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 M_{12} = M_{31} = 1, other elements 0, and N_{11} = N_{33} = 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)^{2}M = 0. But we can prove that if K is a 2 x 2 matrix and K^{3} = 0, then K^{2} = 0.

Suppose K_{11} = a, K_{12} = b, K_{21} = c, K_{22} = d. Then since K^{3} = 0, we have: a^{3} + 2abc + bcd = 0, (1); a^{2}b + abc + b^{2}c + bd^{2} = 0, (2); a^{2}c + bc^{2} + acd + cd^{2} = 0, (3); abc + 2bcd + d^{3} = 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 K^{2} = 0.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001