Proposition 3.1.8 (Linear independence). If $\{x_{1},...,x_{n}\}\subset X$ are linearly independent, $\{y_{1},...,y_{n}\}\subset Y$ are arbitrary and $$0=\sum\limits_{i=1}^{n}x_{i}\otimes y_{i}\in X\odot Y,$$ then $y_{1}=y_{2}=...=y_{n}=0.$ (Here, $X$ and $Y$ are arbitrary vector spaces and $X\odot Y$ denotes the algebraic tensor product.)
Let $A$ and $B$ be the C*-algebras, the involution on all of $A\odot B$ is defined by $$\sum\limits_{i}a_{i}\otimes b_{i}\mapsto \sum\limits_{i}a_{i}^{*}\otimes b_{i}^{*}$$ and to prove that this is well defined, it suffices to show that if $\sum_{i}a_{i}\otimes b_{i}=0$, then $\sum_{i}a_{i}^{*}\otimes b_{i}^{*}=0$ as well. Expanding the $a_{i}$'s out in terms of a basis for $A$ and playing around with the tensor calculus and linear independence will show this to be true.
Well, it is a quotation of a book above. My question is how to expand the $a_{i}$ out in terms of a basis and how to use the linear independence to show the well defined of the map?
Choose a basis $\{v_1, v_2, \ldots, v_k\}$ of the subspace $\text{span}\{b_1,b_2,\ldots, b_n\}$ and write $$ b_i =\sum_j \alpha_{i,j}v_j $$ If $\sum_i a_i\otimes b_i = 0$, it follows that $$ \sum_i (\sum_j\alpha_{i,j}a_i)\otimes v_i = 0 $$ Since the $\{v_i\}$ are linearly independent, your lemma proves that $$ \sum_j \alpha_{i,j}a_i = 0 \quad\forall i\Rightarrow \sum_j \overline{\alpha_{i,j}} a_i^{\ast} = 0 $$ Hence, $$ \sum_i a_i^{\ast}\otimes b_i^{\ast} = \sum_i (\sum_j \overline{\alpha_{i,j}}a_i^{\ast})\otimes v_i^{\ast} = 0 $$