alternative definition of the spatial norm on tensor products In lecture, for $C^*$-algebras $A$ and $B$, the minimal tensor norm on the *-algebraic tensor product $A\odot B$ was defined as follows: 
$$\|\sum\limits_{i=1}^na_i\otimes b_i\|_{\min}=\sup\{\|\pi_A\otimes \pi_B)(\sum\limits_{i=1}^na_i\otimes b_i)\|: \pi_a:A\to B(H_1),\pi_B:B\to B(H_2)\;\ast-\text{representations}\}.$$
Since all non degenerate $*$-representations $\pi :A\to B(H)$ can be decomposed in a direct sum of cyclic $*$-representations, I want to prove that 
$\sup\{\|\pi_A\otimes \pi_B)(\sum\limits_{i=1}^na_i\otimes b_i)\|: \pi_a,\pi_B\;\ast-\text{representations}\}=\sup\{\|\eta_A\otimes \eta_B)(\sum\limits_{i=1}^na_i\otimes b_i)\|: (\eta_A,v_{\eta_A}), \;(\eta_B,v_{\eta_B})\;\;\text{cyclic *-representations}\}$ without using the fact that $\|\enspace \|_{min}$ is the smallest tensor norm.
The estimation $\sup\{\|\pi_A\otimes \pi_B)(\sum\limits_{i=1}^na_i\otimes b_i)\|: \pi_a,\pi_B\;\ast-\text{representations}\}\ge\sup\{\|\eta_A\otimes \eta_B)(\sum\limits_{i=1}^na_i\otimes b_i)\|: (\eta_A,v_{\eta_A}), \;(\eta_B,v_{\eta_B})\;\;\text{cyclic *-representations}\}$ is clear, but I don't know how to prove the other estimation $\le$. I can decompose all the *-representations $\pi_a,\pi_B$ into cyclic representations, but nevertheless then it's not clear for me why $\sup\{\|\pi_A\otimes \pi_B)(\sum\limits_{i=1}^na_i\otimes b_i)\|: \pi_a,\pi_B\;\ast-\text{representations}\}\le\sup\{\|\eta_A\otimes \eta_B)(\sum\limits_{i=1}^na_i\otimes b_i)\|: (\eta_A,v_{\eta_A}), \;(\eta_B,v_{\eta_B})\;\;\text{cyclic *-representations}\}$ should be true. How to prove this?
 A: Fist, show that $(\oplus_{i\in I} H_i) \otimes (\oplus_{j\in J} K_j$) is canonically isomorphic to $\oplus_{i\in I, j\in J} H_i\otimes K_j$, 
as Hilbert spaces. You can see this exercise $3.2.2$ in Brown&Ozawa.
As you said, any $*$-homomorphism can be decomposed as a direct of degenerate part and cyclic $*$-homomorphisms. Once you know that, it suffices to show that for $\sigma_1: A\to B(H_1), \sigma_2: A\to B(H_2), \rho_1:B\to B(K_1)$ and $ \rho_2: B\to B(K_2)$ we get:
$||((\sigma_1\oplus \sigma_2)\otimes (\rho_1\oplus \rho_2))(\sum_{i=1}^n a_i\otimes b_i)||=sup_{i,j\in\{1,2\}}||(\sigma_i\otimes \rho_j)(\sum_{i=1}^n a_i\otimes b_i)||$.
As $(\sigma_1\oplus \sigma_2)\otimes (\rho_1\oplus \rho_2):A\odot B\to B(H_1\oplus H_2)\odot B(K_1\otimes K_2)\hookrightarrow B((H_1\oplus H_2)\otimes (K_1\oplus K_2))$
(I'm sure you know that there is an inclusion, because you need it as part of the proof that $||\ ||_{min}$ is indeed a norm. Anyway, you can find it in Brown&Ozawa Lemma $3.3.9.)$
So, by the isomorphism, and after identifications: 
$(\sigma_1\oplus \sigma_2)\otimes (\rho_1\oplus \rho_2): A\odot B\to B((H_1\otimes K_1)\oplus (H_1\otimes K_2)\oplus (H_2\otimes K_1)\oplus (H_2\otimes K_2))$.
Now, follow the isomorphism to verify that you get 
$((\sigma_1\oplus \sigma_2)\otimes (\rho_1\oplus \rho_2))(\sum_{i=1}^n a_i\otimes b_i)= \oplus_{k,j\in \{1,2\}}\sigma_k\otimes \rho_j(\sum_{i=1}^n a_i\otimes b_i)$, and it implies the desired result.
Of course, you can show it just for $\sigma_1,\sigma_2,\rho_1$, because it is the same idea...
I hope this helps!
