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

[P3X-018]

Homework Statement

I have to show that for f:\mathbb{R}^k\rightarrow \mathbb{C} the following holds

f(x) = f(Ax)\qquad \Leftrightarrow \qquad \|x\| = \|y\|\quad \Rightarrow\quad f(x) = f(y)

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

The Attempt at a Solution

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

Assuming f(Ax) = f(x), then since the linear map A:\mathbb{R}^k\rightarrow \mathbb{R}^k is bijective, we can say that for every y\in\mathbb{R}^k there is an x\in\mathbb{R}^k such that Ax=y, since A is an orthonormal matrix we also have that \|y\|=\|Ax\| = \|x\|.
Now since f=f\circ A we have f(x) = f(Ax) = f(y)
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 \|x \| = \|y\| then there exists and orthonormal matrix A such that Ax = y? Is so then we already have that f(x) = f(y) and since f(y) = f(Ax), we have f(x) = f(Ax), and this ends the proof??

There is a hint saying that if [e_1,\ldots,e_k] and [f_1,\ldots,f_k] are 2 bases for R^k there exists only 1 nonsingular matrix A, such that Ae_i = f_i, 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?

 

[Dick]

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.