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
 A: 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 :)
A: Let $V$ be a vector space over the field $\mathbb{F}$. A norm
$$\| \cdot \|: V \longrightarrow \mathbb{F}$$
on $V$ satisfies the homogeneity condition
$$\|ax\| = |a| \cdot \|x\|$$
for all $a \in \mathbb{F}$ and $x \in V$. So the metric
$$d: V \times V \longrightarrow \mathbb{F},$$
$$d(x,y) = \|x - y\|$$
defined by the norm is such that
$$d(ax,ay) = \|ax - ay\| = |a| \cdot \|x - y\| = |a| d(x,y)$$
for all $a \in \mathbb{F}$ and $x,y \in V$. This property is not satisfied by general metrics. For example, let $\delta$ be the discrete metric
$$\delta(x,y) = \begin{cases} 1, & x \neq y, \\ 0, & x = y. \end{cases}$$
Then $\delta$ clearly does not satisfy the homogeneity property of the a metric induced by a norm.

To answer your edit, call a metric
$$d: V \times V \longrightarrow \mathbb{F}$$
homogeneous if
$$d(ax, ay) = |a| d(x,y)$$
for all $a \in \mathbb{F}$ and $x,y \in V$, and translation invariant if
$$d(x + z, y + z) = d(x,y)$$
for all $x, y, z \in V$. Then a homogeneous, translation invariant metric $d$ can be used to define a norm $\| \cdot \|$ by
$$\|x\| = d(x,0)$$
for all $x \in V$.
A: One possible way of showing that a metric does not arise from a norm is to show that it is bounded, as it then cannot be homogeneous.

Proof:
Take $d$ a bounded metric on a space $X$: i.e. $\exists D \in \mathbb{R}_+$ such that $\forall (x,y) \in X^2, d(x,y) \leq D$. Suppose now for contradiction that $d$ arises from a norm, i.e. the exists a norm $ \|\cdot\|$ such that $d(x,y) = \|x - y\|$. Recall that the distance must then must be homogeneous, for we have $d(\lambda x, \lambda y) = \|\lambda x - \lambda y\| = |\lambda| \cdot \|x - y\| = |\lambda| d(x,y)$.
Take now  $(x_0, y_0) \in X\times X$ such that $d(x_0, y_0)\neq0$ , then we must have that $d\left( \frac{D+1}{d(x_0, y_0)} \cdot x_0, \frac{D+1}{d(x_0, y_0)} \cdot y_0 \right) = \frac{D+1}{d(x_0, y_0)} \cdot d(x_0, y_0) = D + 1 > D $ which contradicts the upper bound of the metric.
A: Here is another interesting example: Let $|x-y|$ denote the usual Euclidean distance between two real numbers $x$ and $y$. Let $d(x,y)=\min\{|x-y|,1\}$, the standard derived bounded metric. Now suppose we look at $\Bbb{R}$ as a vector space over itself and ask whether $d$ comes from any norm on $\Bbb{R}$. Then if there is such a norm say $||.||$, we must have the homogeneity condition: for any $\alpha \in \Bbb{R}$ and any $v \in \Bbb{R}$, 
$$||\alpha v || = |\alpha| ||v||.$$
But now we have a problem: The metric $d$ is obviously bounded by $1$, but we can take $\alpha$ arbitrarily large so that $||.||$ is unbounded. It follows that $d$ does not come from any norm.
A: As Henry states above, metrics induced by a norm must be homogeneous. You can see that they must also be translation invariant: $d(x+a,y+a)= d(x,y).$ So any metric not satisfying either of those can not come from a norm.
On the other hand, it turns out that these two conditions on the metric are sufficient to define a norm that induces that metric: $d(x,0)=\| x \|.$ 
A: Every homogeneous metric induces a norm via:
$$\|x\|:=d(x,0)$$
and every norm induces a homogeneous and translation-invariant metric:
$$d(x,y):=\|x-y\|$$
The clue herein lies in wether the induced norm really represents the metric as:
$$d(\cdot,\cdot)\to\|\cdot\|\to d(\cdot,\cdot)$$
which is the case only for the translation-invariant metrics:
$$d'(x,y)=d(x-y,0)=d(x,y)$$
whereas the induced metric always represents the norm as:
$$\|\cdot\|\to d(\cdot,\cdot)\to\|\cdot\|$$
as a simple check shows:
$$\|x\|'=\|x-0\|=\|x\|$$
