Norm of the one dimension real space I read that apart from a constant factor the absolute value is the only norm on the vector space $\mathbb{R}^1$. Why is that so?
 A: This is a little more general perspective that I hope will be helpful with your intuition about norms. From this perspective the answer to your $1$D question will be obvious.
For every norm $\lVert\cdot\rVert$, there is a "unit ball" consisting of all the points that have norm less than or equal to $1$:
$$B = \{x : \lVert x\rVert \leq 1\}$$
In fact, this unit ball contains all the information there is to know about the norm.
Why? Well, if you want to find the norm of a vector, all you have to do is uniformly scale the unit ball up until it just barely touches the vector, then that scaling factor is the norm of the vector. This follows from the scaling property of norms. (See Minkowski functional for this statement in more technical wording.) 
This leads to a natural question:
Question: What sort of shapes can a unit ball take?
Answer: The unit ball must be a balanced, convex, absorbing set. 
This means,


*

*Balanced: the set is symmetric around zero, in the sense that if $u$ is in the ball, $-u$ must be too. This requirement comes from the scaling requirement for norms, using scaling factor $-1$.

*Convex: the set does not have any holes or "dents" in it, in the sense that for any two points in the set, the line segment between the points is also in the set. This requirement comes from the triangle inequality (after a bit of work).

*(strictly) Absorbing: the set can be uniformly expanded to contain any vector in the space. Also the set can be shrunk so that it it doesn't contain any particular point (except zero). This requirement again comes from the scaling law.


So this means unit balls can be things like spheres centered at zero, cubes centered at zero, ellipsoids centered at zero, and things like that.

Now in $1$D, the only convex sets are line segments, and the only balanced convex sets are line segments centered at zero. Furthermore, the absorbing requirement prevents the "degenerate" cases where a line segment is the single point at zero, or the whole real line. 
Thus the only possible unit balls in $1$D are line segments that are centered at zero, and the only distinction possible between these unit balls is a constant scaling factor.
A: Let $V$ be a Vector Space over the field $F$ where $F$ is a subfield of $\Bbb C$. One of the axioms that specify a Norm on a space is

*

*If $a \in F$ then $||a \cdot \underline v|| = |a| \cdot ||\underline v||$ for every $\underline v \in V$
Now consider the cases when $a = 1$ and $a = -1$. See how this related to the definition of the Absolute Value Function in $\Bbb R$.
