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If two matrices A, B verify the relationship AB = BA Prove that: \[AB^{n} = B^{n}A\] for any integer n.
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The proposition is ture for n=1, by the given condition. Let the proposition be true for some integer k>1. Then AB^k=B^kA Now AB^(k+1)=AB^kB=(AB^k)B=B^kAB=B^kBA=B^(k+1)A/ Hence the proposition is true for k+1 whenever it is true for k. Hence by the method of induction....
Thanks, nevertheless, My professor said : "Using induction is forbidden.",
So let me rethink
AB^n=AB...B=(BA)B...B=B(AB)B...B=B(BA)B...B=BB(AB)B...B=BB(BA)B...B=...=BB...B(AB)=BB...B(BA)=B^nA
What did you do ? the so many B are confusing.
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just used the property BB...B (n times)=B^n
and the given property AB=BA multiples times
Nice.
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