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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.

Rodrigo de Azevedo
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Ishan Banerjee
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    What happens when you take the trace of both sides? –  Jan 23 '13 at 09:22
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    If you have, please answer your own question so it does not keep getting bumped to the main page. –  Jan 23 '13 at 09:48

6 Answers6

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

Myshkin
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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.

Glorfindel
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user1551
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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.

Kyle Yip
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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.

Batominovski
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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.

Jack D'Aurizio
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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, $$|AB-BA|=|I| \implies 0=1, \text{ which is a contradiction }$$

Atul Anurag Sharma
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  • that is not quite right, two matrices having same eigenvalues does not mean their difference must have a zero eigenvalue, you can see this from simple 2x2 diagonal matrices, example, diag([1,2]) and diag([2,1]) – water stone Nov 07 '22 at 18:29
  • $A:=\left[\begin{matrix}1 & 2\7 & 9\end{matrix}\right]$ and $B:= \left[\begin{matrix}1 & 8\1 & 6\end{matrix}\right]\implies AB-BA = \left[\begin{matrix}-54 & -54\-27 & 54\end{matrix}\right] $ which has non-zero determinant, contradicting your assertion. – user8675309 Dec 28 '23 at 19:30