# 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 $$ = a_0b_0+a_1b_1.$$ How to show this is an inner product by showing that this function is the "same" as the standard inner product in $$\mathbb R^2$$?

• V and $\mathbb R^4$ are isomorphic... their canonical inner products are in one-to-one correspondence – phaedo Oct 31 '18 at 2:51
• @phaedo Their inner products are bijective? It sounds strange to me. – user398843 Oct 31 '18 at 2:56
• they are in one-to-one correspondence with each other: $<\cdot,\cdot>_V \leftrightarrow <\cdot,\cdot>_{\mathbb R^4}$ – phaedo Oct 31 '18 at 2:59
• @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. – user398843 Oct 31 '18 at 3:05
• 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 – phaedo Oct 31 '18 at 3:12

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: $$_V = \sum_{i=1}^n x_i y_i$$ 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 $$$$ simplify to 0 when $$i\neq j$$ and 1 otherwise... you're done!

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

• You don't need Gram-Schmidt since we are operating on canonical bases already – phaedo Oct 31 '18 at 3:43
• @phaedo What canonical basis do you have on $V$? – edm Oct 31 '18 at 3:46
• Maybe it's fine as long as we choose a basis no matter it is orthogonal or not. – user398843 Oct 31 '18 at 3:48
• @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 $\mathbb R^n$ – phaedo Oct 31 '18 at 3:49
• @phaedo Sorry, I didn't notice that OP over-edited the question. – edm Oct 31 '18 at 4:27