Inverse of nth power of a linear transformation 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.  


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*First rotate, then invert 


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*Rotating $5$ times gives $10^{\circ}\times 5=50^{\circ}$.   

*Taking the inverse gives $-50^{\circ}$


*First invert, then invert 


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*Inverting gives $-10^{\circ}$.   

*Rotating $5$ times gives $-10^{\circ}\times 5=-50^{\circ}$

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?
