notation question (bilinear form) So I have to proof the following: 
for a given Isomorphism $\phi : V\rightarrow V^*$ where $V^*$ is the dual space of $V$ show that $s_{\phi}(v,w)=\phi(v)(w)$ defines a non degenerate bilinear form.
My question : Does $\phi(v)(w)$ denote the map from $v$ to a linear function $w$? (in this case i had serious trubles in showing linearity in the second argument,really confusing.
Or
maybe it just means $\phi(v)$ times $w$ where $w$ is the scalar value ( we get $w$ by applying  $v$ in the linear function it is mapped to)
I just started today with dual spaces and try my best with the notation , but i couldn't figure it out , please if you have any idea please help me with the notation , i will solve the problem on my own. 

Note that $\phi$ is a map from $V$ to $V^\ast$. So for each $v \in V$, we get an element $\phi(v) \in V^\ast$. Now $V^\ast$ is the space of linear functionals on $V$, i.e.
$$V^\ast = \{\alpha: V \longrightarrow \Bbb R \mid \alpha \text{ is linear}\}.$$
So each element of $V^\ast$ is a function from $V$ to $\Bbb R$. Then for $v, w \in V$, the notation
$$\phi(v)(w)$$
means
$$(\phi(v))(w),$$
i.e. the function $\phi(v): V \longrightarrow \Bbb R$ takes $w \in V$ as its argument and we get an element of $\Bbb R$.
So $s_\phi$ is really a map of the form
$$s_\phi: V \times V \longrightarrow \Bbb R,$$
$$(v, w) \mapsto (\phi(v))(w).$$
