Some days ago, I was thinking on a problem, which states that $AB-BA=I$ does not have a solution in $M_{n\times n}(\mathbb R)$ and $M_{n\times n}(\mathbb C)$. (Here $M_{n\times n}(\mathbb F)$ denotes the set of all $n\times n$ matrices with entries from the field $\mathbb F$ and $I$ is the identity matrix.)

Although I couldn't solve the problem, I came up with this problem, that, does there exist a field $\mathbb F$ for which that equation $AB-BA=I$ has a solution in $M_{n\times n}(\mathbb F)$?

I'd really appreciate your help.

  • 1
    $\begingroup$ possible duplicate of Matrices with real entries such that $(I -(AB-BA))^{n}=0$ $\endgroup$ – Andrés E. Caicedo Jan 15 '12 at 3:23
  • 11
    $\begingroup$ In short: This is not possible, because the traces of $AB$ and $BA$ coincide. $\endgroup$ – Andrés E. Caicedo Jan 15 '12 at 3:24
  • 13
    $\begingroup$ @AndresCaicedo What about over $\mathbb{F}_n$, where $tr(I_{n\times n})=0$? I suspect the answer is still that it is not possible, but another approach is needed. $\endgroup$ – Alex Becker Jan 15 '12 at 3:27
  • 3
    $\begingroup$ There are fields in which $\operatorname{tr} I = n = 0$, though. Maybe there's something to talk about yet. $\endgroup$ – Dylan Moreland Jan 15 '12 at 3:29
  • $\begingroup$ Ah, Ok, sure, I had overlooked that option. $\endgroup$ – Andrés E. Caicedo Jan 15 '12 at 3:31

Let $k$ be a field. The first Weyl algebra $A_1(k)$ is the free associative $k$-algebra generated by two letters $x$ and $y$ subject to the relation $$xy-yx=1,$$ which is usually called the Heisenberg or Weyl commutation relation. This is an extremely important example of a non-commutative ring which appears in many places, from the algebraic theory of differential operators to quantum physics (the equation above is Heisenberg's indeterminacy principle, in a sense) to the pinnacles of Lie theory to combinatorics to pretty much anything else.

For us right now, this algebra shows up because

an $A_1(k)$-modules are essentially the same thing as solutions to the equation $PQ-QP=I$ with $P$ and $Q$ endomorphisms of a vector space.


  • if $M$ is a left $A_1(k)$-module then $M$ is in particular a $k$-vector space and there is an homomorphism of $k$-algebras $\phi_M:A_1(k)\to\hom_k(M,M)$ to the endomorphism algebra of $M$ viewed as a vector space. Since $x$ and $y$ generate the algebra $A_1(k)$, $\phi_M$ is completely determined by the two endomorphisms $P=\phi_M(x)$ and $Q=\phi_M(y)$; moreover, since $\phi_M$ is an algebra homomorphism, we have $PQ-QP=\phi_1(xy-yx)=\phi_1(1_{A_1(k)})=\mathrm{id}_M$. We thus see that $P$ and $Q$ are endomorphisms of the vector space $M$ which satisfy our desired relation.

  • Conversely, if $M$ is a vector space and $P$, $Q:M\to M$ are two linear endomorphisms, then one can show more or less automatically that there is a unique algebra morphism $\phi_M:A_1(k)\to\hom_k(M,M)$ such that $\phi_M(x)=P$ and $\phi_M(y)=Q$. This homomorphism turns $M$ into a left $A_1(k)$-module.

  • These two constructions, one going from an $A_1(k)$-module to a pair $(P,Q)$ of endomorphisms of a vector space $M$ such that $PQ-QP=\mathrm{id}_M$, and the other going the other way, are mutually inverse.

A conclusion we get from this is that your question

for what fields $k$ do there exist $n\geq1$ and matrices $A$, $B\in M_n(k)$ such that $AB-BA=I$?

is essentially equivalent to

for what fields $k$ does $A_1(k)$ have finite dimensional modules?

Now, it is very easy to see that $A_1(k)$ is an infinite dimensional algebra, and that in fact the set $\{x^iy^j:i,j\geq0\}$ of monomials is a $k$-basis. Two of the key properties of $A_1(k)$ are the following:

Theorem. If $k$ is a field of characteristic zero, then $A_1(k)$ is a simple algebra—that is, $A_1(k)$ does not have any non-zero proper bilateral ideals. Its center is trivial: it is simply the $1$-dimensional subspace spanned by the unit element.

An immediate corollary of this is the following

Proposition. If $k$ is a field of characteristic zero, the $A_1(k)$ does not have any non-zero finite dimensional modules. Equivalently, there do not exist $n\geq1$ and a pair of matrices $P$, $Q\in M_n(k)$ such that $PQ-QP=I$.

