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Homework Help: F(x) = f(Ax) iff |x| = |y| => f(x) = f(y), how?

  1. Nov 4, 2007 #1
    1. The problem statement, all variables and given/known data
    I have to show that for [itex] f:\mathbb{R}^k\rightarrow \mathbb{C} [/itex] the following holds

    [tex] f(x) = f(Ax)\qquad \Leftrightarrow \qquad \|x\| = \|y\|\quad \Rightarrow\quad f(x) = f(y)[/tex]

    For every orthonormal n x n-matrices A and [itex] x,y\in\mathbb{R}^k[/itex]

    3. The attempt at a solution

    This problems seems kinda trivial but I can't seem to show this rigorously.

    Assuming [itex] f(Ax) = f(x) [/itex], then since the linear map [itex] A:\mathbb{R}^k\rightarrow \mathbb{R}^k [/itex] is bijective, we can say that for every [itex] y\in\mathbb{R}^k [/itex] there is an [itex] x\in\mathbb{R}^k [/itex] such that [itex] Ax=y [/itex], since A is an orthonormal matrix we also have that [itex] \|y\|=\|Ax\| = \|x\| [/itex].
    Now since [itex] f=f\circ A [/itex] we have [itex] f(x) = f(Ax) = f(y) [/itex]
    So this proves the implication to the right.

    To prove the implication to the left, can I then argue the same way saying that if [itex] \|x \| = \|y\| [/itex] then there exists and orthonormal matrix A such that [itex] Ax = y [/itex]? Is so then we already have that [itex] f(x) = f(y) [/itex] and since [itex] f(y) = f(Ax) [/itex], we have [itex] f(x) = f(Ax) [/itex], and this ends the proof??

    There is a hint saying that if [itex] [e_1,\ldots,e_k] [/itex] and [itex] [f_1,\ldots,f_k] [/itex] are 2 bases for R^k there exists only 1 nonsingular matrix A, such that [itex] Ae_i = f_i [/itex], i = 1,..,k. And if both bases are orthonormal then A is orthonormal.
    How do I use this hint? Or did I use it while talking about the existence of y = Ax?
    Last edited: Nov 4, 2007
  2. jcsd
  3. Nov 4, 2007 #2


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    You use the hint to show that if |x|=|y| then there is an A such that Ax=y. Construct your orthonormal bases such that c*e_1=x and c*f_1=y for some constant c.
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