How to show two inner products are the same? My professor told us there is a theorem which states that given a finite dimensional vector space $V$ with inner product $<|>_V$, we can have some sort of relationship between the inner product $<|>_V$ and the standard inner product of $\mathbb R^n$. What is this theorem? Where can I find this theorem?
For example, $V$ is $\mathbb R_1[t]$, first order polynomials of the variable $t\in \mathbb R$, with the standard basis. Suppose we have a function from $\mathbb R_1[t] \times\mathbb R_1[t] \to \mathbb R$ defined by $$<a_0+ a_1t \mid b_0+b_2 t> = a_0b_0+a_1b_1.$$ How to show this is an inner product by showing that this function is the "same" as the standard inner product in $\mathbb R^2$?
 A: You use Gram–Schmidt process to obtain an orthonormal basis $\{v_1,v_2,\dots,v_n\}$ for $V$. There is a unique linear map $L$ from $V$ to $\Bbb R^n$ that maps $v_i$ to $e_i$. This map is a linear isometry as well, which preserves the inner product in the sense that $\langle Lv,Lw\rangle_{\Bbb R^n}=\langle v,w\rangle_V$
A: This is your answer:
Every n-dimensional vector space is isomorphic to $\mathbb R^n$
RE inner product, as @edm stated, use any orthonormal basis of V: for any two vectors v,w in V, there are unique vectors x,y of $\mathbb R^n$ that are the coordinates of v,w in the basis.  By definition of orthonormal basis:
$$ <v,w>_V = \sum_{i=1}^n x_i y_i $$
which is nothing but the standard inner product of $\mathbb R^n$
To prove the above equation, write $ v = \sum_i x_i e_i $ where $(e_1, \cdots, e_n)$ is the orthonormal basis, and similarly for $w$ using j as index, then compute their inner product as double sum using bilinear property... all inner products $<e_i,e_j>$ simplify to 0 when $i\neq j$ and 1 otherwise... you're done!