Proof. Suppose $M$ is a finite dimensional $A_1(k)$-module. Then we have an algebra homomorphism $\phi:A_1(k)\to\hom_k(M,M)$ such that $\phi(a)(m)=am$ for all $a\in A_1(k)$ and all $m\in M$. Since $A_1(k)$ is infinite dimensional and $\hom_k(M,M)$ is finite dimensional (because $M$ is finite dimensional!) the kernel $I=\ker\phi$ cannot be zero —in fact, it must hace finite codimension. Now $I$ is a bilateral ideal, so the theorem implies that it must be equal to $A_1(k)$. But then $M$ must be zero dimensional, for $1\in A_1(k)$ acts on it at the same time as the identity and as zero. $\Box$

This proposition can also be proved by taking traces, as everyone else has observed on this page, but the fact that $A_1(k)$ is simple is an immensely more powerful piece of knowledge (there are examples of algebras which do not have finite dimensional modules and which are not simple, by the way :) )

Now let us suppose that $k$ is of characteristic $p>0$. What changes in term of the algebra? The most significant change is

Observation. The algebra $A_1(k)$ is not simple. Its center $Z$ is generated by the elements $x^p$ and $y^p$, which are algebraically independent, so that $Z$ is in fact isomorphic to a polynomial ring in two variables. We can write $Z=k[x^p,y^p]$.

In fact, once we notice that $x^p$ and $y^p$ are central elements —and this is proved by a straightforward computation— it is easy to write down non-trivial bilateral ideals. For example, $(x^p)$ works; the key point in showing this is the fact that since $x^p$ is central, the left ideal which it generates coincides with the bilateral ideal, and it is very easy to see that the left ideal is proper and non-zero.

Moreover, a little playing with this will give us the following. Not only does $A_1(k)$ have bilateral ideals: it has bilateral ideals of finite codimension. For example, the ideal $(x^p,y^p)$ is easily seen to have codimension $p^2$; more generally, we can pick two scalars $a$, $b\in k$ and consider the ideal $I_{a,b}=(x^p-a,y^p-b)$, which has the same codimension $p^2$. Now this got rid of the obstruction to finding finite-dimensional modules that we had in the characteristic zero case, so we can hope for finite dimensional modules now!

More: this actually gives us a method to produce pairs of matrices satisfying the Heisenberg relation. We just can pick a proper bilateral ideal $I\subseteq A_1(k)$ of finite codimension, consider the finite dimensional $k$-algebra $B=A_1(k)/I$ and look for finitely generated $B$-modules: every such module is provides us with a finite dimensional $A_1(k)$-module and the observations above produce from it pairs of matrices which are related in the way we want.

So let us do this explicitly in the simplest case: let us suppose that $k$ is algebraically closed, let $a$, $b\in k$ and let $I=I_{a,b}=(x^p-a,y^p-b)$. The algebra $B=A_1(k)/I$ has dimension $p^2$, with $\{x^iy^j:0\leq i,j<p\}$ as a basis. The exact same proof that the Weyl algebra is simple when the ground field is of characteristic zero proves that $B$ is simple, and in the same way the same proof that proves that the center of the Weyl algebra is trivial in characteristic zero shows that the center of $B$ is $k$; going from $A_1(k)$ to $B$ we have modded out the obstruction to carrying out these proofs. In other words, the algebra $B$ is what's called a (finite dimensional) central simple algebra. Wedderburn's theorem now implies that in fact $B\cong M_p(k)$, as this is the only semisimple algebra of dimension $p^2$ with trivial center. A consequence of this is that there is a unique (up to isomorphism) simple $B$-module $S$, of dimension $p$, and that all other finite dimensional $B$-modules are direct sums of copies of $S$.

Now, since $k$ is algebraically closed (much less would suffice) there is an $\alpha\in k$ such that $\alpha^p=a$. Let $V=k^p$ and consider the $p\times p$-matrices $$Q=\begin{pmatrix}0&&&&b\\1&0\\&1&0\\&&1&0\\&&&\ddots&\ddots\end{pmatrix}$$ which is all zeroes expect for $1$s in the first subdiagonal and a $b$ on the top right corner, and $$P=\begin{pmatrix}-\alpha&1\\&-\alpha&2\\&&-\alpha&3\\&&&\ddots&\ddots\\&&&&-\alpha&p-1\\&&&&&-\alpha\end{pmatrix}.$$ One can show that $P^p=aI$, $Q^p=bI$ and that $PQ-QP=I$, so they provide us us a morphism of algebras $B\to\hom_k(k^ p,k^ p)$, that is, they turn $k^p$ into a $B$-module. It must be isomorphic to $S$, because the two have the same dimension and there is only one module of that dimension; this determines all finite dimensional modules, which are direct sums of copies of $S$, as we said above..

This generalizes the example Henning gave, and in fact one can show that this procedure gives all $p$-dimensional $A_1(k)$-modules can be constructed from quotients by ideals of the form $I_{a,b}$. Doing direct sums for various choices of $a$ and $b$, this gives us lots of finite dimensional $A_1(k)$-modules and, then, of pairs of matrices satisfying the Heisenberg relation. I think we obtain in this way all the semisimple finite dimensional $A_1(k)$-modules but I would need to think a bit before claiming it for certain.

