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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 $\mathbb R^2$?

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  • $\begingroup$ V and $\mathbb R^4$ are isomorphic... their canonical inner products are in one-to-one correspondence $\endgroup$ – phaedo Oct 31 '18 at 2:51
  • $\begingroup$ @phaedo Their inner products are bijective? It sounds strange to me. $\endgroup$ – user398843 Oct 31 '18 at 2:56
  • $\begingroup$ they are in one-to-one correspondence with each other: $ <\cdot,\cdot>_V \leftrightarrow <\cdot,\cdot>_{\mathbb R^4} $ $\endgroup$ – phaedo Oct 31 '18 at 2:59
  • $\begingroup$ @phaedo Sorry, I haven't taken an algebraic structure course yet. Aren't these inner products functions? I know what's a bijecive function from one set to another, but I don't know what is a bijection between two functions. $\endgroup$ – user398843 Oct 31 '18 at 3:05
  • $\begingroup$ can you clarify what you are looking for? you are saying you are looking for a theorem, but you did not state what must be proven $\endgroup$ – phaedo Oct 31 '18 at 3:12
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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!

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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 $\langle Lv,Lw\rangle_{\Bbb R^n}=\langle v,w\rangle_V$

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  • $\begingroup$ You don't need Gram-Schmidt since we are operating on canonical bases already $\endgroup$ – phaedo Oct 31 '18 at 3:43
  • $\begingroup$ @phaedo What canonical basis do you have on $V$? $\endgroup$ – edm Oct 31 '18 at 3:46
  • $\begingroup$ Maybe it's fine as long as we choose a basis no matter it is orthogonal or not. $\endgroup$ – user398843 Oct 31 '18 at 3:48
  • $\begingroup$ @edm the original question was about V = vector space of polynomials of order >= 3, for which the canonical basis is (1, X, X^2, X^3). The edited question is more general, but you do not need an orthogonal basis to define a mapping between V and $ \mathbb R^n $ $\endgroup$ – phaedo Oct 31 '18 at 3:49
  • $\begingroup$ @phaedo Sorry, I didn't notice that OP over-edited the question. $\endgroup$ – edm Oct 31 '18 at 4:27

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