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Given that a matrix $X$ satisfies $X^2 = X$ it is clear that $X^{100}=X$ by repeated multiplication of $X$. Algebraically, we might write:

$$X^{100} = (X^2)^{50}=X^{50}=(X^2)^{25}=X^{25}=X(X^2)^{12} = \dots = (X^2)^2 = X $$

But this seems like too much work for such a simple fact. Is there a short algebraic proof?

Daniel Pietrobon
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3 Answers3

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It's easier to just prove that $X^n = X$ by induction; if $n = 1$, this is clear. Else, assume that the result is true for some $n \geq 1$ and conclude that

$$X^{n + 1} = X^{n} X = X X = X^2 = X$$

as desired.

4

HINT:

$$X^{n+2}=X^n\cdot X^2=X^n\cdot X=X^{n+1}$$

Now, if we have $f(n+1)=f(n),$ we can directly say $f(n+1)=f(1)$ for $n\ge0$

lab bhattacharjee
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Well, since $X^2 = X$, i.e., $X$ is idempotent, the standard inductive proof for idempotents in any ring works: the inductive hypothesis is $X^k = X$, whence $X^{k + 1} = XX^k = XX = X^2 = X$. Take $X^2 = X$ the base case and we're done, we have $X^n = X$ whether $n = 10$, $n = 10^6$, or $n$ is one-gazillion or even more! What fun! Cheers!

Robert Lewis
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  • I suppose this is the right way to go about things. However, I still consider this to be a "long" algebraic proof as we are essentially arguing that $X^2=X \Rightarrow X^3 = X \Rightarrow X^4 = X \Rightarrow \dots \Rightarrow X^{100} = X$. – Daniel Pietrobon Aug 27 '13 at 07:15