If A and B are n-th row square matrices, $AB=BA$ and $B^2=0$ prove that $(A+B)^k=A^k+kA^{k-1}B$

Question

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?

Answer 1

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 $(A+B)(A+B)^k=A^{k+1}+(k+1)A^kB$.

Answer 2

Since $A$ and $B$ commute, you can use the binomial formula: for $k\ge 1$, $$(A+B)^k=A^k +\binom k1 A^{k-1}B +\sum_{i=2}^k\binom kiA^{k-i}\underbrace{B^i}_{=\,0}.$$