# 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.

• 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 $$-10^{\circ}\times 5=-50^{\circ}$$

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.
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?