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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

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    @user1551: You are right. Thank you for the answer.2011-11-03

2 Answers 2

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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.

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    @jpv Thanks. I will update the answer if the answer is salvageable at all.2011-11-02
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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.

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    This assumes you have m>n, the opposite of what your question says. As @user1551 points out, with m, Σ has to have some diagonal entries being 0.2011-11-02