Reason for defining a vector as a $(1,0)$ tensor I'm reading Jeevanjee's "An Introduction to Tensors and Group Theory for Physicists" and have come across the following definition of a $(r,s)$ tensor: a map $T:V\times\ldots\times V$ ($r$ times) $\times V^* \times\ldots\times V^*$ ($s$ times) $\to \mathbb{C}$ that is linear in each argument. I can understand that dual vectors are $(1,0)$ tensors, but

vectors can be viewed as $(0,1)$ tensors as follows:
  $$v(f) \equiv f(v)$$where $v \in V$ and $f \in V^*$.

What's the intuition or reasoning behind this roundabout definition? Why forcibly define a vector as a tensor? I guess I'm confused because I'm very much used to the idea of a dual vector as something that eats a vector and outputs a scalar. What's the need to turn it around and define a vector as something that eats a dual vector? Does this have anything to do with the equivalence between a vector and its double dual counterpart, and we're just implicitly assuming the element from $V^{**}$ that corresponds to $v$ as $v$ itself?
 A: You are right in your suspicion that this works only for finite-dimensional $V$, where the natural map $V \rightarrow V^{**}$ is an isomorphism. In general, this map is injective, but $V^{**}$ is much bigger.
Concerning the profit of this definition, have a look at the construction of rational numbers $\mathbb Q$ from integers $\mathbb Z$ - we then say that an integer is now a rational number of the form ${n\over 1}$, that is, integers get naturally embedded into rationals, which is extremely useful. The same happens here: we define tensors in terms of vectors and linear forms (dual vectors), but then it happens that both vectors and forms are instances of tensors. Thus, you can talk about tensors in general, and this will apply to vectors, too (e.g. define tensor fields, Lie or covariant derivatives, etc).

To be honest, I consider this definition of tensors (as multilinear maps) somewhat...outdated. A slightly more modern approach talks about tensor products of vector spaces $V \otimes U$, and then a tensor over a vector space $V$ of covariant degree $n$ and contravariant degree $m$ is simply an element of $V \otimes \dots \otimes V \otimes V^* \otimes \dots \otimes V^*$ (where $V$ appears $m$ times and $V^*$ appears $n$ times). Thus, if $n=0$ and $m=1$, we get simply $V$ - the original space of vectors.
