TITLE: Determinant inequality and positive definite matrix
QUESTION [10 upvotes]: 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.

REPLY [2 votes]: Let me give a proof using the canonical form theory of Hermitian matrices.
For simplicity, let $A^*$ denote the conjugate transpose of $A$, i.e., $A^* = \bar{A}^T$.
Since $B$ is Hermitian and $B > 0$, there exists a non-singular matrix $P$ such that $B = P^*P$ (for instance, $P$ can be taken as the square root of $B$). Now notice that the matrix $(P^*)^{-1}CP^{-1}$ is Hermitian, thus there exists a unitary matrix $U$ such that 
$$(P^*)^{-1}CP^{-1} = U\text{diag}(\mu_1, \ldots, \mu_n)U^*.$$
Since $C \geq 0$, the above expression implies $\mu_i \geq 0, i = 1, \ldots, n$. Let $Q \equiv P^*U$, then $Q$ is non-singular, and 
\begin{align*}
& B = QQ^*, \\
& C = Q\text{diag}(\mu_1, \ldots, \mu_n)Q^*. \tag{*}
\end{align*}
Hence $B + C = Q\text{diag}(1 + \mu_1, \ldots, 1 + \mu_n)Q^*$, consequently,
$$\det(B + C) = \det(QQ^*)\prod_{i = 1}^n(1 + \mu_i) = \det(B)\prod_{i = 1}^n(1 + \mu_i) \geq \det(B).$$
As $\mu_i \geq 0, i = 1, \ldots, n$, the above equality holds if and only if $\mu_1 = \cdots = \mu_n = 0$, i.e., when $C$ is a zero matrix.
Using the representation $(*)$, it can be easily seen that 
$$B^{-1} - (B + C)^{-1} = (Q^*)^{-1}\text{diag}\left(\frac{\mu_1}{1 + \mu_1}, \ldots, \frac{\mu_n}{1 + \mu_n}\right)Q^{-1} \geq 0,$$
where we again used the condition $\mu_i \geq 0, i = 1, \ldots, n$.