P3X-018
Nov4-07, 08:42 AM
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?
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?