# $AB-BA=I$ having no solutions

The following question is from Artin's Algebra.

If $$A$$ and $$B$$ are two square matrices with real entries, show that $$AB-BA=I$$ has no solutions.

I have no idea on how to tackle this question. I tried block multiplication, but it didn't appear to work.

• What happens when you take the trace of both sides?
– user38268
Jan 23, 2013 at 9:22
• If you have, please answer your own question so it does not keep getting bumped to the main page.
– user38268
Jan 23, 2013 at 9:48

Let a $n\times n$ matrix. Take trace of both sides $$\operatorname{trace}(AB-BA)= \operatorname{trace}(I)\Rightarrow \operatorname{trace}(AB)- \operatorname{trace}(BA) =n\Rightarrow0=n$$

As you should have known by now, for real matrices, the equation $AB-BA=I$ has no solution because the LHS has zero trace but the RHS is not traceless. The same conclusion holds for complex matrices. For other fields, an in-depth discussion can be found in the answers to a related question. In particular, it has been proven that a square matrix $M$ is a commutator (i.e. $M=AB-BA$ for some square matrices $A$ and $B$) if and only if $M$ is traceless:

A.A. Albert and Benjamin Muckenhoupt (1957), "On matrices of trace zeros". Michigan Math. J., 4(1):1-3.

It follows that $AB-BA=I_n$ has a solution if and only if $I_n$ is traceless over the underlying field.

Another way is to use the fact that $AB$ and $BA$ have the same set of eigenvalues.

Rewrite the equation as $AB=BA+I$, then it follows that $\lambda$ is an eigenvalue of $AB$ iff $\lambda$ is an eigenvalue of $BA+I$, or equivalently, $\lambda-1$ is an eigenvalue of $BA$.

Repeating the argument, we can see that if $\lambda$ is an eigenvalue of $AB$, then for any postive integer $n$, $\lambda-n$ is an eigenvalue of $AB$, this means $AB$ have infinitely many distinct eigenvalues, a contradiction.

Assume that $A$ and $B$ are $n$-by-$n$ real or complex matrices for some integer $n\geq 1$. If $A\,B-B\,A=I$, then \begin{align}\frac{\text{d}}{\text{d}t}\,\exp(+t\,A)\,B\,\exp(-t\,A)&=\exp(+t\,A)\,(A\,B-B\,A)\,\exp(-t\,A) \\&=\exp(+t\,A)\,I\,\exp(-t\,A)=I\,. \end{align} Hence, $$\exp(+t\,A)\,B\,\exp(-t\,A)=C+t\,I$$ for some constant matrix $C$. Clearly, when $t=0$, we get $C=B$. Therefore, $$\exp(+t\,A)\,B\,\exp(-t\,A)=B+t\,I\,.$$ Take the determinant on both sides to get $$\det(B)=\det(B+t\,I)$$ for all $t\in\mathbb{R}$, but this is absurd because $\det(B+t\,I)$ is a monic polynomial in $t$ of degree $n\geq1$, whence $\det(B+t\,I)$ is nonconstant.

Expanding a now deleted hint: if we assume that $A,B$ have real or complex entries, from $AB=I+BA$ it follows that $\text{Spec}(AB)=1+\text{Spec}(BA)$. This leads to a contradiction since no finite set in $\mathbb{C}$ is invariant with respect to $z\mapsto z+1$, and $AB,BA$ are conjugated hence cospectral.

Eigenvalues of $$AB \text{ and }BA$$ are equal.Therefore, 0 must be the eigenvalue of $$AB-BA$$. The product of all eigenvalues is the determinant of the operator. Hence, $$|AB-BA|=|I| \implies 0=1, \text{ which is a contradiction }$$