# linear algebra – Why this trick to derive \$ formula[A^n,B]\$ in terms of repeated switches work so well?

This is a known result that since generic non-switch operators $$A, B$$, we have
$$A ^ n B = sum_ {k = 0} ^ n binom {n} {k} operatorname {ad} ^ k (A) (B) A ^ {n-k}. Tag A$$
This can be proven for example via induction with little work.

However, while trying to better understand this formula, I realized that there is a much simpler way to derive it, at least on a formal and intuitive level.

### L & # 39; trick

Let $$hat { mathcal S}$$ and $$hat { mathcal C}$$ (meaning "shift" and "commute", respectively) designate operators that act on expressions of the form $$A ^ k D ^ j A ^ ell$$ (designating for simplicity $$D ^ j equiv operatorname {ad} ^ j (A) (B)$$) as following:

begin {align} hat { mathcal S} (A ^ k D ^ jA ^ ell) & = A ^ {k-1} D ^ jA ^ { ell + 1}, \ hat { mathcal C} (A ^ {k-1} D ^ {j + 1} A ^ ell) & = A ^ {k-1} D ^ jA ^ { ell + 1}. end {align}
In other words, $$hat { mathcal S}$$ "moves" the center $$D$$ block left, while $$hat { mathcal C}$$ the fact "eat" the neighbor $$A$$ postman.

It's not hard to see that $$hat { mathcal S} + hat { mathcal C} = mathbb 1$$, which is only another way to define the identity
$$A[A,B]=[A,B]A +[A[AT[UNE[A[A,B]].$$
Moreover, especially, $$hat { mathcal S}$$ and $$hat { mathcal C}$$ commute.
For this reason, I can write

$$A ^ n B = ( hat { mathcal S} + hat { mathcal C}) ^ n (A ^ n B) = sum_ {k = 0} ^ n binom {n} {k} hat { mathcal S} ^ {nk} hat { mathcal C} ^ {k} (A ^ n B),$$
which gives me immediately (A) without any need for recursion or other tips.

### The question

Now everything is fine and dandy, but that leaves me wondering as to why this kind of thing works?
It seems to me that somehow I get around the inconvenience of having to deal with non-commuter operations by switching to a "super-operator" space, in which the same operation can be expressed in terms of commute "Super-operators."

I do not even know how we could formalize these "super-operators" $$hat { mathcal S}, hat { mathcal C}$$because they seem to be objects acting more on the "chains of operators" than on the elements of the algebra of the operators themselves.

Is there a way to formalize this way of managing expressions? Is this a well-known method in this context (I had never seen it but I do not know this type of manipulation well)?