Gram-Schmidt procedure for new base, complete positivity Let say that $v_1,v_2,...v_n$ be vectors in a $m$-dimensional Euclidian space $V$. There exists a natural number $k$ and an isometry  $T:V \Rightarrow R^k$ such that $$Tv_1,Tv_2,..., Tv_n \in  R^k_+=\{x \in R^n|x_i \geq 0\text{ for }i=1,\dots,n\}.$$
This is true, if $Gram(v_1,v_2,...)$ is completely positive. Here $k$ is the number of columns of $B$ and for each $i=1,2,...,n$, $Tv_i$ is the $i$-th row of $B$.
Why the vectors can be embedded in a nonnegative orthant of some Euclidian space?
Which statement provide us existence of $T$? Is it monotonous?
I know that I can get new base by Gram-Schmidt procedure.
Please, I really need an explanation about this.
 A: We want to show that if $A := \operatorname{Gram}\{v_1,v_2,\dots,v_n\}$ is completely positive, then there exists a $k$ and an isometry $T$ such that $Tv_i \in \Bbb R^k_+$.
Let $P$ denote the matrix with columns $v_1,\dots,v_n$; note that $A = P^TP$.  Because $A$ is completely positive, there exists a matrix $B$ for which $B^TB = A$.  Let $b_1,\dots,b_n$ denote the columns of $B$. Let $V$ denote the span of $v_1,\dots,v_n$, and define the linear map $T:V \to \Bbb R^k$ by $T(v_k) = b_k$.
Because $P^TP = B^TB$, we have $v_i^Tv_j = b_i^Tb_j$. It now suffices to show the following:
Claim: $T$ is an isometry over $V$.
Proof: Any vector $v \in V$ can be written in the form
$$
v = c_1 v_1 + \cdots + c_n v_n.
$$
We note that
$$
\begin{align}
\|T(v)\|^2 &= (T(v))^TT(v) = \left(T\sum_i c_i v_i\right)^T \left(T\sum_{j}c_j(v_i)\right)
\\ & = \sum_{i,j} c_ic_j T(v_i)^TT(v_j)
= \sum_{i,j} c_ic_j v_i^Tv_j
\\ & = \left(\sum_i c_i v_i\right)^T \left(\sum_{j}c_j(v_i)\right)
= \|v\|^2.
\end{align}
$$
So, $\|T(v)\| = \|v\|$, which is to say that $T$ is an isometry, as desired.
