$V$ is a real vector-space and $f: V \rightarrow V $ an automorphism. Show that $f$ is an isometry iff $f^* = f^{-1}$.
This has been explained in our textbook for finite-dimensional situations however the exam question demands proof without using the textbook explanations which has left me stumped on how to approach this. Does anyone have a tip where to start with this proof? How does the dimension affect this so dramatically?
EDIT: For the ones asking what the course note proof is:
$\langle v, f^{*}(v) \rangle = \langle f(v), v \rangle = \langle f(v), f(f^{-1}(v))\rangle = \langle v, f^{-1}(v)\rangle$
Which implies $f^{*}=f^{-1}$
Assuming that $V$ is an inner product space with a norm defined by $\lVert x \rVert = \sqrt{\langle x, x\rangle}$, we can proceed as follows.
Firstly, assume $f^*=f^{-1}$. Then for any $x \in V$, $$\lVert fx \rVert^2 = \langle fx, fx\rangle = \langle f^*fx, x\rangle = \langle x, x\rangle = \lVert x \rVert^2. $$ Secondly suppose that $f$ is an isometry. Then it preserves the inner product (since $\langle x, y \rangle = \frac{1}{2}\left(\lVert x + y\rVert^2 - \lVert x\lVert^2 - \lVert y\lVert ^2\right)$. Therefore, $$\langle x,y \rangle = \langle fx, fy \rangle=\langle f^*fx,y\rangle,$$ for all $x,y \in V$, from which it follows that $f^*f = \operatorname{id}$, thus $f^*=f^{-1}$. Note that the last step assumes that $f$ we already know $f$ is invertible.
If we do not assume that $f$ was invertible, and if the vector space is infinitedimensional, we cannot conclude from $f^*f=\operatorname{id}$ that $f$ is invertible.
I don't know in what sense you could consider this proof different from your textbook proof. It's certainly longer and very similar.