# How to prove this inequality for determinant of Hermitian block matrix?

I am given an Hermitian positive definite matrix $$D=\left(\begin{matrix}A&\overline{C}^T\\C&B\end{matrix}\right)$$ $A$ and $B$ are square matrices. The task is to prove the following inequalities:

$\det(D)\leq\det(A)\det(B)\\\det(D)\leq\prod_{i=1}^{n}(d_{ii})$

All I can see is that $A,B$ are also Hermitian, and that determinant is equal to product of eigenvalues for $D, A, B$ since they are Hermitian, and those are positive since matrix is positive definite, and I have no further idea. Maybe you could give an advice or kind of hint.

• sure, they are the same. just edited – nakajuice Jun 5 '13 at 5:54

Hints:

(1) If $D$ is positive semidefinite, so are $A$ and $B$.

(2) If your first inequality holds, the second follows from it by induction. (The second inequality is known as Hadamard's inequality and can be proved more easily than the first.)

(3) Using the Schur complement, $\det(D)=\det(A) \det(B - C^*A^{-1}C)$ and $B-C^*A^{-1}C$ is positive semidefinite.

(4) If $B=B_1+C_1$ where $B_1$ and $C_1$ are positive semidefinite, then $\det(B) \ge \det(B_1)$.

• Could you please give some explanation on (4)? – nakajuice Jun 5 '13 at 9:38
• If $B$ is positive definite and $B-D$ is positive semidefinite, then $I-B^{-1/2}DB^{1/2}$ is psd. So each eigenvalue of $B^{-1/2}DB^{1/2}$ is less than 1, and so $\det(B^{-1/2}DB^{1/2}) \le 1$. – Chris Godsil Jun 5 '13 at 11:38
• $B$ and $D$ have different size – nakajuice Jun 5 '13 at 12:24
• For $D$ read, e.g., $M$. – Chris Godsil Jun 5 '13 at 13:22
• Sorry, but I still don't understand how does to refer to my case. What is left is to prove that $\det(B)\geq\det(B-CA^{-1}C*)$. Actually I have difference of pos. def. matrices in parentheses, so what should I do further? – nakajuice Jun 6 '13 at 18:48