Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I take it $A$ is symmetric but $C$ may not be? – Rahul Apr 29 '12 at 4:28
$A=[1,0;2,1]$ is PSD, but not symmetric since $(u,v)A(u,v)^T=(u+v)^2\ge0$. – sai Apr 29 '12 at 4:58
up vote 3 down vote accepted

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.

share|cite|improve this answer
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. – Keivan Apr 30 '12 at 7:01
Yes, that's right. – Robert Israel Apr 30 '12 at 7:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.