Difference between metric and norm made concrete: The case of Euclid

This is a follow-up question on this one. The answers to my questions made things a lot clearer to me (Thank you for that!), yet there is some point that still bothers me.

This time I am making things more concrete: I am esp. interested in the difference between a metric and a norm. I understand that the metric gives the distance between two points as a real number. The norm gives the length of a a vector as a real number (see def. e.g. here). I further understand that all normed spaces are metric spaces (for a norm induces a metric) but not the other way around (please correct me if I am wrong).

Here I am only talking about vector spaces. As an example lets talk about Euclidean distance and Euclidean norm. Wikipedia says:

A vector can be described as a
directed line segment from the origin
of the Euclidean space (vector tail),
to a point in that space (vector tip).
If we consider that its length is
actually the distance from its tail to
its tip, it becomes clear that the
Euclidean norm of a vector is just a
special case of Euclidean distance:
the Euclidean distance between its
tail and its tip.

What confuses me is that they seem to be having it backwards: The Euclidean metric induces the Euclidean norm: You measure the distance between tip and tail and get the length out of that. What makes my confusion complete is that $L^2$ distance is also called the Euclidean norm (see here).

I would very much appreciate it if somebody could clear the haze.


The metric $d(u,v)$ induced by a vector space norm has additional properties that are not true of general metrics. These are:

Translation Invariance: $d(u+w,v+w)=d(u,v)$

Scaling Property: For any real number $t$, $d(tu,tv)=|t|d(u,v)$.

Conversely, if a metric has the above properties, then $d(u,0)$ is a norm.

More informally, the metric induced by a norm “plays nicely” with the vector space structure. The usual metric on $\mathbb{R}^n$ has the two properties mentioned above. But there are metrics on $\mathbb{R}^n$ that are topologically equivalent to the usual metric, but not translation invariant, and so are not induced by a norm.

Source : Link , Question Author : vonjd , Answer Author : André Nicolas

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