# Self-similar matrices? [closed]

Does anyone know anything about self-similar (infinite) matrices, with more or less fractal(-like) structure and admitting meaningful matrix-algebra operations?

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## closed as not a real question by Mariano Suárez-Alvarez♦, Reid Barton, Anton GeraschenkoJan 5 '10 at 23:23

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Also, could the moderators or/and high-reputation users kindly add here a tag "fractals" or something like that? –  Igor Korepanov Jan 4 '10 at 12:45
How are they "self-similar" or "fractal"? What Hilbert space are these operators acting on? –  john mangual Jan 4 '10 at 12:47
Well, John, give me some time to think how to explain this... Or, in the case if you can read some Russian, here are two short texts with examples of such matrices: csc.ac.ru/ej/file/4381 and csc.ac.ru/ej/file/4641 . And thanks to Dmitri for creating the fractals tag! –  Igor Korepanov Jan 4 '10 at 13:29
This question should be closed, I think. If you can turn it into something more concrete with maybe a motivation and a small explanation, then I would be quite interested in reading it! As it stands, the answer is probably yes, as apparently someone knows something about that kind of matrices. –  Mariano Suárez-Alvarez Jan 4 '10 at 19:22
"nobody could say anything" perhaps because the question was too vague? Maybe they even had some information you're looking for, but they had no way to know it! –  Reid Barton Jan 5 '10 at 22:41

You should learn about the hyperfinite II_1 factor, which is the limit of the inclusions

M_1 --> M_2 --> M_4 --> M_8 --> ....


(here M_k is the k by k matrices over $\mathbb{C}$) where each inclusion is given by tensoring with the identity matrix in M_2. Every 'finite' element is "self-similar" in a sense.

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For the record, btw, I'm in the camp that thinks this is a poor question: too vague, and it's unclear what you mean by "self-similar". Moreover, not knowing whether this answer is basic stuff you already know, or somewthing new, I have little motivation to expand on it. –  Scott Morrison Jan 5 '10 at 22:28
Scott, thanks for you answer. I am just studying what this mathoverflow is, and your answer is helpful for this. Perhaps I will explain in the next version of my question that just tensor products are not "fractal enough" - or, otherwise, I may decide that this was a wrong site to ask serious questions. But thank you anyhow! –  Igor Korepanov Jan 5 '10 at 22:37
Igor, this is "the wrong site" to ask questions that are so vague that it's not clear what an answer would even look like! In a comment above you equate creativity with vagueness, but I don't see how creativity prevents you from defining the terms that you use in the question. –  HJRW Jan 5 '10 at 23:06
Igor, "anything vaguely resembling self-similarity would work" Except that Scott's answer vaguely resembles self-similarity, and apparently it doesn't work! You could have saved Scott the trouble of giving an answer you didn't need if you'd given more details in the question. –  HJRW Jan 5 '10 at 23:26
Actually, mostly no --- I saw the high noise/signal discussion going on on meta, and gave an answer in order to more clearly highlight the fact that your question was too vague to receive a useful answer. :-) –  Scott Morrison Jan 6 '10 at 9:09