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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.

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    $\begingroup$ What happens when you take the trace of both sides? $\endgroup$ – user38268 Jan 23 '13 at 9:22
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    $\begingroup$ If you have, please answer your own question so it does not keep getting bumped to the main page. $\endgroup$ – user38268 Jan 23 '13 at 9:48
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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.

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

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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.

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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.

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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.

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Another approach can be to consider the equations resulting from $$\underset{{\bf A \text{ or } B}}{\min}{ {\bf \|AB-BA-I}\|_2}$$

This can be investigated using vectorization and Kronecker products.


Or we can write $\bf AB = BA + I$ and transposing both sides ${\bf B}^T{\bf A}^T = {\bf A}^T{\bf B}^T + \bf I$ and see by identification that ${\bf B}^T$ is another $\bf A$ and ${\bf A}^T$ another $\bf B$. On the other hand, if $\bf A$ and $\bf B$ switch places, what happens then? Can maybe this (anti-)symmetry be used somehow?

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  • $\begingroup$ That is not what I meant. Edited to try clarifying a bit. $\endgroup$ – mathreadler Sep 5 '16 at 16:45
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    $\begingroup$ Well I still don't see what you mean. Instead of making a hint to a question that has been fully resolved (in a much simpler fashion) I suggest you try to write exactly what you mean instead of trying to be cryptic. $\endgroup$ – Winther Sep 5 '16 at 16:51
  • $\begingroup$ I am just trying to add some approaches which weren't mentioned. Symmetries of various kinds are very powerful and useful concepts to many problems. $\endgroup$ – mathreadler Sep 5 '16 at 16:56
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    $\begingroup$ Ok, so what you added is just speculation? Not what I like to see in answers. $\endgroup$ – Winther Sep 5 '16 at 17:05
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    $\begingroup$ What discovery? You have simply done some random manipulations and then speculating that it might work. However there is no idea here to go with. You just say symmetry like that is a magical word! btw “symmetry” is how the simplest answer does it; $tr(AB) = tr(BA)$. If this type of answer was OK then why not also add another one saying “If we add $7$ to each side then what happen? Can we exploit the fact that 7 is prime somehow?” and justify it with “primes are the most important things in mathematics”. This is of course silly, but I don’t see any difference with what you are doing here. $\endgroup$ – Winther Sep 5 '16 at 17:19

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