# Non-commuting matrices and nilpotence

Is it possible to find $n\times n$ matrices $M,N$  such that $(MN)^n=0$ but $(NM)^n\ne 0$?

I can see that  $(MN)^n=0\implies (NM)^{n+1}= 0$ by associativity. But I don't see that it is necessarily true for $(NM)^n$.

I can also see that $MN=0\not\implies NM=0$ but I cannot find an example for $(MN)^n=0$ but $(NM)^n\ne 0$.

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No. As it turns out, $MN$ and $NM$ must have the same nonzero eigenvalues (although the multiplicities need not agree), so if $(MN)^n = 0$ then $MN$ has only zero eigenvalues, so the same must be true of $NM$, which must therefore have characteristic polynomial $t^n$.

One elegant proof uses the following lemma: in any ring, $1 - xy$ is invertible if and only if $1 - yx$ is invertible. Now use the fact that $\lambda$ is a nonzero eigenvalue of $MN$ if and only if $1 - \frac{MN}{\lambda}$ is not invertible.

Another proof proceeds as follows: let $M, N$ have entries $m_{ij}, n_{ij}$ which are formal variables in a polynomial ring $R = \mathbb{Z}[m_{ij}, n_{ij}]$. Observe that $$\det(Nt - NMN) = \det(N) \det(t - MN) = \det(t - NM) \det(N).$$

Since $R$ is an integral domain, we conclude that $$\det(t - MN) = \det(t - NM)$$

as polynomials in $R$. So in fact $MN$ and $NM$ must have the same characteristic polynomial. (Note, however, that the first proof is much more general; among other things, it remains valid in infinitely many dimensions.)

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+1, Nice answer! – wxu May 21 '12 at 14:18
On the other hand it is possible for smaller exponents than $n$ the size of the matrices, e.g. $MN$ might be zero while not $NM$. – hardmath May 21 '12 at 20:26
@hardmath: yes, the OP mentions this so I assume he is already aware. – Qiaochu Yuan May 21 '12 at 20:28
See also here on the proof of $\$ Sylvester's identity $\rm\ \ det(1+AB) = det(1+BA)\$ that proceeds by taking $\rm\ det\$ of $\rm\ (1+A\ B)\ A\ =\ A\ (1+B\ A)\$ then universally (generically) cancelling $\rm\ det(A)\:,\$ and see here for more examples of such universal proofs. – Bill Dubuque May 23 '12 at 22:13
See here for much more on $\rm\:1-xy\:$ invertible iff $\rm\:1-yx\:$ invertible, in particular Halmos' tantalizing problem to "explain" the very slick power series proof. – Bill Dubuque May 23 '12 at 22:19

Here's another proof of the fact which works over any field $F$ of characteristic $0$, where the following lemma applies:

Lemma. Let $u$ be an endomorphism of a finite dimensional $F$-vector space $V$. Then $u$ is nilpotent iff for all $k=1,\dots,n=\dim V$, $\mathrm{Tr}(u^k)=0$.

Suppose $M,N$ are $n\times n$ matrices with coefficients in a field of characteristic $0$ such that their product is nilpotent, i.e. $(MN)^n=0$. For any $k\geq 1$ $$0=\mathrm{Tr}((MN)^k)=\mathrm{Tr}((MN)^{k-1}MN)=\mathrm{Tr}(N(MN)^{k-1}M)=\mathrm{Tr}((NM)^k)$$ The lemma then implies that $NM$ is nilpotent aswell.

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You can combine your observation that $(MN)^n=0\implies (NM)^{n+1}=0$ with the result of this other question, showing that $(NM)^{n+1}=0\implies (NM)^n=0$.

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