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. 

This kind of notation occurs sometimes. In certain contexts you'll have a function from some set to a set of functions. In your case, $\phi : V \to V^{\ast}$ associates with each vector $v$ a function $\phi(v)$, so since the latter is a function $\phi(v) \in V^{\ast}$ it is just $\phi(v): V \to \mathbb{F}$ where $\mathbb{F}$ is the field, so it's perfectly good to write $\phi(v)(w)$ for the action of $\phi(v)$ on the vector $w$.
So in general, suppose $A$ is any set and $\mathcal{F}$ is a set of functions from some other set $B$ to some other set $C$, then if we define $\phi : A \to \mathcal{F}$ for each $a \in A$ we have $\phi(a) : B \to C$ so that the action on some $b \in B$ is just written as $\phi(a)(b)$. Note that some people prefer writing $(\phi(a))(b)$, however the first notation is more common.
