My question is very simple: A type $(m,n)$ tensor is an element of $V^{\otimes m}\otimes (V^*)^{\otimes n}$. Is there a reason/motivation, beyond more general definitions, to consider the dual space of $V$ in this definition? Or is it just convention?
I don't expect some mindblowing answer, so to speak, but maybe a clarification of the why the tensor product of $V$ with it's dual would be so interesting.
On
Let $V$ a finite dimensional vector space and let $e:=(e_1, \cdots e_n)$ one of its bases. Let $\tilde{e}$ another bases for $V$. The two bases are related to each other by a linear transformation. i.e. there is a $n \times n$ matrix $A$ such that $$ \tilde{e}=Ae $$ or $\tilde{e_i}=\sum_j A^j_i e_j$ Consider now the dual cobases of $V^*$ i.e. $e^*:=({e^*}^1, \cdots {e^*}^n)$ such that $$ {e^*}^i(e_j)= \delta_j^i $$ Let $B$ the matrix of the change of bases of the dual bases, we have $$ \operatorname{Id}_n= \tilde{e^*}(\tilde{e})= B e^*(A e)=BA e^*(e)=BA \operatorname{Id}=BA $$ so $B=A^{-1}$
A bases for the space of $(1,1)$ tensors is given by $$ e_i \otimes {e^*}^j $$ for $i,j \in \{1, \cdots n\}$. So it change bases as $$ \tilde{e_i} \otimes \tilde{{e^*}^j}=\sum_k \sum_h A^k_i (A^{-1})^j_h e_k \otimes {e^*}^h $$ while e.g a $(2,0)$ tensor change bases as $$ \tilde{e_i} \otimes \tilde{e_j}=\sum_k \sum_h A^k_i A^h_i e_k \otimes e_h $$ This is quite used in physics
As you probably know, for finite-dimensional spaces, $V$ is isomorphic to $V^\ast$ but such isomorphisms are not canonical. It is often important to work with canonical isomorphisms, for instance when doing vector bundles over manifolds. On the other hand, there are certain canonical isomorphisms such as that between $Hom(V,V)$ and $V^\ast\otimes V$, a fact important for example in Riemannian geometry. That's why one needs to work with the more general tensor products that you mentioned.