Positive-definite matrix inequality $\begin{vmatrix} A & B\\ B^{\prime} & D \end{vmatrix} \leqslant\mid A\mid\mid D\mid $

Question

Prove$$ \begin{vmatrix} A & B\\ B^{\prime} & D \end{vmatrix} \leqslant\mid A\mid\mid D\mid $$ Where $\begin{bmatrix} A & B\\ B^{\prime} & D \end{bmatrix}$ is an $n$-order positive-definite matrix and $\dim (A) =r$ and equality holds iff $B=0$

Answer 1

Since by the Schur decomposition

$$\left[\begin{array}{cc}A&B\\B'&D \end{array}\right]= \left[\begin{array}{cc}I&0\\B'A^{-1}&I \end{array}\right] \left[\begin{array}{cc}A&0\\0&D -B'A^{-1}B\end{array}\right] \left[\begin{array}{cc}I&A^{-1}B\\0&I \end{array}\right],$$ enough is to show that $\det (D -B'A^{-1}B)\leq \det D.$ Now $C=B'A^{-1}B$ is semi positive definite (SDP) and $D_1=D-C$ is positive definite. Write $$D=D_1+C=D_1^{1/2}(I+D_1^{-1/2}CD_1^{-1/2})D_1^{1/2}$$ denote by $\lambda_1,\ldots,\lambda_n$ the eigenvalues of the SDP matrix $C_1=D_1^{-1/2}CD_1^{-1/2}$ and we get $$\det D=\det D_1\times \prod(1+\lambda_i)\geq \det D_1.$$ Equality occurs if and only if $C=0$, or $B'A^{-1}B=0.$ If $B_1,\ldots, B_{n-r}$ are the columns of $B$ we have $B'_iA^{-1}B_i=0$ for all $i$ and hence $B=0.$