Let $B$ and $C$ be $n \times n$ Hermitian matrices, with $B$ positive definite and $C$ positive semidefinite.
Show that $B+C$ is positive definite
Show that $\det(B) \leq \det(B+C)$. What is the equality case?
Show that $B^{-1}-(B+C)^{-1}$ is positive semidefinite.
I proved (1). And for (2), I tried using Cholesky decomposition but does not gain anything.
Any hints, ideas? Thanks.
Best Answer
Hints for 2: because $B$ is positive definite, you can write $$ B+C=B^{1/2}(I+B^{-1/2}CB^{-1/2})B^{1/2}\implies\det(B+C)=\det(B)\det(I+B^{-1/2}CB^{-1/2}). $$ Now argue that the eigenvalues of $I+B^{-1/2}CB^{-1/2}$ are no less than $1$. What then can you say about the product of those eigenvalues?
Hints for 3: Let's prove a more general problem.
(a) First show that if $A$ is psd, then $I-A$ is psd implies $A^{-1}-I$ is psd.
Proof: $$ v'(A^{-1}-I)v=(A^{-1/2}v)'(I-A)(A^{-1/2}v)\geq0.\quad\square $$
(b) Suppose that $X$ and $Y$ are positive definite and $X-Y$ is psd. We claim that $Y^{-1}-X^{-1}$ is psd.
Proof: Use (a) along with the following observation: $$ 0\preceq X-Y=X^{1/2}(I-X^{-1/2}YX^{-1/2})X^{1/2}\implies 0\preceq I-X^{-1/2}YX^{-1/2}.\quad\square $$
Can you find the appropriate $X$ and $Y$ for your problem?