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