Matrix multiplied by itself n times equals identity
Given a $A \in \mathbb{R}^{100\times 100}$ and $A^{6} = I_{100}$ and $A^{14} = I_{100}$, ist say $A^{2} = I_{100}$ too?
And what would then be the relationship governing whether $A^{2} = I_{100}$ or even $A^{n}$ is valid given two different conditions as above?
$\endgroup$3 Answers
$\begingroup$Hint: If $A^m = A^n = I$ for two integers $m, n$, then $A^{am+bn} = I$ for any $a, b \in \mathbb{Z}$.
Can you express $2$ in the form $6a + 14b$?
$\endgroup$ $\begingroup$Since $A$ is invertible, you have $A^{-6} = I$. Hence $A^2 = A^{14} A^{-6} A^{-6} = I$.
$\endgroup$ $\begingroup$$\mathbb Z\to \mathbb R^{100\times 100}$, $n\mapsto A^n$ is a group homomorphism and its kernel must be a subgroup of $\mathbb Z$ and it contains $14$ and $6$. The smallest such subgroup is $2\mathbb Z$.
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