### 53rd Putnam 1992

**Problem B6**

Let M be a set of real n x n matrices such that: (1) 1 ∈ M; (2) if A, B ∈ M, then just one of AB, - AB is in M; (3) if A, B ∈ M, then either AB = BA or AB = -BA; (4) if 1 ≠ A ∈ M, then there is at least one B ∈ M such that AB = - BA. Prove that M contains at most n^{2} matrices.

**Solution**

For example, we might take the set:

I = (1 0) A = (0 1) B = (1 0) C = (0 -1)
(0 1) (1 0) (0 -1) (1 0)

We find A^{2} = B^{2} = -C^{2} = I, AB = -BA =C, AC = -CA = B, CB = -BC = A.
We show first that if A is in M, then A^{2} = ± I. Take any B. Then BA = kAB where k^{2} = 1, so BA^{2} = k ABA = k^{2} A^{2}B = A^{2}B. So A^{2} commutes with every element of M. Obviously -A^{2} does also. But by (2) one of A^{2} and -A^{2} must be in M and (4) then guarantees a non-commuting element unless A^{2} = ± I.

If there are more than n^{2} elements, then we can find a linear combination of elements which is zero: ∑ λ_{A}A = 0 (*). If we multiply on both sides by B we get ∑ ±λ_{A}A = 0. We could add this to (*) to get a linear combination with fewer terms provided BA = AB for at least one A in (*) and BA = -AB for at least one A in (*).

The only way to guarantee a + sign is if A = I. But we can do that by multiplying by A, where A appears in the combination. The A term then becomes an I term. Guaranteeing a - sign is then easy. There must be at least one other term (apart from I). Suppose it is A. Then choose B such that AB = - BA.

So if there are more than n^{2} terms we can find a linear relation which we can shorten without limit. Contradiction. So there are at most n^{2} terms.

53rd Putnam 1992

© John Scholes

jscholes@kalva.demon.co.uk

7 Jan 2001