Ideals of $M_2(A)$ , $A$- non commutative non unital Let $A$ be a non unital, non commutative $C^{\ast}-$ Algebra. Let $J$ be an ideal of $M_2(A)$. Assume $$J= \begin{bmatrix}P&Q\\R &S\end{bmatrix}$$ It is easy to prove that $Q$ is an ideal of $A$ and $R=Q^{\ast}$. 

Is it true that $P=QQ^{\ast} $

I have proved that $QQ^{\ast} \subset P$, I am unable to prove the converse. 
 A: You have $P=Q=R=S=I$ for some ideal $I$ of $A$. That is, any ideal of $M_2(A)$ is of the form $M_2(I)$ for $I$ an ideal of $A$. 
First, for any $p\in P$ and $a,b\in A$, you get 
$$
\begin{bmatrix} a&0\\0&0\end{bmatrix}\begin{bmatrix} p&0\\0&0\end{bmatrix}\begin{bmatrix} b&0\\0&0\end{bmatrix}=\begin{bmatrix} apb&0\\0&0\end{bmatrix}\in J, 
$$
so $apb\in P$. Together with the same computation for sums, and limits if you want, you get that $P$ is an ideal in $A$. 
For any $q\in Q$ and $a,b\in A$, $p\in P$, $r\in R$, $s\in S$ such that $\begin{bmatrix} p&q\\r&s\end{bmatrix}\in J$,
$$
\begin{bmatrix} a&0\\0&0\end{bmatrix}\begin{bmatrix} p&q\\r&s\end{bmatrix}\begin{bmatrix} 0&0\\b&0\end{bmatrix}=\begin{bmatrix} aqb&0\\0&0\end{bmatrix}.
$$
So $aqb\in P$ for all $q\in Q$ and $a,b\in A$. Using an approximate identity, we get $Q\subset P$. For any $p\in P$ and $q,r,s$ such that $\begin{bmatrix} p&q\\r&s\end{bmatrix}\in J$,
$$
\begin{bmatrix} a&0\\0&0\end{bmatrix}\begin{bmatrix} p&q\\r&s\end{bmatrix}\begin{bmatrix} 0&b\\0&0\end{bmatrix}=\begin{bmatrix} 0&apb\\0&0\end{bmatrix}.
$$
So $apb\in Q$ for all $a,b\in A$ and $p\in B$. Again using approximate identities we get $p\in Q$, so $P\subset Q$ and thus $P=Q$. In particular $Q$ is selfadjoint, so $R=Q$. A similar computation shows that $S=Q=P$. 
