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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.

Thanks in advance!

Rodrigo de Azevedo
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nakajuice
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1 Answers1

4

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 $\det(B) \ge \det(B_1)$.

Chris Godsil
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  • Could you please give some explanation on (4)? – nakajuice Jun 05 '13 at 09:38
  • 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 05 '13 at 11:38
  • $B$ and $D$ have different size – nakajuice Jun 05 '13 at 12:24
  • For $D$ read, e.g., $M$. – Chris Godsil Jun 05 '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 06 '13 at 18:48
  • $B-CA^{-1}C^$ is positive semidefinite (by the theory of the Schur complement. Also $CA^{-1}C^$ is positive semidefinite. As $B=(B-CA^{-1}C^)+(CA^{-1}C^)$ and by (4) this implies that $\det(B)\ge\det(B-CA^{-1}C^*)$. – Chris Godsil Jun 06 '13 at 19:38
  • Wow that's a cool feature of Schur's complement, I haven't heard about it, but prove by definition is a bit painful. Thank you, I'll accept your answer. – nakajuice Jun 06 '13 at 20:06
  • It's stated on the wikipedia page for the Schur complement, and is an easy consequence of the development. – Chris Godsil Jun 06 '13 at 21:29