A question about corners of $C^\ast$-algebras

Question

Let $\mathcal{A}$ be a $C^\ast$-algebra, $p\in M_n(\mathcal{A})$ a projection, is there a $k\in\mathbb{N}$ such that $pM_n(\mathcal{A})p\cong M_k(\mathcal{A})$ ?

Thanks a lot!

Answer 1

Not necessarily. Let $\mathcal A=C ([0,1]\cup [2,3]) $, $n=2$, $q=1_{[0,1]} $, and $$p=\begin {bmatrix}q&0\\0&0\end {bmatrix}. $$ Then $$pM_2 (\mathcal A)p =\begin {bmatrix}\mathcal Aq&0\\0&0\end {bmatrix}\simeq C ([0,1]), $$ which is projectionless, so not isomorphic to $M_k (\mathcal A) $ for any $k $.

Answer 2

Have you tried a single example before asking? The following is the first one can think of: $A=\mathbb{C}^2$, $p=(1,0)$, $n=1$. And it works.