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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.

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    What definition of orthogonal are you working with?2012-04-29
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    Is $h$ the inner product? And what is the precise definition of orthogonal you are using?2012-04-29
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    Yes h is the inner product, sorry I haven't been very clear. FGH2012-04-29
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    i.e. b) is suggesting that the quadratic form of M is preserved.2012-04-29
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    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".2012-04-29
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    Orthogonal: I haven't been given a precise definition, the only definition that has been given isAn orthogonal transformation f between two orthonormal bases in M is a linear transformation f:M→M such that h(fx,fy)=h(x,y) for all vectors x,y∈M.2012-04-29
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    But is that not what we are trying to prove? Sorry! I'm confused.2012-04-29
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    to prove a) to b) do I compute f(x+y).f(x+y)-f(x-y).f(x-y)?2012-04-29
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    Might it be this? Two nonzero null vectors x and y in M are orthogonal iff they are parallel. i.e. iff there is a t in Real Numbers s.t. x=ty2012-04-29
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    That can't be the original definition you were given, because then (a) and (b) would be equivalent "by definition". Perhaps the definition was $f^*=f^{-1}$ ? Check your notes/textbook carefully, especially the part before you get to this result.2012-04-29
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    I think I have found it: If h is an inner product on V, then two vectors m and n for which h(m,n)=0 are said to be orthogonal is there is no ambiguity as to which inner product is intended.2012-04-29
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    @user30243: The definition of "orthogonal vectors" is not at issue. The question is the definition of orthogonal **transformation**.2012-04-29
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    It doesn't seem to explicitly say! The last thing that it could be is: If {e1,e2,e3,e4} and {e*1,e*2,e*3,e*4} are two orthonormal bases for M then there is a unique linear transformation f:M->M s.t. f(eA)=e*A for each A=1,2,3,4.2012-04-29
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    No, that's not a definition of orthogonal transformation. What is your book, and where did you get the problem you are looking at?2012-04-29
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    The Geometry of Minkowski Spacetime by Gregory L. Naber, Springer-Verlag 19912012-04-29

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