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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.

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    $\begingroup$ It is even not possible for a positive scalar. Why are you asking? $\endgroup$ – user251257 Dec 20 '17 at 19:52
  • $\begingroup$ You know that such a $B$ would need to be nonsingular and singular at the same time. $\endgroup$ – Algebraic Pavel Dec 20 '17 at 22:01

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