Norm "maintaining" matrices Let $A$ be an $m\times n$ matrix such that $m < n$. I would like to know the conditions on $A$ such that the following is true:
$$\|Ax\| \leq \|Ay\| \implies \|x\| \leq \|y\|$$
It can easily be shown that if $\kappa(A)=1$ (condition number) then this property is satisfied. I am looking for the most general type of matrices that satisfy this condition.
Any help is much appreciated.
Thanks,
Phanindra
 A: EDIT: This answer mistakenly assumes that $A$ is a square $n \times n$ matrix.
I assume we're working over $\mathbb R$. We claim that $A$ must be a scalar multiple of an orthogonal matrix. 
First, we prove that if $\| x \| =  \| y \|$, then $\| A x \| = \| A y \|$. Towards a contradiction, assume that $\| x \| = \|y \|$ and $\| A x \| \neq \| A y \|$. Without loss of generality, we can assume that $\| Ax \| < \|A y \|$. Moreover, it is clear that both $x$ and $y$ are nonzero. (Why?) Now, fix a number $\beta$ such that $$
1 < \beta < \frac{\| A y \|}{\| Ax \|}.
$$
Then, defining $z = \beta x$, it is clear that


*

*$\| A z \| = \beta \| A x \| < \| A y \|$.

*$\| z \| = \beta \| x \| = \beta \| y \| > \| y \|$. 
This is a contradiction to the hypothesis (since $\| A z \| < \| A y \|$ but $\| z \| > \| y \|$). Hence, if $\| x \| = \| y \|$, then $\| A x \| = \| A y \|$. 
It now remains to show that $A$ is a multiple of an orthogonal matrix. Fix a unit vector $u$. Then since $\| x \| = \| (\| x \| u) \|$, it follows from (2.) that $\| A x \| = \| A (\| x \| u) \| = \| x \| \cdot \| A u \|$. Now, if $\| A u \| = 0$, then $A$ must be the zero matrix (why?) and we are already done. On the other hand, assuming $\| A u \| > 0$, it is easy to see that the matrix $$ B := \frac{1}{\| A u \|} A $$ is a linear isometry and hence orthogonal.  
A: You just want the SVD.  There exist unitary matrices U and V and a diagonal matrix Σ with non-negative real entries such that:$$A=U\Sigma V^* \qquad \|Ax\| = \|U\Sigma V^*x\| = \|\Sigma V^*x\|$$
So take x amongst the columns of V to get that all entries of Σ have equal value, so that A is more or less a scalar multiple of a unitary matrix, just possibly rank deficient since it is not square:
$$A = \lambda UV^*$$
Here you can require λ to be non-negative, but this is just absorbing complex scalars of absolute value 1 into the unitary matrices.
