I understand that a norm is defined as: $$||\space.||:V\to \mathbb{R}$$ which is a notion of distance defined on a vector space to give the magnitude of a vector (distance from the origin). Is it true that this distance can be any type of distance you want to define it to be? For example, taking the norm to be the Manhattan (Taxi Metric) distance, where $$||x||=\sum_{i=1}^{n}{|x_{i}|^{2}} \text{ where }x\in\mathbb{R}^{n}$$ or you could simply define the norm to be the absolute value of the first component of $x$.

I also understand that distance is a function $$d:V\times V\to \mathbb{R}$$ which measures between two points. So as the Euclidean distance is defined to be: $$dist(x,y)=\bigg(\sum_{i=1}^{n}{(x_{i}-y_{i})^2}\bigg)^{\frac{1}{2}}$$ and the Euclidean Norm is defined to be $$|x|=\bigg(\sum_{i=1}^{n}{x_{i}^2}\bigg)^{\frac{1}{2}}$$ then the norm is simply the Euclidean distance between a vector $x$ and $0$ so can you essentially think of them as the same thing? Is it true to say that a norm is a type of distance? Or would it be the other way around to say that the distance is a type of norm?


  • $\begingroup$ The norm is the distance from the origin, and the distance is the norm of the difference. $\endgroup$ – Berci Oct 21 '18 at 10:35
  • 2
    $\begingroup$ It is a bit delicate: If you have a norm, then you can define a distance as $d(x,y) = \|x-y\|$. The converse is NOT true, meaning that if you have a distance $d$, the function $|x| = d(x,0)$ is NOT always a norm $\endgroup$ – Exodd Oct 21 '18 at 10:37
  • $\begingroup$ I think your second displayed formula (for the Manhattan distance) is wrong. At least, as the Manhattan distance is usually understood. $\endgroup$ – kimchi lover Oct 21 '18 at 13:05

Firstly, note that norms only can be defined in vector spaces, but every metric space has a distance function. If we only talk about vector spaces, every norm determines a metric, and some metrics determine a norm.

For all norms $\lVert \cdot\rVert$ in vector space $V$, the function $d(x,y) = \lVert x-y \rVert$ is a distance function on $V$ (which satisfies the conditions of distance function. see here).

For all distance functions $d(\cdot, \cdot)$ in vector space $V$, the function $\lVert x\rVert=d(x, 0)$ is a norm on vector space $V$, if following conditions are satisfied (see here):

  • $d(x,y)=d(x+a, y+a)$ (translation invariance)
  • $d(cx,cy)=\lvert c\rvert d(x,y)$ (homogenity)

So, because the Euclidean distance function is homogenous and translation-invariant, it determines a norm on $\mathbb{R}^n$.

But, for example suppose the discrete distance function:

$d^\prime(x,y)=\begin{cases}1 & \text{if}\ x\ne y\\0 & \text{if}\ x= y\end{cases}$,

which can be shown that is translation invariant but not homogenous ( $\forall x\ne y\ d(2x, 2y) = 1 \ne 2 = 2d(x,y)$ ); then the function: $f(x)=d^\prime(x,0)$ is not a norm function, because $f(cx)\ne\lvert c\rvert f(x)$ (see the properties of norm function here).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.