Can i define an inner product so my basis will be orthogonal? Can every basis of a finite dinensional vector space be orthogonal with respect to some inner product? If no, than why?
 A: At least for a finite-dimensional vector space, the answer is "yes". 
Given a basis $b_1, \ldots, b_n$, if you have two vectors $v$ and $w$, define their inner product as follows: 


*

*Write each in terms of the basis:
$$
v = x_1 b_1 + \ldots + x_n b_n \\
w = y_1 b_1 + \ldots + y_n b_n
$$

*Compute the inner product as $x_1y_1 + \ldots + x_n y_n$. 
I leave it to you to check that this is, indeed, an inner product, and that under this product the basis is orthonormal. 
A: Sure.  Here's why. To calculate the inner product of any two vectors, write each as a linear combination of the basis vectors and use the ordinary formula.
That will define an inner product, and the basis you started with is orthonomrmal.
This all works because the basis gives you a linear transformation that maps the (finite dimensional) vector space bijectively to $\mathbb{R}^n$ and you use the standard inner product there.
A: Yes, that's true for finite dimensional vector spaces.
Let $K$ be $\mathbb{R}$ or $\mathbb{C}$ and let $V$ be a finite dimensional $K$-vector space. If $B=\{v_1,\cdots,v_n\}$ is a basis for $V$, consider $T\colon V\to K^n$ given by $T(v_i)=e_i$ (here $\{e_1,\cdots,e_n\}$ is the canonical basis for $K^n$). Clearly $T$ is an isomorphism.
Besides, we can define an inner product in $V$ by the formula: for $u,v\in V$,
$$\langle u,v\rangle_T = \langle T(u), T(v)\rangle$$
(Show this is in fact an inner product)
With such construction, $(V,\langle\cdot,\cdot\rangle_T)$ has $B$ as an orthonormal basis.
