$(V^*)^{\otimes n} \cong (V^{\otimes n})^*$

Question

We assume that $V$ is finite dimensional. Make $\theta: (V^*)^n\to (V^{\otimes n})^*$ by $$ \theta(\alpha_1,\cdots,\alpha_n)(v_1 \otimes \cdots \otimes v_n ) := \prod_{i=1}^n \alpha_i(v_i). $$ Then, since $\theta$ is multilinear, $\theta$ extends to a linear function from $(V^*)^{\otimes n}$ to $(V^{\otimes n})^*$. I want to show that $\theta$ is an iso. Since the dimensions of both spaces are the same, we only have to show that $\ker \theta = 0$. How can this be done ?

Answer 1

Suppose $V$ is a $K$-vector space.

Let $\lbrace v_1, \ldots v_k \rbrace$ be a basis of $V$, and $\lbrace \delta_1, \ldots \delta_k \rbrace$ the dual basis.

We know that $\lbrace \delta_{i_1}\otimes \ldots \otimes \delta_{i_n} \rbrace_{i_1 , \ldots i_n \in \lbrace 1 \ldots k \rbrace}$ is a basis for $(V^{*})^{\otimes n}$.

Then if $\alpha \in (V^*)^{\otimes n}$ we have $$\alpha = \sum_{i_1 , \ldots i_n \in \lbrace 1 \ldots k \rbrace}(r_{i_1 , \ldots i_n })\delta_{i_1}\otimes \ldots \otimes \delta_{i_n} \ \ \ r_{i_1 , \ldots i_n } \in K \ \ \ \forall r_{i_1 , \ldots i_n }$$ Thus if $\theta(\alpha) = 0$ then $$0 = \theta(\alpha)(v_{j_1}\otimes \ldots \otimes v_{j_n} ) =\sum_{i_1 , \ldots i_n \in \lbrace 1 \ldots k \rbrace}(r_{i_1 , \ldots i_n })\prod_{h=1}^{h=n} \delta_{i_h}(v_{j_h}) = r_{j_1 , \ldots j_n}$$ This is true for all $\lbrace j_1, \ldots j_n \rbrace$ and so $\alpha = 0$