Strict positiveness on a C*-algebra given by generators and relations. Let $A$ be a C*-algebra with generators $a_1,a_2,\ldots,a_n$ and some (non-important) relations (the relations imply that $\|a_i\|\leq 1$, so that $A$ exists). Among the given relations we have that $$ a_1 = \sum_{1 \leq i \leq n} a_i a_i^*. $$ This implies that $a_1$ is positive; I would like to prove that $a_1$ is strictly positive. So far my attempts to prove this have been in vain. If $\phi$ is a multiplicative linear positive functional, then $\phi(a_1)=0$ implies $\phi(a_i)=0$ and then $\phi = 0$; I've no clue on the general case. Thanks in advance.

Let $\phi$ be a state. Suppose that $\phi(a_1)=0$. Then, for some fixed $j$, $$ 0\leq\phi(a_ja_j^*)\leq\phi(a_1)=0, $$ so $\phi(a_ja_j^*)=0$. Now, since $\phi$ is completely positive, it satisfies the Kadison-Schwarz inequality; thus $$\tag{1} 0\leq\phi(a_j)\phi(a_j)^*\leq\phi(a_ja_j^*)=0, $$ so $\phi(a_j)\phi(a_j)^*=0$, which implies $\phi(a_j)=0$. Since we have equality in $(1)$, this means that $a_j$ is in the multiplicative domain of $\phi$. As we can do this for all $j$, $\phi$ is multiplicative and so $\phi=0$.