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Is the following equation impossible?

$AB-BA=I_n$, where $A,B\in M_n(\mathbb C)$

If it does, then how do I prove it? Thanks in advance!

zwim
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Anonymus
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2 Answers2

5

Hint:

Take trace on both sides.

Notice that $tr(AB)=tr(BA)$.

Siong Thye Goh
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An alternative proof, not using the trace:

Let show by induction $\mathcal P(n):A^nB-BA^n=nA^{n-1}$

$\mathcal P(0): IB-BI=B-B=0\quad$ ok

$\mathcal P(1): AB-BA=I\quad$ ok

assuming $\mathcal P(n)$ then

$\begin{cases} A(A^nB-BA^n)=A^{n+1}B-ABA^n=A(nA^{n-1})=nA^n\\ (A^nB-BA^n)A=A^nBA-BA^{n+1}=(nA^{n-1})A=nA^n\\ \end{cases}$

$\begin{array}{ll} \overset{(+)}{\implies}A^{n+1}B-BA^{n+1}&=2nA^n+ABA^n-A^nBA\\ &=2nA^n-A(A^{n-1}B-BA^{n-1})A\\ &=2nA^n-A((n-1)A^{n-2})A\\ &=2nA^n-(n-1)A^n\\ &=(n+1)A^n\end{array}$

By linearity we get for every polynomial $p$ the relation $p(A)B-Bp(A)=p'(A)$

This is in particular true for the minimal polynomial $m(A)=0$, and this implies that $m'(A)=0$.

But in $M_n(\mathbb C)$ which has finite dimension $\deg(m')<\deg(m)$ contradicting the minimality of $m$, thus $AB-BA=I$ is not possible.


Note: It is possible to have this relation in infinite dimension.

For instance in $\mathbb C[X]$ let's have $f(P)=P'$ and $g(P)=XP$

$(f\circ g-g\circ f)(P)=(XP)'-X(P')=XP'+P-XP'=P$

$\implies f\circ g-g\circ f=id$

zwim
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    We have escaped the trace! It's a good answer. Note that, in the infinite dimensional case, assume that $A,B$ are continuous. Then $(n+1)||A^n||\leq 2||A^n||||A||||B||$ implies that there is $p$ s.t. $A^p=0$; in the same way, there is $q$ s.t. $B^q=0$. In your counter-example, $f$ is not continuous. –  Nov 09 '17 at 14:10
  • Instructive remark, thanks. – zwim Nov 09 '17 at 14:26