# Under what conditions is the difference between a rank $1$, postive semidefinite matrix and a hermitian matrix positive semidefinite?

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.
Thanks user1551. As a follow up, is it possible to prove that a necessary and sufficient condition in terms of the eigenvalues doesn't exist? Are there any further constraints on the form of $A$ and $B$ that would engineer such a condition? – Stan Dec 11 '12 at 10:06
@Stan This is no such necessary and sufficient condition because the eigenvalues of $A$ and $B$ do not characterize the eigenvalues of $A-B$. For instance, Let $A=\mathrm{diag}(1,0)$, $B=\mathrm{diag}(1/2,0)$ and $B'=\mathrm{diag}(0,1/2)$. Then $B$ and $B'$ have identical spectrum, both the pairs $A,B$ and $A',B'$ satisfy the necessary condition as specified in my answer, yet $A-B=0$ is positive semidefinite but $A-B'=\mathrm{diag}(1,-1)$ is not. – user1551 Dec 12 '12 at 1:00
This example shows that you not only need to look at the eigenvalues, but also need to look at how the eigenvectors of $A$ and $B$ align. But then if you want to find a necessary and sufficient condition in terms of both eigenvalues and eigenvectors, things would become more complicated and it may be simpler to directly determine whether $A-B$ itself is positive semidefinite. – user1551 Dec 12 '12 at 1:04
Ok, that makes sense. In the overall problem I'm considering $B$ is an $nxn$ matrix and I have $n$ such $A$ ($A_i$). I am trying to find the most general $B$ such that $A_i - B$ is positive semidefinite for all $i$. As they are rank $1$ projectors I can write $A_i=v_i v_i^T$ so I guess I'd want to be able to write $B$ in terms of these $v_i$ and their inner products such that it satisfies the constraints. Does it sound plausible to find a necessary and sufficient condition on the form of $B$ along those lines? – Stan Dec 12 '12 at 9:30
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. – user1551 Dec 12 '12 at 10:00