# Determinant inequality and positive definite matrix

Let $B$ and $C$ be $n \times n$ Hermitian matrices, with $B$ positive definite and $C$ positive semidefinite.

1. Show that $B+C$ is positive definite

2. Show that $\det(B) \leq \det(B+C)$. What is the equality case?

3. 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.

• OP: any feedback on my answer below? – Kim Jong Un Jul 30 '15 at 5:43

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?

• I got it thank you very much – nerd Jul 30 '15 at 14:58
• wait. Why are the eigenvalues of $I+B^{-1/2}CB^{-1/2}$ greater than 1? Is it because $B,C$ are positive definite, so their square roots and inverse have positive eigenvalues. So adding $I$ their eigenvalues increases by 1. – nerd Jul 30 '15 at 15:11
• @nerd Yes, the eigenvalues of $I+B^{-1/2}CB^{-1/2}$ are just $1$ plus the eigenvalues of $B^{-1/2}CB^{-1/2}$. – Kim Jong Un Jul 30 '15 at 15:19
• Why would you name your account after a horrible ruthless dictator?! – Akiva Weinberger Oct 11 '15 at 2:59

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 $\mu_i \geq 0, i = 1, \ldots, n$.