Not every metric is induced from a norm I have studied that every normed space $(V, \lVert\cdot \lVert)$ is a metric space with respect to distance function
$d(u,v) = \lVert u - v \rVert$, $u,v \in V$. 
My question is whether every metric on a linear space can be induced by norm? I know answer is  no but I need proper justification. 
Edit: Is there any method to check whether a given metric space is induced by norm ?
Thanks for help

Personal Note:
Good question, I remember wondering the same thing myself back when I new to real analysis.  Here's a counter example:

Counter Example
\begin{equation}
d(x,y):=I_{\{(x,x)\}}(x,y):=\begin{cases}
1 &:\,  \,x=y\\
0 &:\,  \,x\neq y.
\end{cases}
\end{equation}
This is indeed a metric (check as an exercise).  
Where $I_A$ is the indicator function of the set $A$; where here the set $(x,x)\in V\times V$.  
If $r\in \mathbb{R}-\{0\}$, then $d(rx,ry)\in \{0,1\}$ hence $rd(x,y)\in \{0,r\}$ which is not in the range of $d:V\times V \rightarrow \mathbb{R}$.  
So, the metric $d$ fails the property that any metric induced by a norm must have, ie:
\begin{equation}
\mbox{it fails to have the property that: } \|rx-ry\|  =r\|x-y\|.
\end{equation}

Interpretation & Some Intuition:
The problem is that the topology is too fine, so all this topology can do is distinguish between things being the same or different.
As opposed to a norm topology which distinguised between object being, or not being on the same line (for some appropriate notion of line) (as well as them being different).  
Hope this helps :)
