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If $A^4$ = $I_n$ , but $A \neq I_n$, $A^2 \neq I_n$, and $A^3 \neq I_n$, which powers of A are equal to A^-1? I can list examples like:

$A^4 = I_n$

$A^{-1} * A^4 = A^{-1} * I_n$

$(A^{-1} * A) * A^3 = A^{-1}$

Therefore $A^3 = A^{-1}$.

How do I make a proof out of this?

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    then multiply for $A^{4k}$ – 2017-02-16

3 Answers 3

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Note that $A^3A = A^4 = I_n$, so A^3 is the left inverse of A. But you can also write $AA^3 = A^4 = I_n$ and you get that $A^3$ is the right inverse of A.

Then, you have that $A^{-1} = A^3$.

Furthermore, as $A^4 = I_n$, $A^{-1} = A^3 = A^3I_n = A^3A^{4k} = A^{4k + 3}$, for every $k \in \mathbb{N}$.

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Hint:

note that $a^4=A^3\cdot A=A^4 \cdot A^4=A^7A=A^4\cdot A^4 \cdot A^4=\cdots =I_n$

and remember that the inverse, if it exists, is unique.

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$A^4 = A(A^3) = I = (A^3)A$, so $A^3$ is the right and left inverse of $A$.

Note the simple pattern that $A^n = A^{-1}$ as long as $n \equiv 1 \pmod 4$ and you get the generalization that $A^{-1} = A^{4k + 3}, k \in \mathbb{N} \cup 0$