If A and B are n-th row square matrices, $AB=BA$ and $B^2=0$ prove that $$(A+B)^k=A^k+kA^{k-1}B$$
Any ideas?
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Proof by induction on $k$. I leave the base case ($k=1$) for you.
Note that, using the inductive hypothesis, we have $$(A+B)(A+B)^k=(A+B)(A^k+kA^{k-1}B)=A^{k+1}+kA^kB+BA^k+kBA^{k-1}B$$
Now use $BA=AB$ repeatedly to conclude that $BA^k=A^kB$, and $kBA^{k-1}B=kA^{k-1}B^2=0$.
Hence $(A+B)(A+B)^k=A^{k+1}+(k+1)A^kB$.
vadim123
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I'm sorry but I can't seem to get it. I've got: $$AA^k +kA^kB+BA^k+kBA^{k-1}B$$ $$AA^k +kA^kB+BA^k+kBAA^kB$$ $$AA^k +kA^kB+BA^k+kABA^kB$$
So I haven't really got anything. Where am I supposed to go from here?
– crylikeacanary Mar 28 '18 at 21:58 -
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Write $BA^k$ as $BAAAAAAA$, then move the $B$ along one step at a time $BAAAAAAA=ABAAAAAAA=AABAAAAAA=AAABAAAAA$ etc. – vadim123 Mar 28 '18 at 22:36
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Since $A$ and $B$ commute, you can use the binomial formula: for $k\ge 1$, $$(A+B)^k=A^k +\binom k1 A^{k-1}B +\sum_{i=2}^k\binom kiA^{k-i}\underbrace{B^i}_{=\,0}.$$
Bernard
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Which is why we're supposed to figure out a way to do it without using it – crylikeacanary Mar 28 '18 at 22:18
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for k=1 $$A+B=A+B$$
for k=k+1 $$(A+B)^k(A+B)=A^k(A+(k+1)B)$$
– crylikeacanary Mar 28 '18 at 21:49