Q:

Maps from $V\otimes V^*$ to the underlying field

Let $V\in \operatorname{Vect}(k)$ I was asked to consider if the evaluation map was the only map from  $V^*\otimes V$ to $k$.  My question is, is it correct for me to rephrase this question to be, can you find a map $\phi:V^*\otimes V \to k$ that is independent of any choice of basis?   An answer to my rephrased question is something I think I can solve. 

A:

Maps $V^*\otimes V \to k$ are the same thing as elements of $(V^*\otimes V)^*$. Since this is not the zero vector space (assuming $V \neq 0$), it has more than one element.  If your field is infinite, say $k = \mathbb R$ or $k = \mathbb C$, then $(V^*\otimes V)^*$ is also infinite.  This proves that the evaluation map is not the only map. 