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Let $A$ be a positive semi-definite matrix, and let $C$ be a rank 1 matrix. Prove that $A-C$ has at most one negative eigenvalue.

PS: It's easy to show that if $A$ and $C$ commute, then the statement is true, but most of the time they do not commute.

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  • $\begingroup$ I take it $A$ is symmetric but $C$ may not be? $\endgroup$ – Rahul Apr 29 '12 at 4:28
  • $\begingroup$ $A=[1,0;2,1]$ is PSD, but not symmetric since $(u,v)A(u,v)^T=(u+v)^2\ge0$. $\endgroup$ – sai Apr 29 '12 at 4:58
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Suppose not. Then $B=A-C$ has two linearly independent eigenvectors $u_i$ for negative eigenvalues. Thus for any nonzero $u \in V = \text{Span}(u_1, u_2)$, $u^T B u < 0$. Now we can write $C = a b^T$ for some vectors $a$ and $b$ , and there is some nonzero $v \in V$ with $b^T v = 0$. Then $C v = 0$ so $v^T A v = v^T B v < 0$, contradicting the statement that $A$ is PSD.

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  • $\begingroup$ Thanks @Robert. Very helpful and clear answer. I just want to clarify that $b^Tv=0$ has a nonzero solutions in $V$, since the solution space of $b^Tx=0$ is $n-1$ dimensional and $V$ is 2 dimensional, so the intersection is nontrivial. $\endgroup$ – Keivan Apr 30 '12 at 7:01
  • $\begingroup$ Yes, that's right. $\endgroup$ – Robert Israel Apr 30 '12 at 7:06

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