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?

  • 4
    $\begingroup$ 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, 2013 at 23:22
  • 1
    $\begingroup$ @StefanSmith I don't think it "almost certainly refers to the Euclidean norm". Actually any vector norm will do. $\endgroup$
    – user1551
    Nov 30, 2013 at 0:27
  • 1
    $\begingroup$ @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"). $\endgroup$ Nov 30, 2013 at 1:10
  • $\begingroup$ I believe this can be made explicit by adding a subscript 2 for l2 norm (another term for Euclidean norm). $\endgroup$ Nov 13, 2021 at 2:57

1 Answer 1


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).

  • 1
    $\begingroup$ I understand that Ax in this example can be the vector (1,2 3), however I don't see what x would be in that example or more importantly what ||x|| is? $\endgroup$ Jan 17, 2020 at 17:39