### 42nd Putnam 1981

**Problem B4**

A is a set of 5 x 7 real matrices closed under scalar multiplication and addition. It contains matrices of ranks 0, 1, 2, 4 and 5. Does it necessarily contain a matrix of rank 3?

**Solution**

Answer: no.

The 5 x 7 is something of a red herring. Note that we would expect the answer to be no, because addition and scalar multiplication do not impose any mixing on the matrix elements.

Consider the 5 x 5 matrix with a in the first 4 positions on the diagonal, c is the last position, b at positions (4,5) and (5,4) and zeros elsewhere. Taking (a, b, c) = (1, 0, 1), (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, 0) gives matrices of rank 5, 4, 2, 1, 0 respectively. If a = 0, then all the entries in rows 1, 2 and 3 are zero, so the rank is at most 2. If a is non-zero, then there is certainly a 4 x 4 unit submatrix, so the rank is at least 4. Thus no member of the set has rank 3.

If we add two columns of zeros to every member of the set, then we get a counter-example for the 5 x 7 case.

42nd Putnam 1981

© John Scholes

jscholes@kalva.demon.co.uk

16 Jan 2001