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If $B,C$ are $n$ rowed square matrices and if $A=B+C, BC=CB, C^2=O$, then show that for every $n \in \mathbb N$,
$$A^{n+1}=B^n(B+(n+1)C)$$

I tried to prove it using mathematical induction. But I could not get $P(1)$ to be true. $$P(1): A^2 = B(B+2C)$$ I couldn't equate them.

Please offer some assistance. Thank you. :)

chndn
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3 Answers3

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$$A^2=(B+C)^2=B^2+BC+CB+C^2$$

Now use $CB=BC$ and $C^2=0$.

N. S.
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Try P(0) it wll be easier. But P(1) should work too.

gvo
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  • I need to use P(1). Can you show me how to get it true? – chndn May 22 '13 at 13:49
  • Well, N. S. answered it before me. But why do you need P(1)?? If you have P(0) and your mathematical induction, it will be also true for P(1) and P(2), P(3), ... P(n) – gvo May 22 '13 at 13:52
  • My text book states that first to prove p(1)is true, then assuming k to be true, prove that k+1 is true – chndn May 22 '13 at 14:17
  • If you start with P(1) it is proved for $n \geq 1$, if you start with P(0) it is proved for $n \geq 0$ – gvo May 22 '13 at 14:38
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Note that this is just the Binomial Theorem truncated to the first two terms, i.e.

$$ (B\!+\!C)^{n}\! = B^{n}\! + n B^{n-1} C + C^2(\cdots)\, \equiv\, B^n + n B^{n-1} C\ \pmod{C^2}$$

It is true in any ring where $\,\color{#c00}{BC = CB}.\,$ An inductive proof is easy. The base case $\,n= 0\,$ is true, being $\, 1 = 1.\,$ Suppose as inductive hypothesis that it is true for $\, n = k.\,$ The inductive step is

$$\begin{eqnarray}(B+C)^{k+1} &=\,& (B+C)^{k}(B+C)\\ &=\,& (B^k+kB^{k-1}C)(B+C)\quad {\rm by\ the\ induction\ hypothesis} \\ &=\,& B^{k+1}+ B^k C + kB^{k-1}\color{#c00}{CB} + kB^{k-1}C^2 \\ &=\,& B^{k+1}+(k\!+\!1) B^k C\ \ \ {\rm by}\ \ \color{#c00}{CB = BC},\ \ C^2\equiv 0\end{eqnarray}$$

Key Ideas
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