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

1. Nov 4, 2007

### P3X-018

1. The problem statement, all variables and given/known data
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 $x,y\in\mathbb{R}^k$

3. 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?

Last edited: Nov 4, 2007
2. Nov 4, 2007

### 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.