I'm trying to prove the $8$ axioms of a vector space to the following space:
Let $V$ and $W$ be two finite-dimensional vector spaces over the field $\mathbb F$. We define the tensor product between $V$ and $W$ as the set:
$$ V \otimes W \,\,\colon = \{ \phi: V^* \times W^* \rightarrow \mathbb F \,|\, \phi( \cdot , f) \in V^* \,\,\, \text{and} \,\,\,\phi(g, \cdot) \in W^* \,\, \forall g \in V^* , \,\, \forall f \in W^* .\}$$
Given $v \in V$ and $w \in W$, we denote as $v \otimes w$ the element of $V \otimes W$ defined as:
$$(v \otimes w) (g,f) := g(v)f(w), \forall (g,f) \in V^* \times W^* $$
Prove that $V \otimes W$ is a $\mathbb F$ vector space and
$$ V \otimes W = Span( \{ v \otimes w : v \in V, w \in W \}) $$
I already did:
Suppose that $\phi, \psi, \theta \in V \otimes W$, and $\alpha$, $\beta \in \mathbb F$:
1- $(\psi + \phi)(g,f) = \psi(g,f) + \phi(g,f) = \phi(g,f)+\psi(g,f) = (\phi + \psi)(g,f)$
2- $(\phi + \psi)(g,f) + \theta(g,f) = \psi(g,f) + \phi(g,f) + \theta(g,f) = \phi(g,f) + (\psi + \theta)(g,f)$
3- I dont know how to show the existence of the aditive neutral
4- Since i dont know who is the neutral aditive, i cant show the aditive inverse
5- $(\alpha \beta)\phi(g,f) = \phi(\alpha \beta g,f) = \alpha \phi(\beta g,f) = \alpha(\beta \phi(g,f))$
6a- $(\alpha + \beta) \phi(g,f) = \phi((\alpha + \beta)g,f) = \phi(\alpha g,f) + \phi(\beta g,f)= \alpha \phi(g,f) + \beta \phi(g,f)$
6b- $(\alpha + \beta) \phi(g,f) = \phi(g,(\alpha + \beta)f) = \phi(g, \alpha f) + \phi(g, \beta f)= \alpha \phi(g,f) + \beta \phi(g,f)$
7a- $\alpha (\psi + \phi)(g,f) = (\psi + \phi)(\alpha g,f) = \psi(\alpha g,f) + \phi(\alpha g,f) = \alpha \phi(g,f) + \alpha \psi(g,f)$
7b- $\alpha (\psi + \phi)(g,f) = (\psi + \phi)(g ,\alpha f) = \psi(g, \alpha f) + \phi(g, \alpha f) = \alpha \phi(g,f) + \alpha \psi(g,f)$
8a- $1_{\mathbb F} \phi(g,f) = \phi(1g,f) = \phi(g,f)$
8b- $1_{\mathbb F} \phi(g,f) = \phi(g,1f) = \phi(g,f)$.
I never worked with bilinear forms before, so i'm really not sure about if i'm correct on my attempts, and i ask for help on the remaining items.
Thanks in advance!
The way you've defined the multiplication by scalars is unconventional but correct. As for the neutral it is given by the map $(f,g)\mapsto 0 $ where $0$ denotes the zero in $\mathbb F$. It follows that the inverse of $\psi $is given by the map $(f,g) \mapsto -(\psi(f,g))$ where we took the inverse of $\psi(f,g)$ in $\mathbb F.$
As a final comment don't forget to check that the $\psi + \phi,0,-\psi,\alpha \phi$ are actually functions in $V \otimes W$. That is, you must check that these maps are bilinear. You must check this because its an essential property of vector spaces. Remember that a vector space $V$ has a addition $+$ which is a map $ V \times V \rightarrow V$. So given $\psi,\phi \in V$ it is crucial for $\psi +\phi$ to be again an element in $V$.
Adressing comments :
Consider the following definition of a vector space.
In your exercise you are given the following set
$$ V \otimes W = \{\phi : V^* \times W^* \rightarrow \mathbb F \; \vert \; \phi \text{ is bilinear} \}.$$
To prove that this is a vector space you must define maps
$$ + : (V\otimes W) \times (V \otimes W) \rightarrow (V \otimes W)$$ $$ \cdot : \mathbb F \times (V\otimes W) \rightarrow (V \otimes W)$$ such that these maps verify the $8$ conditions. The first step to solve your problem is therefore to define the maps $+$ and $\cdot$. Only then can you start trying to prove the properties 1,...,8 Let us first focus on defining the addition.
