Reason for $(0,1)$ tensor as a vector? Assume, we have a finite-dimensional vector space $V = (R^3,+,-)$ over the field $(R,+,-)$ and the dual space is the set of all the linear maps, $V^* = \{f:R^3 \to R| f \text{ is linear}\}$. We define $(r,s)$ tensor as multilinear map: $V \times ... \times V(\text{r times})\times V^* \times ... \times V^*(\text{s times}) \to R$.   I understand that we can view the dual space, $V^{*}$ as a $(1,0)$ tensor since it maps $R^3 \to R$. But how can we view $(0,1)$ tensor as a vector since a $(0,1)$ tensor will be a map between $V^* \to R$.
 A: As the vector space $V$ is finite-dimensional, there is an isomorphism $\Phi:V\to V^{**}$, given by
$$[\Phi(v)](f)=f(v)$$
for $v\in V$ and $f\in V^*$.
A: One can construct maps $V^*\to R$ by evaluation of each linear functional $f$ in a fixed element $v_0$ of $V$, that is
$$f\mapsto f(v_0),$$
such a map is linear, since for a linear combination $Af+Bg$ in $V^*$
one can see
$$(Af+Bg)(v_0)=Af(v_0)+Bg(v_0),$$
where $A,B\in R$.
Taking different $v_0$'s and if you call them $\Phi_{v_0}$ those maps above, you can see that the set
$\{\Phi_{v_0}\}$ is a vector space under the operations

*

*$\Phi_{v_1}+\Phi_{v_2}=\Phi_{v_1+v_2}$ and

*$k\Phi_{v_0}=\Phi_{kv_0}$ for $k\in R$.

A basis for this space is got by considering a basis for $V$: If $\{b_1,..., b_n\}$ is such one then
$$\{\Phi_{b_1},..., \Phi_{b_n}\},$$
will give you a basis for this vector space.
A: I'm a bit surprised by the terminology you're using.
In general, a tensor product of vector spaces $A,B$ is a vector space $A \otimes B$ together with a bilinear map $\Phi \colon A \times B \to A \otimes B$ that satisfies the universal property
$$
\forall \ \text{bilinear } \Psi \colon A \times B \to C \quad \exists ! \ \text{linear } \overline{\Psi} \colon A \otimes B \to C 
\qquad \text{such that } \Psi = \overline{\Psi} \circ \Phi.
$$
One can check that such a space exists, and all possible choices are isomorphic. In this sense, we can call $A \otimes B$ the tensor product of $A$ and $B$.
In a similar way, one defines the tensor product of any family $A_i$ of vector spaces by considering multilinear maps.

Let $V$ be now a real vector space. The space of all $(r,s)$-tensors - as you describe it - is the space of all multilinear maps
$$
V \times ... \times V(\text{r times})\times V^* \times ... \times V^*(\text{s times}) \to \mathbb{R}.
$$
According to the definition, this is isomorphic to the space of linear maps
$$
V \otimes ... \otimes V(\text{r times})\otimes V^* \otimes ... \otimes V^*(\text{s times}) \to \mathbb{R},
$$
or in other words, to the dual space
$$
(V \otimes ... \otimes V(\text{r times})\otimes V^* \otimes ... \otimes V^*(\text{s times}))^*,
$$
which in turn is isomorphic to
$$
V^* \otimes ... \otimes V^*(\text{r times})\otimes V^{**} \otimes ... \otimes V^{**}(\text{s times}).
$$
In this context, the space of $(0,1)$-tensors is $V^{**}$, which in the special case of finite dimensional spaces is naturally isomorphic to $V$.
However, if we stick to the definition of tensor product, then we see that the tensor product of $V$ is just $V$ ($1$-linear maps are just linear maps). Your definition actually involves taking the dual space twice, which of course makes no difference as long as $V^{**}=V$.
