### 51st Putnam 1990

Problem A5

M and N are n x n matrices such that (MN)2 = 0. Must (NM)2 = 0?

Solution

Answer: yes for n = 2; no for n > 2.

For n ≥ 3, Take M12 = M31 = 1, other elements 0, and N11 = N33 = 1, other elements 0. Then (MN)31 = 1 and its other elements are 0. So (MN)2 = 0. (NM)31 = (NM)12 = 1 and its other elements are 0. So (NM)2 is non-zero (its 3,2 element is 1). So we have a counter-example for n ≥ 3.

Note that if (MN)2 = 0, then (NM)3 = N(MN)2M = 0. But we can prove that if K is a 2 x 2 matrix and K3 = 0, then K2 = 0.

Suppose K11 = a, K12 = b, K21 = c, K22 = d. Then since K3 = 0, we have: a3 + 2abc + bcd = 0, (1); a2b + abc + b2c + bd2 = 0, (2); a2c + bc2 + acd + cd2 = 0, (3); abc + 2bcd + d3 = 0, (4). Taking (1) - a/b (2) gives: a + d = 0 or bc = ad. We show that in fact a + d = 0 and bc = ad. For suppose a + d = 0, then (3) gives bc = ad. Conversely, if bc = ad, then (1) implies a(a + d)2 = 0, so a = 0 or a + d = 0. If a = 0, then (4) implies d = 0 and so a + d = 0.

But a + d = 0 and bc = ad, gives immediately that K2 = 0.