Let $\phi,\psi \in V \otimes W$ and let us define a new function
$$ \phi + \psi : V^* \times W^* \rightarrow \mathbb F : (f,g) \mapsto \phi(f,g)+ \psi(f,g).$$
This defines a map
$$ + : (V \otimes W) \times (V \otimes W) \rightarrow \{ h : V^* \times W^* \rightarrow \mathbb F \} : (\phi,\psi) \mapsto \phi + \psi $$ If we manage to prove that $\phi + \psi \in V \otimes W$ then the map $+$ will actually be a map $$ + : (V \otimes W) \times (V\otimes W) \rightarrow (V \otimes W)$$ (which is exactly what we are looking for in order to define a vector space structure on $V \otimes W$.) We now prove that $\phi + \psi \in V\otimes W$
Recall the definition of the set $V \otimes W$. By definition $V\otimes W$ is the set of functions $ V^* \times W^* \rightarrow \mathbb F$ which are bilinear. Hence to prove $\phi + \psi \in V\otimes W$ we have to prove that $\phi + \psi $ is a bilinear map. That is we must prove that $\phi + \psi$ is linear in each component. Let $f',f'' \in V^*, g \in W^*$ then
\begin{align*} (\phi + \psi)(f'+f'',g) &= \phi(f'+f'',g) + \psi(f'+f'',g) \\ &= \phi(f',g)+ \phi(f'',g)+ \psi(f',g)+ \psi(f'',g) \\ &= \phi(f',g) + \psi(f',g) + \phi(f'',g)+ \psi(f'',g) \\ &= (\phi + \psi)(f',g) + (\phi+ \psi)(f'',g) \end{align*}
where from line 1 to line 2 we used the bilinearity of $\psi$ and $\phi$ (note that they are bilinear since $\phi,\psi \in V\otimes W$.) From line 2 to line 3 we used the commutativity in $(\mathbb F,+)$. Then from line 3 to line 4 we used the definition of $\phi + \psi$. \begin{align*} (\phi + \psi)(\lambda f,g) &= \phi(\lambda f,g) + \psi(\lambda f,g) \\ &= \lambda \cdot \Big( \phi(f,g) \Big) + \lambda \cdot \Big( \psi(f,g) \Big)\\ &= \lambda \cdot \Big( \phi(f,g) + \psi(f,g) \Big) \\ &= \lambda \cdot \Big( (\phi + \psi)(f,g) \Big) \end{align*} Where we used the bilinearity of $\phi,\psi \in V\otimes W$. Note that the the multiplication $\lambda \cdot \Big( \phi(f,g) \Big)$ is done in $(\mathbb F,\cdot)$.
This proves that $\phi + \psi$ is linear in the first component. I'm sure you're capable to prove the linearity in the second component.
With this proof we now have a map
$$ + : (V \otimes W) \times (V \otimes W) \rightarrow (V\otimes W): (\phi, \psi) \mapsto \phi + \psi $$
For the scalar multiplication I suggest you define the scalar multiplication as follows. Let $\phi \in V \otimes W$ and for any $\lambda \in \mathbb F$ define the function
$$ \lambda \phi : V^* \times W^* \rightarrow \mathbb F : f,g \mapsto \lambda \cdot \Big( \phi(\lambda f,g)\Big)$$
Now you prove that $\lambda \phi$ is bilinear (i.e. $\lambda\phi \in V \otimes W) so that you get a map
$$ \cdot : \mathbb F \times (V\otimes W) \rightarrow (V \otimes W): \lambda,\phi \mapsto (\lambda \phi) $$
Now you can start proving the conditions one to eight. This should be easy but there is a subtle point I believe you have not understood. Consider the property $2$
As I've told you the map you should consider is the map
$$ 0_{V\otimes W}: V^* \times W^* \rightarrow \mathbb F : (f,g) \mapsto 0 \in \mathbb F$$
For the property 2 to be valid you must check that the map $0_{V\otimes W}$ is in the set $V \otimes W$. That is you must check that this map is bilinear
Similarly the property $3$ tells you that the additive inverse of an elements in a vector space belongs to said vector space. Hence you must check that $-\psi \in V\otimes W$. That is you must check that the map
$$ -\psi : V^* \times W^* \rightarrow K : f,g \mapsto - \Big(\psi(f,g) \Big)$$ is bilinear.
This is what I meant by my comment that you should check that $\psi + \phi, 0, -\psi, \lambda \psi \in V \otimes W$.