I have a formula, which I have no idea how to solve, because I don't know that double vertical-line sign: $\|{\rm Ax} \|$?

$${\rm x} \ne 0 \in \Bbb R^n, \quad 0 < m \le \frac {\| {\rm Ax} \|} {\| {\rm x} \|} \le M, \quad cond(A) \le \frac M m .$$

What does it mean? How should I solve this?


migrated from mathematica.stackexchange.com Nov 29 '13 at 23:56

This question came from our site for users of Wolfram Mathematica.

  • 2
    $\begingroup$ vector norm. en.wikipedia.org/wiki/Norm_%28mathematics%29 How should I solve this What are solving for? what is the known and what is the unknown? $\endgroup$ – Nasser Nov 29 '13 at 23:22
  • $\begingroup$ @StefanSmith I don't think it "almost certainly refers to the Euclidean norm". Actually any vector norm will do. $\endgroup$ – user1551 Nov 30 '13 at 0:27
  • $\begingroup$ @user1551 : I removed my comment, but if someone writes $\|\mathbf{x}\|$, when $\mathbf{x}\in \mathbb{R}^n$, and they don't provide any additional context, then unless they are being deliberately uncooperative, they are using the Euclidean norm. Unfortunately, the OP did not give us enough info to be 100% sure what norm is being used here. (I will take your word for it that any norm works here and file it under "interesting facts to be investigated later"). $\endgroup$ – Stefan Smith Nov 30 '13 at 1:10

Double bars (or sometimes even single bars) tend to denote a norm in Mathematics. Most likely, the double bars here are denoting the Euclidean norm. This is just the length of the vector. So for example, the vector (I shall write it horizontally for compactness) $(1,2,3)$ has length $$ \|(1,2,3) \|=\sqrt{1^2+2^2+3^2}=\sqrt{14} $$ and the vector $$ \|(3,-1,2) \|=\sqrt{3^2+(-1)^2+2^2}=\sqrt{14} $$ Notice that $A\mathbf{x}$ is just a vector, so $\|A\mathbf{x}\|$ is just the length of the vector. $\|\mathbf{x}\|$ is just the length of $\mathbf{x}$. So here you are looking for scaling of $\mathbf{x}$ under transformation by $A$ to be between $m$ and $M$. (Look at $\frac{\|A\mathbf{x}\|}{\|\mathbf{x}\|}$ and think about what it means 'pictorially' to see what I am talking about).