Of course, this only deals with the simplest case. The algebra $A_1(k)$ has non-semisimple finite-dimensional quotients, which are rather complicated (and I think there are pretty of wild algebras among them...) so one can get many, many more examples of modules and of pairs of matrices.

  • 2
    $\begingroup$ I'm pretty late, but it's never too late to praise so great an answer. Do you have references for the non-simple finite-dimensional quotients? $\endgroup$ – PseudoNeo Jun 4 '15 at 8:10

As noted in the comments, this is impossible in characteristic 0.

But $M_{2\times 2}(\mathbb F_2)$ contains the example $\pmatrix{0&1\\0&0}, \pmatrix{0&1\\1&0}$.

In general, in characteristic $p$, we can use the $p\times p$ matrices $$\pmatrix{0&1\\&0&2\\&&\ddots&\ddots\\&&&0&p-1\\&&&&0}, \pmatrix{0\\1&0\\&\ddots&\ddots\\&&1&0\\&&&1&0}$$ which works even over general unital rings of finite characteristic.

  • $\begingroup$ Your update is, of course, the same as that given by @Ted - "great minds think alike"? $\endgroup$ – user16299 Jan 15 '12 at 5:57
  • $\begingroup$ I believe that this example is analyzed in Kaplansky's book on Lie Algebras. $\endgroup$ – Geoff Robinson Jan 15 '12 at 7:25
  • $\begingroup$ @Yemon: Either that, or there are not that many essentially different solutions. (I looked briefly for a proof that all sets of solutions are similar, but that is not the case -- my general solution has matrices of rank $p-1$, but you can add $I$ to one or both of them to get a solution where one or both of the matrices are invertible). $\endgroup$ – Henning Makholm Jan 15 '12 at 14:47

In characteristic 0, we have an infinite-dimensional example over the vector space $\mathbb{F}[x]$: Let $A$ be multiplication by $x$, and $B$ be differentiation with respect to $x$.

This can be used to construct a finite-dimensional example over characteristic $p$. Let $\mathbb{F}$ be a field of characteristic $p$. Let $V$ be the $p$-dimensional space of polynomials over $\mathbb{F}$ of degree < $p$. Let $A$ be multiplication by $x$ on the elements $1, x, \ldots, x^{p-2}$ but send $x^{p-1}$ to 0. Let $B$ be differentiation with respect to $x$.

This example was constructed by noticing that multiplying $x^{p-1}$ by $x$ and then differentiating with respect to $x$ gives 0 anyway in characteristic $p$, so we can have $A$ send $x^{p-1}$ to 0 with no effect on the desired outcome. This allows us to truncate the infinite-dimensional example in characteristic 0 to a finite-dimensional example in characteristic $p$.

Edit: As Yemon Choi mentions in the comment below, this actually gives $AB-BA=-I$. To get $AB-BA=I$, either interchange $A$ and $B$ above, or (somewhat nonstandardly) interpret $A$ and $B$ as right operators instead of left operators.

  • 1
    $\begingroup$ Very nice! A nice instance of using an infinite-dimensional idea to get a finite-dimensional construction. Small nitpick: I think with the particular choice of $A$ and $B$ you state, we get $AB-BA= -I$ $\endgroup$ – user16299 Jan 15 '12 at 5:23

HINT $\ $ Extending a 1936 result of Shoda for characteristic $0,$ Benjamin Muckenhoupt, a 2nd year graduate student of A. Adrian Albert, proved in the mid fifties that in the matrix algebra $\rm\ \mathbb M_n(F)\ $ over a field $\rm\:F\:$, a matrix $\rm\:M\:$ is a commutator $\rm\ M\: = \: A\:B - B\:A\ $ iff $\rm\:M\:$ is traceless, i.e. $\rm\ tr(M) = 0\:.$

From this we infer that $1$ is a commutator in $\rm\mathbb M_n(\mathbb F_p)\ \iff\ n = tr(1) = 0\in \mathbb F_p \iff\ p\ |\ n \:.$

Muckenhoupt and Albert's proof is short and simple and is freely acessible at the link below
A. A. Albert, B. Muckenhoupt. On matrices of trace zero, Michigan Math. J. 4, #1 (1957), 1-3.

Tracing citations to this paper reveals much literature on representation by (sums of) commutators and tracefree matrices. For example, Rosset proved that in a matrix ring $\rm\:\mathbb M_n(R)\:$ over a commutative ring $\rm\:R\:,\:$ every matrix of trace zero is a sum of two commutators.

  • 1
    $\begingroup$ I always like your precise, compact and still most informative answers. Thanks and +1! $\endgroup$ – Tapu Jan 16 '12 at 4:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.