Prove equivalency of orthogonal transformation, $h(f(x),f(y))=h(x,y)$ and $f$ maps an orthonormal basis to another? Please help! How do I go about proving this please? 
Let $f: M \longrightarrow M$ be a linear transformation. Then the following are equivalent:
a) $f$ is an orthogonal transformation.
b) $h(f(x),f(y))=h(x,y)$.
c) $f$ takes any orthonormal basis for $M$ onto another orthonormal basis for $M$.
Do I use Gram-Schmidt Process?
Thanks, FGH
Sorry, should have specified M is Minkowski Space-time.
 A: I will use the following common definition of "orthogonal transformation": a transformation $f$ is orthogonal if and only if $f^*=f^{-1}$, where $f^*$ is the unique transformation such that for all $x$ and $y$, $h(x,f(y)) = h(f^*(x),y)$.
To go from (i) to (ii), note that $h(f(x),f(y)) = h(f^*(f(x)),y) = h(f^{-1}(f(x)),y) = h(x,y)$.
To go from (ii) to (iii), note that if $\{u_i\}$ is orthonormal, then so is $\{f(u_i)\}$: because $h(f(u_i),f(u_j)) = h(u_i,u_j) = \delta_{ij}$ by assumption. Also, $f$ is one-to-one, since $f(x)=\mathbf{0}$ implies $0 = h(f(x),f(x)) = h(x,x)$, so $x=\mathbf{0}$. Since your vector space is finite dimensional, this implies $f$ is invertible. If $x$ is orthogonal to all $f(u_i)$, then $h(u_i,f^{-1}(x)) = h(f(u_i),x)=0$ for all $u_i$, so $f^{-1}(x)=\mathbf{0}$ (since the $u_i$ are an orthonormal basis), so $x=\mathbf{0}$. Thus, $\{f(u_i)\}$ are a maximal orthonormal set, hence an orthonormal basis.
To go from (iii) to (i), let $\{e_i\}$ be the an orthonormal basis. For each $i$, we have
$$h(f^*f(e_i),e_j) = h(f(e_i),f(e_j)) = \delta_{ij}$$
since $f$ takes orthonormal bases to orthonormal bases. Therefore, we have
$$f^*f(e_i) = \sum_j(f^*f(e_i),e_j)e_j  = e_i$$
for each $i$, so $f^*f = \mathrm{id}$.
