When is the following negative semi-definite?

Question

I would like to know under what conditions on $c,A$ would the following be (symmetric) negative semi-definite,

$$\Delta = (A + c c^T)^{-2} - A^{-2} $$

where $A$ is a (symmetric) positive definite matrix where $c$ is a column vector.

Answer 1

Suppose $c\ne0$. The statement $(A+cc^T)^{-2}\preceq A^{-2}$ is equivalent to $A^2\preceq(A+cc^T)^2$, which in turn is equivalent to $D=\|c\|^2cc^T + (Ac)c^T + c(Ac)^T\succeq0$. Let $\{c,c^\perp\}$ be an orthogonal basis of a vector subspace containing $c$ and $Ac$. Let also $Ac=\lambda c+\mu c^\perp$. Then $D$ is positive semidefinite iff $\pmatrix{\|c\|^2+2\lambda&\mu\\ \mu&0}\succeq0$, meaning that $\lambda\ge-\frac{\|c\|^2}2,\ \mu=0$ and $Ac=\lambda c$. However, as $A$ is positive definite, if $(\lambda,c)$ is an eigenpair, $\lambda$ must be positive. Therefore the condition $\lambda\ge-\frac{\|c\|^2}2$ is redundant.

Thus we conclude that $(A+cc^T)^{-2}\preceq A^{-2}$ if and only if $c=0$ or $c$ is an eigenvector of $A$.