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$$ AB-BA=I\implies \forall k\in\mathbb Z^{+} \quad A^kB-BA^k=kA^{k-1} $$

note: $A,B$ are some linear operators over $V$ i.e. a linear mapping from $V$ to $V$ or $A:V\to V$

thanks @Mvaldi for pointing out that $AB-BA=I$ can't be satisfied for linear operators on finite-dimensional vector spaces.

I want to prove it by induction. But got stuck on the first step.

if we have proved that $A^kB-BA^k=kA^{k-1}$, then
$$ A^{k+1}B=A(BA^k+kA^{k-1})=ABA^k+kA^k $$

and $$ B^{k+1}A=(A^kB-kA^{k-1})A=A^kBA-kA^k $$

thanks @Semiclassical for fixing the equation.

what can I do then.

perhaps we need to find some matrix equation that is equivalent to $AB-BA=I$..

hehelego
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4 Answers4

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For a matrix $M$ denote $$D(M) = M B - B M$$ Let's show that $D$ behaves like a derivation, that is $$ D(M\cdot N) = D(M)\cdot N + M \cdot D(N)$$ LHS is $M N B - B M N$, RHS is $(M B- B M)N + M( N B - B N)$, checks OK.

Then let's show that $$D(A^k) = k A^{k-1}$$ by induction on $k$. Just check for $k=1$, and then use $$D(A^{k+1}) = D(A^k \cdot A) = D(A^k) \cdot A + A^k \cdot D(A)$$

orangeskid
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  • thanks, and it seems that the problem is sort of related to differential operator? – hehelego Jan 03 '21 at 07:15
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    @hehelego: Commutation relation... You have such an example if $A = \frac{d}{d x}$ and $B$ is "multiplication by $x$". But they must act on infinite dimensional spaces ( $B$ "increases the degree" ); yes, it is – orangeskid Jan 03 '21 at 07:33
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Hint: Multiply $A$ from right $+$ Multiply $A$ from left.

C.F.G
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thanks @C.F.G for providing hint.

(Induction). Suppose that we have $AB-BA=I$ and $A^kB-BA^k=kA^{k-1}$. Then:

$$ \begin{aligned} A^k(AB-BA)&=A^{k+1}B-A^kBA=A^k\\ (A^kB-BA^k)A&=A^kBA-BA^{k+1}=kA^k\\ \implies &A^{k+1}B+(-A^kBA+A^kBA)-BA^k=(k+1)A^k \end{aligned} $$

hehelego
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Unfortunately, there is NO pair of 2 $n \times n$ matrix A,B such that $AB-BA=I_{n}$. To verify it, we can use property of trace. In other words, tr(AB)=tr(BA) for all square matrix A,B.

Mvaldi
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  • Sorry, this is not the original problem statement. – hehelego Jan 03 '21 at 06:53
  • The original problem is stated as A,B are linear operators over $V$. not matrices – hehelego Jan 03 '21 at 06:54
  • thanks for pointing out the fact that this can't be true for matrices. – hehelego Jan 03 '21 at 06:56
  • Maybe you already learned that all linear operator can be identified as matrix, so if this contradiction is occured in matrix situation, so does linear operators. – Mvaldi Jan 03 '21 at 06:56
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    But not true for linear operators on vector spaces that have infinitely many dimensions. – hehelego Jan 03 '21 at 06:58
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    For linear operators, the simplest example is just $(D_x x-xD_x )f=f$. – Semiclassical Jan 03 '21 at 07:00
  • Uh... For infinite dimension, you cannot get A,B such that AB-BA=I_{n} for such an positive integer n. [Identify to matrix and compare size of AB and BA.] – Mvaldi Jan 03 '21 at 07:00
  • @semiclassical What is operator $D_{x}$? – Mvaldi Jan 03 '21 at 07:02
  • Differentiation with respect to $x$. – Semiclassical Jan 03 '21 at 07:03
  • Is $(D_{x}x-x D_{x})f$ means that $(x f(x))'-x f'(x)$? Ah... taking multiplying x as $T_{1}$ and differentiating as $T_{2}$ are operators such that $T_{1}T_{2}-T_{2}T_{1}$ is identity operator defined in polynomial space. – Mvaldi Jan 03 '21 at 07:17
  • @Semiclassical So the original problem is handling for infinite dimensional situation, right? – Mvaldi Jan 03 '21 at 07:19
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    This should probably be a comment, not an answer. In any case the claim is wrong without further qualification. Over a field of finite characteristic $p$ and any positive multiple $n$ of $p$, there are matrices $A, B$ such that $AB - BA = I_n$: In this case $\operatorname{tr}(I_n) = 0$, so the nonexistence argument in this answer does not apply. See Henning Makholm's example here: https://math.stackexchange.com/a/99180/155629 The claim is true over fields of characteristic $0$. – Travis Willse Jan 03 '21 at 08:25