods.
Nov 15, 2014 at 0:01 comment added Scott Andrews What sort of properties of these matrices are you looking for? I would think that, despite perhaps having an infinite number of nonzero entries in some columns, these matrices would in many ways behave like finite dimensional matrices. Also, another interesting infinite matrix setting is upper-triangular matrices. If you simply require that all matrices are upper-triangular, you remove any convergence issues.
Nov 14, 2014 at 23:42 comment added Alex Degtyarev These are endomorphisms of a free module of infinite rank.
Nov 14, 2014 at 23:24 history asked Jack M CC BY-SA 3.0