Vector Space definition My book lists ten axioms that must hold for a set of objects (vectors) $V$ to be called a vector space. One of those axioms is:
$$1\vec{u} = \vec{u}$$
Is there a reason why this axiom must be on the list? What is its purpose, because it seems kind of obvious? Is there a case when a real scalar, 1, multiplied by a vector doesn't return that same vector?
 A: Yes, this axiom is necessary.
Given any vector space $(\mathbb{V}, +, \,\cdot\,)$ over a field $\mathbb{F}$, we can produce another algebraic structure $(\mathbb{V}, +, \odot)$ with "multiplication" map
$$\odot: \mathbb{F} \times \mathbb{V} \to \mathbb{V}$$
the zero map
$$f \odot \mathbf{u} := \mathbf{0}.$$
Then, $(\mathbb{V}, +, \odot)$ satisfies all of the vector space axioms except
$$1 \odot \mathbf{u} = \mathbf{u}.$$
A: In general, vector space $V$ can be over an arbitrary field $F$. And we can define a scalar multiplication between the element of $V$ and element of $F$.  
$1\vec{u} = \vec{u}$ means the multiplicative identity in the field is also the  identity of this scalar multiplication.
A: Let $V$ be the abelian group $F^n$ where $F$ Is a field. Define $av:=0$ for every $a\in F, v\in V$.
You can show that all the vector space axioms hold except the one you speak of. 
The main reason you want 1 to act as the identity is so that you can identify $F$ as a subring of the ring of linear transformations of $V$, sharing the same identity.
