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I am trying to prove the following proposition:

Let $A$ and $B$ square matrix of order $n$. If $AB = BA$, then $A^m B^n=B^n A^m$ .

Using the power definition of matrices and matrices compatible, the proposition is proved, but it is too extensive.

Is there another way?

Victor Huuanca Sullca
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The most straightforward way is probably to show (by induction on $k$) that if matrices $X$ and $Y$ satisfy $XY=YX$, then they satisfy $X^kY = YX^k$.

Then apply this twice: first, with $X=A$, $Y=B$, and $k=m$ to show that $A^mB = BA^m$. Second, with $X=B$, $Y=A^m$, and $k=n$ to show that $B^nA^m = A^mB^n$.

(You could also do the whole thing with a double induction on $m$ and $n$, but nobody likes double induction.)

Misha Lavrov
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