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- Path: sparky!uunet!olivea!spool.mu.edu!darwin.sura.net!convex!constellation!hardy.math.okstate.edu!wrightd
- From: wrightd@math.okstate.edu (David J. Wright)
- Newsgroups: sci.math
- Subject: Question on Putnam B6
- Summary: Trying to prove a stronger statement.
- Keywords: Putnam Exam, Problems
- Message-ID: <BzqE89.LKz@math.okstate.edu>
- Date: 23 Dec 92 21:26:32 GMT
- Organization: Oklahoma State University, Math Department
- Lines: 34
-
-
-
- Well, I've been thinking about B6 now, and can't figure out
- the truth of the matter. So I might as well ask. Of course,
- I understand the solution to the stated problem, but it's the
- unstated problems that are bugging me.
-
- I can prove that if n is odd, then |M| must be 1. This is a
- consequence of each A^2 = +- I and the stated hypotheses.
-
- Also, if M is a valid set of matrices of degree n, then
-
- M' = { ( +- A 0 ( 0 +- A
- 0 A ) and A 0 ) , A \in M }
-
- is a valid set of matrices of degree 2n. Note that
- Card( M' ) = 4 Card( M )
-
- Moreover, if n = 2m is even, then after conjugation
- M must contain ( 0 -I
- I 0 )
- again using m x m block notation.
-
- If I could just show that after another conjugation M also
- contains ( -I 0
- 0 I )
- then I would be done because that would imply that M arises
- from a valid set of matrices of degree m by the above construction.
-
- However, I'm not sure that that last statement is true. Still I
- would believe the maximal cardinality in degree n is
- 2^2k where 2^k exactly divides n.
-
- Anyway, it was an interesting problem.
-