Let $A_{n\times n}$ and $B_{n\times n}$ be positive Hermitian matrices.

Show that $$\det(A+B)\ge 2^n\sqrt{\det(A)\det(B)}.$$

I know that $$\det(A+B)\ge \det(A)+\det(B)$$

But my problem is that I can't,(maybe this is an old reslut,and also I can't find it),

Thank you very much!


This is a corollary of Minkowski's Determinant Theorem: $\det(A+B)^\frac{1}{n}\geq \det(A)^\frac{1}{n}+\det(B)^\frac{1}{n}.$ Apply AM-GM inequality to the right-hand side.

| cite | improve this answer | |
  • $\begingroup$ this can AM-GM inequality? Thank you $\endgroup$ – china math Dec 5 '13 at 11:02
  • $\begingroup$ Sorry, I am not sure what you are asking? $\endgroup$ – Casteels Dec 5 '13 at 11:03

This seems to be a bit late for an answer, but the following theorem is proven in "Linear Algebra" by Lax in chapter 10.

Let $A$ and $B$ be self-adjoint, positive, $n \times n$ matrices. Then for all $0<t<1,$ \begin{align} \det(tA + (1-t)B) \geq (\det A)^{t}(\det B)^{1-t}. \end{align} Your answer follows with $t = \frac{1}{2}$.

| cite | improve this answer | |
  • $\begingroup$ It's never too late for a good answer! $\endgroup$ – Casteels Apr 15 '14 at 7:42
  • $\begingroup$ Can you send me pdf document or links ? Thanks you very much! $\endgroup$ – Road Human Apr 6 '15 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.