1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: F(x) = f(Ax) iff |x| = |y| => f(x) = f(y), how?
  1. F(x, y, z) ? (Replies: 2)

  2. F(x,y) = |x| + |y| (Replies: 23)

  3. Y' = f'(x) proof (Replies: 2)

  4. Limit of f(x,y) (Replies: 10)