Let $A\ge 0$ be a positive definite matrix. Can I always find a matrix B such that $$ BB^\ast=A, \qquad BB^t=0,$$ where $B^t$ is the transpose and $B^\ast $ is the conjugate transpose? If yes, is this decomposition numerically efficient?
I tried construct $B$ as a complex combination of the Cholesky transform of $A$ but was not successful.