Suppose $A$ is a unital $C^*$-algebra,$a$ is a full element in $A$.Show that there exist $n\in \Bbb N$ and $x_1,x_2,\cdots,x_n,y_1,y_2,\cdots y_n \in A$ such that $1_A = \sum_{i=1}^n x_i a y_i$.
Let $I:=\bigg \lbrace \sum_{i=1}^n x_i a y_i : x_i, y_i \in A, \ n\in \mathbb{N} \bigg \rbrace$,$J$ is the closure of $I$,since $a$ is full,then $a$ is not contained in any proper closed two sided ideal in $A$.We can conclude that $J=A$,then $1_A\in J$.How to prove that $1_A$ has the form of $\sum_{i=1}^n x_i a y_i$?
Since the collection of invertible elements of $A$ is open, there is some invertible element $b\in I$, and we may write $$b=\sum_{i=1}^nx_iay_i.$$ Then we have $$1_A=b^{-1}b=\sum_{i=1}^n(b^{-1}x_i)ay_i.$$