# Prove equivalency of orthogonal transformation, $h(f(x),f(y))=h(x,y)$ and $f$ maps an orthonormal basis to another?

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.

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What definition of orthogonal are you working with? – Brett Frankel Apr 29 '12 at 22:07
Is $h$ the inner product? And what is the precise definition of orthogonal you are using? – Arturo Magidin Apr 29 '12 at 22:13
Yes h is the inner product, sorry I haven't been very clear. FGH – user30243 Apr 29 '12 at 22:14
i.e. b) is suggesting that the quadratic form of M is preserved. – user30243 Apr 29 '12 at 22:14
Yes: but there are many possible definitions of "orthogonal transformation" (you are giving two possibilities here!) In order to help you, one thing we need to know is what is your original definition of "orthogonal transformation". – Arturo Magidin Apr 29 '12 at 22:16

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 $f^*f = \mathrm{id}$.