Positive definiteness of difference of inverse matrices












6












$begingroup$


Let $A$ and $B$ be two $n times n$ symmetric and positive definite matrices.



If $A prec B$, then is it true that $B^{-1} prec A^{-1}$?



Here, $A prec B$ means that $B-A$ is positive definite.






linear-algebra matrices inverse positive-definite







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$endgroup$

















    6












    $begingroup$


    Let $A$ and $B$ be two $n times n$ symmetric and positive definite matrices.



    If $A prec B$, then is it true that $B^{-1} prec A^{-1}$?



    Here, $A prec B$ means that $B-A$ is positive definite.






    linear-algebra matrices inverse positive-definite







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      2



      $begingroup$


      Let $A$ and $B$ be two $n times n$ symmetric and positive definite matrices.



      If $A prec B$, then is it true that $B^{-1} prec A^{-1}$?



      Here, $A prec B$ means that $B-A$ is positive definite.






      linear-algebra matrices inverse positive-definite







      share|cite|improve this question











      $endgroup$




      Let $A$ and $B$ be two $n times n$ symmetric and positive definite matrices.



      If $A prec B$, then is it true that $B^{-1} prec A^{-1}$?



      Here, $A prec B$ means that $B-A$ is positive definite.






      linear-algebra matrices inverse positive-definite




      linear-algebra matrices inverse positive-definite






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 22 '17 at 17:47









      Rodrigo de Azevedo

      12.9k41857




      12.9k41857










      asked Aug 22 '17 at 16:36









      Sudipta RoySudipta Roy

      29518




      29518






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          HINT:



          Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $det(B-lambda A)=0$ has all its roots $lambdage 1$ and conversely if $lambdage1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)



          Then, $det(A^{-1}-mu B^{-1})=0$



          $Rightarrow det Bdet(A^{-1}-mu B^{-1})det Age0$ $quad($ as $A$ and $B$ are both p.d., $det A,det B>0)$



          $Rightarrow det(B-mu A)=0$



          So the roots of $det(A^{-1}-mu B^{-1})=0$ are the same as the roots of $det(B-mu A)=0$.



          As $B-A$ is n.n.d. we have $muge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            exercise from what book?
            $endgroup$
            – becko
            Jun 4 '18 at 15:56






          • 1




            $begingroup$
            @becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
            $endgroup$
            – StubbornAtom
            Jun 4 '18 at 16:03










          • $begingroup$
            I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
            $endgroup$
            – StubbornAtom
            Sep 15 '18 at 7:01



















          0












          $begingroup$

          For ease of notation I'll use $ge$ instead of $succeq$ and so on.



          Lemma 1: $A le B Rightarrow C^T A C le C^T B C$ for any conformable matrix $C.$



          Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x ge 0$ for any conformable vector $x$ so that $C^T(B-A)C ge 0.$



          Lemma 2: $I preceq B Rightarrow$ $B$ is invertible and $B^{-1} preceq I.$



          Proof.




          • Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G le G^T B G = D.$ Thus all eigenvalues of $B$ are $ge 1 >0,$ hence $B$ is invertible.

          • Now write $B^{-1} = B^{-1/2}IB^{-1/2} le B^{-1/2}BB^{-1/2} = I,$ where $B^alpha := GD^{alpha}G^T.$


          Proposition: $0 < A le B Rightarrow$ $B$ invertible and $B^{-1} le A^{-1}.$



          Proof.
          begin{align*}
          Ale B &Rightarrow B-A ge 0 \
          &Rightarrow A^{-1/2}(B-A)A^{-1/2} ge 0\
          &Rightarrow A^{-1/2}BA^{-1/2} ge I\
          &Rightarrow A^{1/2}B^{-1}A^{1/2} le I,\
          end{align*}
          hence
          begin{align*}
          B^{-1} &= A^{-1/2}A^{1/2}BA^{1/2}A^{-1/2}\
          &= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\
          &le A^{-1/2}IA^{-1/2}\
          &= A^{-1}.
          end{align*}






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
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            active

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            active

            oldest

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            active

            oldest

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            1












            $begingroup$

            HINT:



            Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $det(B-lambda A)=0$ has all its roots $lambdage 1$ and conversely if $lambdage1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)



            Then, $det(A^{-1}-mu B^{-1})=0$



            $Rightarrow det Bdet(A^{-1}-mu B^{-1})det Age0$ $quad($ as $A$ and $B$ are both p.d., $det A,det B>0)$



            $Rightarrow det(B-mu A)=0$



            So the roots of $det(A^{-1}-mu B^{-1})=0$ are the same as the roots of $det(B-mu A)=0$.



            As $B-A$ is n.n.d. we have $muge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              exercise from what book?
              $endgroup$
              – becko
              Jun 4 '18 at 15:56






            • 1




              $begingroup$
              @becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
              $endgroup$
              – StubbornAtom
              Jun 4 '18 at 16:03










            • $begingroup$
              I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
              $endgroup$
              – StubbornAtom
              Sep 15 '18 at 7:01
















            1












            $begingroup$

            HINT:



            Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $det(B-lambda A)=0$ has all its roots $lambdage 1$ and conversely if $lambdage1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)



            Then, $det(A^{-1}-mu B^{-1})=0$



            $Rightarrow det Bdet(A^{-1}-mu B^{-1})det Age0$ $quad($ as $A$ and $B$ are both p.d., $det A,det B>0)$



            $Rightarrow det(B-mu A)=0$



            So the roots of $det(A^{-1}-mu B^{-1})=0$ are the same as the roots of $det(B-mu A)=0$.



