Prove bilinear form $g$ is a functional Let $V$ be a finite dimensional vector space over the field $K$, and let $g$ be a bilinear form on $V$, written $\langle,\rangle$.
Show that for each $w\in V$, the map $v\to\langle v,w\rangle$ is a functional $L_w$ on $V$, and that the map $w\to L_w$ is a linear map of $V$ into the dual space $V^{*}$
A functional is a mapping of the Vector space into its field of scalars:
$\varphi:V\to K$
$v\in V$
$v\to\langle v,w\rangle\in K$
Therefore $\langle v,w \rangle$ is a functional.
Since $\langle v,w\rangle$ is linear once $\langle v,w_1+\alpha w_2\rangle=\langle v,w_1\rangle+\alpha\langle v,w_2 \rangle$, in which $\alpha\in K$
$w_1+\alpha w_2\to L_{{w_1}+\alpha{w_2}}=L_{w_1}+\alpha L_{w_2}$
Questions:
1) Is the definition of functional($\varphi:V\to K$
) enough to prove this bilinear form (scalar product) is a functional?
2) Can we infer that if $\langle v,w_1+\alpha w_2\rangle=\langle v,w_1\rangle+\alpha\langle v,w_2 \rangle$ is true so it is $w_1+\alpha w_2\to L_{{w_1}+\alpha{w_2}}=L_{w_1}+\alpha L_{w_2}$?
Thanks in advance!
 A: To your first question: it's not quite clear what you're asking.  However, the definition $L_w(v) = \langle v,w \rangle$ is certainly enough to prove that $L_w$ is a function.  In particular: to show that $L_w$ is a functional, you should note that $L_2$ outputs a scalar and show that for $v_1,v_2 \in V$ and $\alpha \in K$, we have
$$
L_w(\alpha v_1 + v_2) = \alpha\,L_w(v_1) + L_2(v_2)
$$
To your second question: you seem to have the right idea.  Note that the equation
$$
L_{w_1 + \alpha w_2} = L_{w_1} + \alpha L_{w_2}
$$
is an equality of two functions (functionals, more specifically).  The equation really means that for every input $v \in V$, we have
$$
L_{w_1 + \alpha w_2}(v) = [L_{w_1} + \alpha L_{w_2}](v)
$$
which is to say that for every $v \in V$,
$$
L_{w_1 + \alpha w_2}(v) = L_{w_1}(v) + \alpha L_{w_2}(v)
$$
So, in order to show that $w \mapsto L_w$ is a linear map, we should show that the above is true. Of course, when you apply the definition of $L_w$, this amounts to the equation that you've written in terms of $\langle \cdot, \cdot \rangle$.
A: 
A functional is a mapping of the Vector space into its field of scalars

Not quite so. The correct definition in this context is that

A (linear) functional is a linear mapping of the vector space into its field of scalars.

So to prove that the mapping $\varphi=L_w$ is a linear functional, besides pointing out that it maps from $V$ to $K$, you must also show that it's linear. Of course, it's pretty obvious in this case, since it's based on a bilinear form, but you do have to address this property in your proof.
As for your second question: yes, that's correct. Although, to make this proof more readable and less confusing, you can phrase it in an element-wise fashion: for any $v\in V$,
$$L_{w_1+\alpha w_2}(v) = \langle v,w_1+\alpha w_2 \rangle = \langle v,w_1 \rangle + \alpha \langle v,w_2 \rangle = L_{w_1}(v)+\alpha L_{w_2}(v).$$
