This Wikipedia article says that tensors can be defined as miltilinear maps or be defined using tensor products. Could anybody explain with a simple example why these two approaches give the same objects?
[Added:] In Abstract Algebra by Dummit and Foote, we have the following theorem,
Let $V$ be a vector space over $\mathbb{R}$ and consider $L=\mathbb{R}$, $M_1=V$, $M_2=V^*$. Then every multilinear map $\varphi:V^*\times V\to\mathbb{R}$ is corresponding to a homomorphism $\Phi: V^*\otimes V\to\mathbb{R}$. According to one of the approaches in the linked Wikipedia article, $\varphi$ is a $(1,1)$-tensor. On the other hand, an element of the tensor product $V^*\otimes V$ is also called a tensor. Does $\varphi$ also corresponds to an element of $V^*\otimes V$? Does one need to identify $V^*\otimes V$ with its dual to have such correspondence?
For example taking two linear maps $f,g:\Bbb R^n\to\Bbb R$ one can construct a bilinear map $f\otimes g:\Bbb R^n\times\Bbb R^n\to\Bbb R$ via $$f\otimes g(v,w)=f(v)g(w).$$ Attached to $f\otimes g$ there is a matrix associated by evaluating $$f\otimes g(b_i,b_j),$$ that is $$\left(\begin{array}{cccc} f(b_1)g(b_1)&f(b_1)g(b_2)&...&f(b_1)g(b_n)\\ f(b_2)g(b_1)&f(b_2)g(b_2)&...&f(b_2)g(b_n)\\ \\ f(b_n)g(b_1)&f(b_n)g(b_2)&...&f(b_n)g(b_n)\\ \end{array}\right),$$ and where the $\{b_i\}$ are basis vectors for $\Bbb R^n$.
An important consequence is the alternating bilinear maps or a.k.a. bi-vectors, 2-vectors, 2-forms... are the maps $f\wedge g:\Bbb R^n\times\Bbb R^n\to\Bbb R$ given by \begin{eqnarray*} f\wedge g(v,w)&=&(f\wedge g-g\wedge f)(v,w)\\ &=&f(v)g(w)-g(v)f(w)\\ &=& \det\left[\begin{array}{cc} f(v)& g(v)\\ f(w)& g(w) \end{array}\right] \end{eqnarray*} are natural to describe.