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Let $A$ be a positive semidefinite matrix of rank $1$. Let $B$ be a general Hermitian matrix.

Under what conditions on $B$ (probably in terms of $A$) is $A-B$ positive semidefinite?

I was thinking that it may be along the lines of the generalized eigenvalue problem but can't quite see how.

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For your $A-B$ to be positive semidefinite,

  • one sufficient condition is $B$ being negative semidefinite;
  • one necessary condition is $\lambda_1^\downarrow(B)\le\lambda_1^\downarrow(A)$ and $\lambda_2^\downarrow(B),\,\lambda_3^\downarrow(B),\ldots,\, \lambda_n^\downarrow(B)\le0$, where $\lambda_k^\downarrow(\cdot)$ means the $k$-th largest eigenvalue; this is a consequence of Weyl's inequality;
  • if you are looking for a necessary and sufficient condition in terms of the eigenvalues of $A$ and $B$, I'm afraid there isn't one.
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    If you only know that each $A_i$ has rank 1, but don't know the exact $v_i$, I think not much can be said about $B$. If you do know what each $v_i$ looks like, then it may be possible to characterize $B$, but there is also a possibility that $B$ does not exist.2012-12-12