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I want to prove this:

Given a vector space $V$ on $\mathbb{R}$ with a positive definite inner product $\left \langle .,. \right \rangle$. Show that there exist a natural number $p$ and a linear map $T:V\rightarrow \mathbb{R}^{p}$ such that $\left \langle u,v \right \rangle=T(u).T(v)$ for all $u,v\in V$ where $.$ is the standard dot product.

My try: Let $V$ be $\mathbb{R}^{n}$. If we pick any $p \geq n$ and choose $T(u)=proj_{\mathbb{R}^{n}}(u)$ where $proj_{\mathbb{R}^{n}}(u)$ is the linear map that projects vectors from $\mathbb{R}^{p}$ to $\mathbb{R}^{n}$ then we have the inner product of elements $T(V)$ reduced to the one desired of $\mathbb{R}^{n}$. Is my proof legitimate? What about if I had been asked for the case $p < n$?

Thanks in advance.

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  • $\begingroup$ Hint: For $p=dim(V)$ the statement is equivalent to the fact that there is an orthormal basis for the inner product space $(V,\langle .,. \rangle)$. $\endgroup$ – Andreas Cap Jun 19 '16 at 8:38
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Let $\;U:=\{u_1,...,u_p\}\;$ be an orthonormal basis of $\;V\;$ ,and let $\;T:V\to\Bbb R^p\;$ be the map $\;Tu_i:=e_i\;$ , with $\;\{e_i\}\;$ the usual, standard (and orthonormal with the usual inner product) basis of $\;\Bbb R^p\;$.

Thus, we can write with resepct to the basis $\;U\;$ above:

$$\begin{cases}u=\sum\limits_{k=1}^n \langle u,u_k\rangle u_k\\{}\\v=\sum\limits_{k=1}^n \langle v,u_k\rangle u_k \end{cases}\;\;\implies\;\;\langle u,v\rangle=\sum_{k,i=1}^n \langle u,u_k\rangle\langle v,u_i\rangle\langle u_k,u_i\rangle$$$${}$$

$$=\sum_{k=1}^n\langle u,u_k\rangle\langle v,u_k\rangle=Tu\cdot Tv\;,\;\;\text{since}\;\;$$

$$Tu=T\left(\sum_{k=1}^n\langle u,u_k\rangle u_k\right)=\sum_{k=1}^n\langle u,u_k\rangle Tu_k=\sum_{k=1}^n\langle u,u_k\rangle e_k$$

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  • $\begingroup$ Thank you! But why is the basis $U$ consisting of $p$ vectors while the vectors $u$ and $v$ go from $1$ to $n$? $\endgroup$ – John11 Jun 19 '16 at 9:45
  • $\begingroup$ Well, I really didn't see that but it you say "if we pick..", then I presume we can pick as we want, so for me $\;p=n=\dim V\;$ , to avoid problems...:) $\endgroup$ – DonAntonio Jun 19 '16 at 9:49

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