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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.

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    I'm a little confused here. A linear map is always independent of a choice of basis (it's just the matrix that depends on it). – 2017-01-24
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    @manthanomen Well, to consider what the evaluation map does to elements of $V^*\otimes V$ you might consider what it does to the basis elements, i.e. the pure tensors. Sorry if this isn't clear, I admit I'm just trying to find a direction to proceed on the original question I was asked. – 2017-01-24
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    Okay I see what you mean. Yes, maps from the tensor product are usually defined by specifying what they do to basis elements. – 2017-01-24
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    Would $\phi = 0$ do? – 2017-01-24

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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.

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    If the field and the dimension of $V$ are both finite, the set $(V^* \otimes V)^*$ is finite. – 2017-01-24
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    @MooS Thanks for the correction. I've fixed it. – 2017-01-24
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    Cool thanks; that answers my question as asked. However, I believe I need to ask for clarification on what I was asked to consider originally. I probably am missing a bit of the context that the questioner assumed I had a handle on. – 2017-01-24