Tensor Products Acting on Vectors Is this always (if at all) true, and if so why? 
$\nu\otimes\omega(v,w)=\nu(v)\omega(w)$
Where $\nu$ and $\omega$ are covectors, and $v$ and $w$ vectors. 
Please tell me if the question or notation is unclear. Thanks! 
 A: Let me give a kind of example:
Asertion: If $V$ is finite dimensional vector space over the field $F$, there is an isomorphism 
$$\hom(V,V)\to V\otimes V^*.$$
The key step is fixing basis. 
Let $V={\rm gen}\{b_1,...,b_n\}$ and $V^*={\rm gen}\{\beta^1,...,\beta^n\}$. We can choose the basis requiring $\beta^k(b_l)=\delta^k{}_l$.
The space $\hom(V,V)$ consists of linear transformations $V\to V$ that can be thought as the matrices who spring in the linear combinations (supposing that $n=3$)
$$Tb_1=T^1{}_1b_1+T^2{}_1b_2+T^3{}_1b_3,$$
$$Tb_2=T^1{}_2b_1+T^2{}_2b_2+T^3{}_2b_3,$$
$$Tb_3=T^1{}_3b_1+T^2{}_3b_2+T^3{}_3b_3,$$
or briefly, à la Einstein-Penrose
$$Tb_k=T^s{}_kb_s\qquad\mbox{$s$- summation}.$$
Let $A=[T^i{}_j]$ be the matrix associated.
In another hand, the space $V\otimes V^*$ can be thought as the vector space span by the basic objects $b_i\otimes\beta^j$ which can be considered as bilinear maps $V^*\times V\to F$ working as
$$b_i\otimes\beta^j(f,v)=f(b_i)\beta^j(v).$$
So a generic element is characterized as a bi-indexed linear combination
$Q^{\mu}{}_{\nu}b_{\mu}\otimes\beta^{\nu}$.
With them now we can map
$$\Phi(T)=T^i{}_jb_i\otimes\beta^j.$$
That this map, $\Phi$, is a linear transformation and is injective is explained when is completed:
1) We have: $\Phi(aT+cS)=a\Phi(T)+c\Phi(S)$ for any pair of scalar $a,c$ and each pair of linear transformations $T,S$.
2) The kernel of $\Phi$ is trivial: We calculate 
$$\Phi(T)=\stackrel{\to}0,$$ 
and this is
iff 
$$T^i{}_jb_i\otimes\beta^j=\stackrel{\to}0$$
iff for any pair $(f,v)$ we have
$$T^i{}_jb_i\otimes\beta^j(f,v)=0.$$
But choosing basics $f=\beta^k$ and $v=b_l$ we get
$$T^i{}_jb_i\otimes\beta^j(\beta^k,b_l)=0.$$
This is 
$$T^i{}_j\beta^k(b_i)\beta^j(b_l)=0.$$
$$T^i{}_j\delta^k{}_i\delta^j{}_l=T^k{}_l=0$$
for all election $k,l$ so the linear map $T=\stackrel{\to}0$.
All this implies that $\Phi$ is an isomorphism.      
