Dual space and inner/scalar product space $V$ is vector space of finite dimension. $〈· , ·〉$ is an inner product on $V$.(Field $F$)
We set transformation $T \colon V \rightarrow V^*$ as the following: $(T(v))(w) = 〈v , w〉$.
Prove that $T$ is Isomorphism.
I don't know how to prove that it is 1 on 1 and onto. 
I mean, the dualic space is confusing me since I don't understand it properly.
For 1-1 : I need to assume that $〈v_1,w_1〉 = 〈v_2,w_2〉$ and show that $v_1=v_2$ and $w_1 = w_2$? I'm not sure what is my domain... $T$ is from $V$ to $V^*$ so should I show that $v_1 = v_2$ only?
For onto: I need to show that for every functional from $V^*$ there is $v$ from $V$ such that it equals? doesn't make sense because the inner product is a scalar from $F$. I feel very helpless about this, could someone help me please  ?
 A: You are wrong about the injectivity claim. Let me translate it correctly for you.

Injectivity of $T: V \rightarrow V^*$ means that given $v \in V$, such that $T(v)=0$ implies that $v=0$.

Now what does $T(v)=0$ mean? Well, $T(v)$ is a linear functional $T(v): V \rightarrow F$, defined by $T(v)(w)=\langle v,w \rangle$, and a linear functional is zero, if and only if it maps everything to zero. So 
$$
T(v)=0
$$
really means that 
$$\langle v,w \rangle=0
$$ for all $w \in V$. But from the properties of an inner product you can deduce that, if for some $v \in V$ we have $\langle v,w \rangle=0$ for all $w \in V$, then necessarily $v=0$ (do you see how?).
As for surjectivity, since $V$ is a finite-dimensional $F$-vector space, $V^*$ has the same dimension as $V$ (dual basis). Now what can you conclude about an injective linear map$^{\dagger}$ between two vector spaces of the same dimension.
$^{\dagger}($Of course you also have to prove that $T$ is a $F$-linear map, but that again follows from the definition of an inner product.)
A: Let $V$ be a finite dimensional vector space over $\mathbb K$ (choose for example $\mathbb R$). 


*

*On the dual space $V^{*}$: why?


The dual space $V^{*}$ is defined as
$$V^{*}:=\{ \varphi: V\rightarrow \mathbb K,~~ \varphi~ \text{linear}\};$$
in other words, the dual space is the space of all linear functionals, i.e. those maps from the original vector space $V$ to the ground field $\mathbb K$ which are linear. 
Why is this new space interesting?
In a certain sense, introducing $V^{*}$ we used all information we had from the beginning to produce a new linear space. In fact, we started just with a pair $(V,\mathbb K)$ and we arrive at a new vector space $V^{*}$ whose elements "connect" $V$ to $\mathbb K$ respecting the non trivial structure on $V$, i.e. the linear space structure.
Moreover, if $V$ is finite dimensional, then $V^{*}$ is finite dimensional as well. 
Let $\{e_i\}_{i=1,\dots,n}$  be a basis of $V$ ($n$ is the dimension of $V$). The dual set
$$\{\varphi_j\}_{j=1,\dots,n}$$
of elements  $\varphi_j\in V^{*}$ s.t.
$$\varphi_j(e_i):=\delta_{ij}$$
is a basis of $V^{*}$. To prove it you can check any book of linear algebra. 
Note that we used only the existence of the pair $(0,1)$, with $0:=\delta_{ij}$ for $i\neq j$ and $1:=\delta_{ij}$ for $i=j$ in the ground field $\mathbb K$ to introduce our basis $\{\varphi_j\}_{j=1,\dots,n}$.
Such pair $(0,1)$ always exists by definition of field: the $0$ is the zero element of the addition, while $1$ denotes the unit element of multiplicative composition.


*

*On $T:V\rightarrow V^{*}$


Let $T:V\rightarrow V^{*}$ be the linear map
$$T(v)(w):=\langle v,w\rangle, $$
denoting by $\langle \cdot,\cdot \rangle $ an inner product on $V$. We want to prove that $T$ is an isomorphism. We need to check that $T$ is injective and surjective.
On injectivity: as $T$ is linear, to prove injectivity is equivalent to show that
$$\operatorname{Ker}(T)=\{v\in V: T(v)=0 ~\text{in}~ V^{*}\}=\{0\}.$$
By definition, $T(v)=0$ in $ V^{*}$ (here "0" denotes the zero map in $V^{*}$!) if
$$\forall w\in V\Rightarrow T(v)(w)=\langle v,w\rangle=0~~(*)$$
in $\in\mathbb K$; $(*)$ holds if and only if $v=0$. In fact, 
in $(*)$ we can choose $w=v$, arriving at
$$T(v)(v)=\langle v,v\rangle=0\Leftrightarrow v=0 $$
by definition of inner product (check it!). In other words we have proved that $T(v)=0\Leftrightarrow v=0$, i.e. $\operatorname{Ker}(T)=\{0\}$.
On surjectivity You need to prove that
$$\forall \varphi\in V^{*}~~ \exists v\in V : T(v)=\varphi$$
in $V^{*}$, i.e.
$$\forall w\in V~~ T(v)(w)=\varphi(w) $$
in $\mathbb K$. This last statement is equivalent to
$$\forall w\in V~~ \langle v,w\rangle=\varphi(w), $$
which can be easily proven using a base on $V$ and the dual basis on $V^{*}$, as above.
I leave it to you. I hope it helps.
A: For injectivity:
You need to show that whenever $T(v_1) = T(v_2)$, then $v_1 = v_2$ for any $v_1, v_2 ∈ V$.
Now $T(v_1)$ and $T(v_2)$ are elements in $V^*$ and therefore linear maps $V → F$.
By definition of equality, those maps they are the same if they do some thing on $V$, i.e. if $∀ w ∈ V:\, T(v_1)(w) = T(v_2)(w)$ or by definition of $T$: $∀ w ∈ V:\, 〈v_1,w〉 = 〈v_2,w〉$.
Now, use that $T$ is a linear map and show that its kernel is zero. So show:
$∀ v ∈ V:\, T(v) = 0 ⇒ v = 0$.
This will do to show injectivity. The fact that any inner product is a non-degenerate bilinear map might come in handy.
For surjectivity use that any non-trivial linear map $ϕ : V → F$ has a kernel of dimension $n-1$, where $n = \dim V$. Now in the inner product space, this means that its orthogonal complement $(\ker ϕ)^⊥$ has dimension 1 and is generated by some nonzero element $v$. Show that $\ker T(v) = \ker ϕ$. Conclude that there is a $λ ∈ F$ such that $T(λv) = ϕ$. (Take $λ = ϕ(v)/〈v,v〉$.)
