Do the eigenvalues of $\left( A^H A \right)^{-1} - \left( A^H A + S \right)^{-1}$ lie in $(0,1)$?

Question

Many numerical examples shows that the eigenvalues of

$$ P := \left( A^H A \right)^{-1} - \left( A^H A + S \right)^{-1} $$

where $A$ is full column rank matrix and $S$ is a nonnegative diagonal matrix, lie in $(0,1)$? I guess it holds in the general case, but I don't know how to prove. Can someone please help?

Yes. And in my problem, the diagonal entries are positive and identical. — ecook, Mar 18 '25 at 00:55
Consider $S=tI$. If $(A^HA)^{-1}$ has an eigenvalue greater than $1$, so does $(A^HA)^{-1}-(A^HA+tI)^{-1}$ when $t>0$ is sufficiently large. — user1551, Mar 18 '25 at 10:07

score 2 · Answer 1 · answered Mar 17 '25 at 11:26

$$A^HA = S = 1/3\implies (A^HA)^{-1}-(A^HA+S)^{-1} = 3/2>1$$ but it's true that $P$ is positive definite due to Loewner ordering: $$ A^HA+S>A^HA\implies (A^HA)^{-1}>(A^HA+S)^{-1} $$

see Loewner ordering of symetric positive definite matrices and their inverse

Do the eigenvalues of $\left( A^H A \right)^{-1} - \left( A^H A + S \right)^{-1}$ lie in $(0,1)$?

1 Answers1