How can i show that the tensor product of vector spaces is a vector space 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!
 A: 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.

A vector space $V'$ is a set of elements called vectors with following maps:
$$ + : V' \times V' \rightarrow V' : (u,v) \mapsto u+v$$
called the addition map and
$$ \cdot: \mathbb F \times V' \rightarrow V' : (\lambda, v) \mapsto \lambda v $$called the scalar multiplication. Moreover these maps verify the following conditions (which are the 8 conditions you cite in your post and I will note write them all out again).

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 $2$

There exists a vector $0 \in V'$ such that $v + 0 = 0 + v = v$ for all $v \in V'$

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 $\psi + \phi, 0, -\psi, \lambda \psi \in V \otimes W$.
