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. 
 A: 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
$$s_\phi: V \times V \longrightarrow \Bbb R,$$
$$(v, w) \mapsto (\phi(v))(w).$$
A: $\phi:V\to V^\ast$
$\phi(v)\in V^\ast$ means that $\phi(v):V\to F$
$\phi(v)(w)$ is just the value of $\phi(v)$ at $w$.
A: 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.
