Why are vectors considered to be rank (0,1) tensors and dual vectors considered to be rank (1,0) tensors? Sean Carrol in his book of general relativity, he defines a tensor to be a multilinear map from a collection of dual vectors and vectors to $\mathbb{R}$:
$T:T^*_p \times...\times T^*_p \times T_p \times...\times T_p \to \mathbb{R}$.
He then goes on to say that a scalar is a type $(0,0)$ tensor, a vector is a type $(1,0)$ tensor and a dual vector is a type $(0,1)$. However, I thought that dual vectors are elements of a dual space, $T^*_p$, so shouldn't it be that dual vectors are rank $(1,0)$ tensors and vectors rank $(0,1)$ tensors?
 A: Given a vector $v$ and a dual vector $f$, you can produce a scalar $f(v)$. This can be viewed as a map $v \mapsto f(v)$ or $f \mapsto f(v)$.
So a vector determines a (linear) map from the space of dual vectors to scalars (i.e. a $(1,0)$-tensor since we have 1 dual vector input and no vector inputs). Likewise a dual vector determines a (linear) map from the space of vectors to scalars (i.e. a $(0,1)$-tensor since we have no dual vector inputs and 1 vector input).
A: We say T is a $(a,b)$ tensor $T:T^*_p \times...\times T^*_p \times T_p \times...\times T_p \to \mathbb{R}$ where $a$ is the number of $T^*_p$s and $b$ is the number of $T_p$s.
A dual vector is an element of $T_p^*$, which means that it gives a map $T_p\rightarrow \mathbb R$, and so it's a $(0,1)$ tensor.
A vector is an element of $T_p$, which means that it gives a map $T_p^*\rightarrow \mathbb R$, and so it's a $(1,0)$ tensor. (The map is the "evaluation map" mentioned by Bill Cook.)
A: Why is a scalar function written like f(x,y)? Because it has 2 inputs and spits out a scalar. Why is a vector a (1,0)tensor? It has 1 input (being 1 co-vector and 0 vectors) and spits out a scalar. That's all ;-)
