Families of vectors in finite dimensional Hilbert space This is an exercise left for the reader in proof of Proposition 1.12 in Pisier's book "Introduction to Operator Space Theory".
Assume that we have $n$ vectors $x_1, \dots, x_n$ in $\ell_2^N$ (where $n>N$) which satisfy inequality $\sum_{k=1}^{n} \|x_k\|^2 \leq 1$. Now, we would like to find $n \times N$ matrix $(b_{jk})$ of norm (considered as an operator from $\ell_2^N$ to $\ell_2^n$) not greater than $1$ and $N$ vectors $y_1, \dots, y_N$ in $\ell_2^N$ which, again, satisfy $\sum_{k=1}^{N} \|y_k\|^2 \leq 1$ and
$$x_j = \sum_{k=1}^{N} b_{jk}y_{k}.$$
I have an idea that maybe one should try to attack this problem in a somewhat abstract fashion, probably using separation theorem or something similar; most certainly I am missing something obvious. Anyway, any help will be appreciated.

First of all, a general comment on books about operator spaces. Pisier's "Introduction to Operator Spaces" book is great but is occasionally terse. It may help to sometimes also look at Effros and Ruan's "Operator Spaces" and Paulsen's "Completely Bounded Maps and Operator Algebras". For example, the following is  Lemma 2.2.1 of Effros and Ruan's book.
We are seeking an isometry $b : \ell_2^N \to \ell_2^n$ and a vector $\tilde y \in \ell_2^N \otimes \ell_2^N$ such that 
$$
(Id_n \otimes b)
(\tilde y)
=
\tilde x.
$$
Let $\{e_i\}_{i=1}^n$ denote the standard orthonormal basis in $\ell_2^n$ and let $\{f_i \}_{i=1}^N$ be the standard orthonormal basis in $\ell_2^N$.
Let 
$$
\eta = \sum_{i=1}^n x_i \otimes e_i \in \ell_2^N \otimes \ell_2^n.
$$
 This can be rewritten as 
$$\eta=\sum_{j=1}^N f_j \otimes \eta_j
$$ 
for vectors $\eta_j \in \ell_2^N$.
Let $F \subseteq \ell_2^N$ be the subspace spanned by $\{ \eta_j \}$. As $\dim F \leq N \leq n$, there exists an isometry $b: \ell_2^N \hookrightarrow \ell_2^n$ such that the image of $b$ contains $F$. For each $j$, there exists a unique vector $\tilde \eta_j$ such that $b(\tilde \eta_j) = \eta_j$. Then if $\tilde \eta =\sum_{j=1}^N f_j \otimes \tilde \eta_j$, we have that 
$$
(Id_n \otimes b) (\tilde \eta) = \eta = \sum_{i=1}^n x_i \otimes e_i .
$$
Rewriting $\tilde \eta$ as $\sum_{i=1}^N y_i \otimes f_i$, we have
$$
(Id_n \otimes b) (\sum_{i=1}^N y_i \otimes f_i) =  \sum_{i=1}^n x_i \otimes e_i,
$$
or 
$$
x_j = \sum_{k=1}^N b_{jk} y_k,
$$
as desired.
