(Stinespring dilation) Let $A$ be a unital $C^{\ast}$-algebra and $\phi: A \rightarrow B(H)$ be a completely positive map. Then, there exist a Hilbert space $H_{1}$, and a $^{\ast}$-representation $\pi: A \rightarrow B(H_{1})$ and operator $V:H \rightarrow H_{1}$ such that $\phi(\cdot)=V^{\ast}\pi(\cdot)V$. In particular, $\Vert \phi \Vert = \Vert V \Vert^{2}= \Vert \phi(1) \Vert$.
In the first step of the proof of this theorem, defining a sesquilinear form $\langle \cdot, \cdot \rangle$ on $A \odot H$ (this is the algebraic tensor product) by $$\langle\sum_{j}~b_{j} \otimes \eta_{j},~\sum_{i}~a_{i}\otimes \xi_{i}\rangle = \sum_{i, j} \langle \phi(a_{i}^{\ast}b_{j})\eta_{j},~\xi_{i}\rangle_{H}~.$$
I can not understand the form of the element in $A \odot H$ (that is $\sum_{i}~a_{i}\otimes \xi_{i}$). why not $a_{i}\otimes \xi_{i}$? I mean how to explain the sum of $a_{i}\otimes \xi_{i}$.
If you only consider elementary tensors, then what would their sum be? Remember that you want $A\odot H$ to be a vector space.