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.
Best Answer
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.