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Let $A\geq0$ be an $n\times n$ positive semi-definite matrix over $\mathbb{C}$, i.e., $x^* Ax\geq 0$ for all $x\in\mathbb{C}^n$. Given $A$, is it always possible to find real numbers $\alpha,\beta\geq0$, a diagonal matrix $D\geq 0$, and a vector $y\in \mathbb{C}^n$ such that $$A=\alpha D+\beta\, yy^*\,? $$

Probably it is not always possible, but I could not find an example.

Poor.Jack
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  • nope....it is not positive semi definite over $\mathbb{C}$, as it is not Hermitian. A counterexample, if exists, must be for $n\geq 3$. – Poor.Jack Jun 02 '16 at 23:49

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Your condition implies that by just changing the diagonal elements of $A$, we can make a rank $1$ operator out of $A$ (in other words, the columns will span a one-dimensional space).

Clearly, this will almost never work; for a concrete counterexample, you could just take a $4\times 4$ matrix with a $2\times 2$ block of full rank in the upper right corner.