How to prove this property for linear operator $A,B$ : $AB-BA=I\implies A^kB-BA^k=kA^{k-1}\quad k\in \mathbb N$.

Question

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

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.

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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.

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

Answer 1

Hint: Multiply $A$ from right $+$ Multiply $A$ from left.

Answer 2

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.

Answer 3

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 $$D(A^{k+1}) = D(A^k \cdot A) = D(A^k) \cdot A + A^k \cdot D(A)$$

Answer 4

thanks @C.F.G for providing hint.

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

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