I am trying to solve Exercise 5.13 of Andreas Gathmann's Commutative Algebra Lecture notes.
"Let $V$ and $W$ be vector spaces over a field $K$. We call $V^∗ := {\rm Hom}_K(V,K)$ the dual vector space to $V$. Moreover, denote by $BLF(V)$ the vector space of bilinear
forms $V \times V \to K$.
(a) Show that there are (natural, i. e. basis-independent) linear maps
$\Phi : V^∗ \otimes W \to {\rm Hom}(V,W)$ such that $\Phi(\phi \otimes w)(v) = \phi(v)·w$.
(There are still two map $\Psi$ and $T$ in the statement of this part but I won't write them here because I can handle with them)
(b) Prove that $\Phi$ is injective.
(d) Assume that $\dim_K V = n < \infty$, so that $V^∗ \otimes V$ is naturally isomorphic to ${\rm Hom}_K(V,V)$ by (c), which in turn is isomorphic to $Mat(n \times n,K)$ in the standard way after choosing a basis of $V$. Using these isomorphisms, $T$ becomes a linear map $Mat(n \times n,K) \to K$ that is invariant under a change of basis. Which one?"
(This exercise has part (c) but I won't write them here because I have no question about it)
For part (a), I have constructed the bilinear map
$f: V^* \times W \to {\rm Hom}(V,W), (\phi,w) \mapsto (v \mapsto \phi(v)·w)$
which induces $\Phi$ by the universal property of tensor products. However, I am not sure if my proof for part (b) has any mistakes or not. This is my proof:
"Denote $\Phi(\phi \otimes w)$ by $\phi w$. Suppose $x=\sum\limits_{i \in I} (\phi_i \otimes w_i) \in V^* \otimes W$ such that $\Phi(x)=0$ (Of course we understand that $\#I<\infty$).
WLOG, we can assume that $\{w_i\}$ is a linearly independent system in W (Otherwise we can simplify $x$ since we can represent some $w_i$'s as linear combinations of some other $w_j$'s). We have
$0=\Phi(x)=\sum\limits_{i \in I} \Phi(\phi_i \otimes w_i)=\sum\limits_{i \in I} \phi_i w_i$
$\Rightarrow 0=\left(\sum\limits_{i \in I} \phi_i w_i\right)(v)=\sum\limits_{i \in I} (\phi_i w_i)(v)=\sum\limits_{i \in I} \phi_i(v) w_i$, $\forall v \in V$
$\Rightarrow \phi_i(v)=0, \quad \forall i \in I, \quad \forall v \in V$
$\Rightarrow \phi_i=0, \quad \forall i \in I$
$\Rightarrow x=\sum\limits_{i \in I} (0 \otimes v_i)=0$.
Hence $\ker \Phi=0$."
Thanks for your help.
(Updated: I have fixed my typo. Thanks Don for realizing it :D)
(Updated: Now I have new question for part (d). I do not understand what "that is invariant under a change of basis" means. Also, I the think part (d) wants us to find a "formula" of $T$ that does not depend on the bases of $V$, but I am not sure if my thoughts are right.)
Best Answer
It is correct, indeed!
Although there's a typo in the line starting with "Denote $\Phi(\phi \otimes w)...$". In the equality $x=\sum \phi_i \otimes v_i$, it should say $w_i$, not $v_i$.
You can also compare your proof with the one I was thinking about; very similar in practice:
Let $\{v_i\}$ and $\{w_j\}$ basis for $V$ and $W$, respectively. Then we have the dual basis $\{v^i\}$ for $V^*$. We know $\{v^i \otimes w_j\}$ is a basis for $V^* \otimes W$, so showing $\Phi$ is injective is equivalent to showing $\{\Phi(v^i \otimes w_j)\}$ is linearly independent.
Let $a_{ij}$ scalars such that $$\sum_{i,j} a_{ij} \Phi(v^i \otimes w_j) = 0.$$ Applying the vector $v_i$ at both sides we get $$\sum_{i,j} a_{ij} w_j = 0$$ for every $i$. Since $\{w_j\}$ is a basis, we get $a_{ij}=0$ as desired.
$\textbf{Remark.}$ I've just realised my proof tacitly assumes $V$ is finite-dimensional, to guarantee the dual basis of $\{v_i\}$ exists. So your proof is more general than mine :)