How to prove this matrix inequality $det(A+B)ge 2^nsqrt{det(A)det(B)}$












6












$begingroup$


Question:



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




Show that
$$det(A+B)ge 2^nsqrt{det(A)det(B)}.$$




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!






linear-algebra matrices inequality







share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    Question:



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




    Show that
    $$det(A+B)ge 2^nsqrt{det(A)det(B)}.$$




    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!






    linear-algebra matrices inequality







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      3



      $begingroup$


      Question:



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




      Show that
      $$det(A+B)ge 2^nsqrt{det(A)det(B)}.$$




      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!






      linear-algebra matrices inequality







      share|cite|improve this question











      $endgroup$




      Question:



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




      Show that
      $$det(A+B)ge 2^nsqrt{det(A)det(B)}.$$




      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!






      linear-algebra matrices inequality




      linear-algebra matrices inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 9:05









      dmtri

      1,5042521




      1,5042521










      asked Dec 5 '13 at 10:44









      china mathchina math

      10.2k631117




      10.2k631117






















          2 Answers
          2






          active

          oldest

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          3












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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



















          7












          $begingroup$

          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 $t = frac{1}{2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          3












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          3












          3








          3





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '13 at 10:59









          CasteelsCasteels

          9,99742234




          9,99742234












          • $begingroup$
            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


















          • $begingroup$
            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
















          $begingroup$
          this can AM-GM inequality? Thank you
          $endgroup$
          – china math
          Dec 5 '13 at 11:02




          $begingroup$
          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




          $begingroup$
          Sorry, I am not sure what you are asking?
          $endgroup$
          – Casteels
          Dec 5 '13 at 11:03











          7












          $begingroup$

          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 $t = frac{1}{2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          7












          $begingroup$

          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 $t = frac{1}{2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          7












          7








          7





          $begingroup$

          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 $t = frac{1}{2}$.






          share|cite|improve this answer









          $endgroup$



          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 $t = frac{1}{2}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 15 '14 at 0:48









          DRichDRich

          45438




          45438












          • $begingroup$
            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


















          • $begingroup$
            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
















          $begingroup$
          It's never too late for a good answer!
          $endgroup$
          – Casteels
          Apr 15 '14 at 7:42




          $begingroup$
          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




          $begingroup$
          Can you send me pdf document or links ? Thanks you very much!
          $endgroup$
          – Road Human
          Apr 6 '15 at 16:07


















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