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Using matrices it is easy to show that doing a linear transformation n times and then taking inverse is same as inverting the linear transformation and then doing it n times: $$(A^n)^{-1}=(AAAA\ldots)^{-1} = \ldots A^{-1}A^{-1}A^{-1}A^{-1}=(A^{-1})^n~~\blacksquare$$

I'm wondering if this can be shown with out reference to matrices, that is by just using linearity properties like f(ax+by)=af(x)+bf(y)?


If what I'm asking is not clear, please consider rotation by $10^{\circ}$ as an example.

  • First rotate, then invert
    • Rotating $5$ times gives $10^{\circ}\times 5=50^{\circ}$.
    • Taking the inverse gives $-50^{\circ}$
  • First invert, then invert
    • Inverting gives $-10^{\circ}$.
    • Rotating $5$ times gives $-10^{\circ}\times 5=-50^{\circ}$
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2 Answers 2

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If $f$ is an invertible map, then$$f^n\circ(f^{-1})^n=f\circ f\circ\cdots\circ\overbrace{f\circ f^{-1}}^{=\operatorname{Id}}\circ f^{-1}\circ\cdots\circ f^{-1}=\operatorname{Id}.$$The fact that $f$ is linear is not relevant.

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You can prove it by induction: for $n=1$ it’s banal. Suppose thesis true for a certain $n$. Then $(f^{n+1})^{-1}=(f\circ f^n)^{-1}=(f^n)^{-1}\circ f^{-1}=(f^{-1})^n\circ f^{-1}=(f^{-1})^{n+1}$ Did you asked for a similar demonstration?

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