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Is a matrix $A$ raised to the power $0$ $ $ just defined to be $I$, or $I$ can actually be derived somehow from matrix multiplication?

Would something like the following suffice to derive it?

$I=A^nA^{-n}=A^{(n-n)}=A^0 $ $ $ $ $ $ $ $ $ $ \Longrightarrow $ $ $ $ $ $ $ $ $ $ A^0=I$

Assuming $n\in\mathbb{N}$.

Or the index manipulation I'm using is not valid for matrices?

Stephen
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    For invertible $A$, your argument looks fine. Now, if $A$ is singular then $A^n$ is well defined for nonnegative $n$ and I guess it still makes sense to define $A^0 = I$. – Surb Nov 25 '17 at 15:17
  • So for singular $A$ it's more of a (sensible) definition than a consequence of matrix arithmetic? – Stephen Nov 25 '17 at 15:32

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Whenever you have an associative prduct with unit (no matter if it is numbers or matrices) you define positive powers as iterated product and you extend the rule to include the zeroth power as the unit and the negative power as the (if it exists) inverse. This allows to use th same rules of power as you have in arithmetic (in other terms that gives you a group homomorphism from $\mathbb Z$ to the subgroup generated by $A$ and $A^{-1}$).

N. Ciccoli
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