            As $B-A$ is n.n.d. we have $muge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              exercise from what book?
              $endgroup$
              – becko
              Jun 4 '18 at 15:56






            • 1




              $begingroup$
              @becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
              $endgroup$
              – StubbornAtom
              Jun 4 '18 at 16:03










            • $begingroup$
              I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
              $endgroup$
              – StubbornAtom
              Sep 15 '18 at 7:01














            1












            1








            1





            $begingroup$

            HINT:



            Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $det(B-lambda A)=0$ has all its roots $lambdage 1$ and conversely if $lambdage1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)



            Then, $det(A^{-1}-mu B^{-1})=0$



            $Rightarrow det Bdet(A^{-1}-mu B^{-1})det Age0$ $quad($ as $A$ and $B$ are both p.d., $det A,det B>0)$



            $Rightarrow det(B-mu A)=0$



            So the roots of $det(A^{-1}-mu B^{-1})=0$ are the same as the roots of $det(B-mu A)=0$.



            As $B-A$ is n.n.d. we have $muge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)






            share|cite|improve this answer









            $endgroup$



            HINT:



            Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $det(B-lambda A)=0$ has all its roots $lambdage 1$ and conversely if $lambdage1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)



            Then, $det(A^{-1}-mu B^{-1})=0$



            $Rightarrow det Bdet(A^{-1}-mu B^{-1})det Age0$ $quad($ as $A$ and $B$ are both p.d., $det A,det B>0)$



            $Rightarrow det(B-mu A)=0$



            So the roots of $det(A^{-1}-mu B^{-1})=0$ are the same as the roots of $det(B-mu A)=0$.



            As $B-A$ is n.n.d. we have $muge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 23 '17 at 15:43









            StubbornAtomStubbornAtom

            5,91811238




            5,91811238












            • $begingroup$
              exercise from what book?
              $endgroup$
              – becko
              Jun 4 '18 at 15:56






            • 1




              $begingroup$
              @becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
              $endgroup$
              – StubbornAtom
              Jun 4 '18 at 16:03










            • $begingroup$
              I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
              $endgroup$
              – StubbornAtom
              Sep 15 '18 at 7:01


















            • $begingroup$
              exercise from what book?
              $endgroup$
              – becko
              Jun 4 '18 at 15:56






            • 1




              $begingroup$
              @becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
              $endgroup$
              – StubbornAtom
              Jun 4 '18 at 16:03










            • $begingroup$
              I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
              $endgroup$
              – StubbornAtom
              Sep 15 '18 at 7:01
















            $begingroup$
            exercise from what book?
            $endgroup$
            – becko
            Jun 4 '18 at 15:56




            $begingroup$
            exercise from what book?
            $endgroup$
            – becko
            Jun 4 '18 at 15:56




            1




            1




            $begingroup$
            @becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
            $endgroup$
            – StubbornAtom
            Jun 4 '18 at 16:03




            $begingroup$
            @becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
            $endgroup$
            – StubbornAtom
            Jun 4 '18 at 16:03












            $begingroup$
            I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
            $endgroup$
            – StubbornAtom
            Sep 15 '18 at 7:01




            $begingroup$
            I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
            $endgroup$
            – StubbornAtom
            Sep 15 '18 at 7:01











            0












            $begingroup$

            For ease of notation I'll use $ge$ instead of $succeq$ and so on.



            Lemma 1: $A le B Rightarrow C^T A C le C^T B C$ for any conformable matrix $C.$



            Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x ge 0$ for any conformable vector $x$ so that $C^T(B-A)C ge 0.$



            Lemma 2: $I preceq B Rightarrow$ $B$ is invertible and $B^{-1} preceq I.$



            Proof.




            • Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G le G^T B G = D.$ Thus all eigenvalues of $B$ are $ge 1 >0,$ hence $B$ is invertible.

            • Now write $B^{-1} = B^{-1/2}IB^{-1/2} le B^{-1/2}BB^{-1/2} = I,$ where $B^alpha := GD^{alpha}G^T.$


            Proposition: $0 < A le B Rightarrow$ $B$ invertible and $B^{-1} le A^{-1}.$



            Proof.
            begin{align*}
            Ale B &Rightarrow B-A ge 0 \
            &Rightarrow A^{-1/2}(B-A)A^{-1/2} ge 0\
            &Rightarrow A^{-1/2}BA^{-1/2} ge I\
            &Rightarrow A^{1/2}B^{-1}A^{1/2} le I,\
            end{align*}
            hence
            begin{align*}
            B^{-1} &= A^{-1/2}A^{1/2}BA^{1/2}A^{-1/2}\
            &= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\
            &le A^{-1/2}IA^{-1/2}\
            &= A^{-1}.
            end{align*}






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              For ease of notation I'll use $ge$ instead of $succeq$ and so on.



              Lemma 1: $A le B Rightarrow C^T A C le C^T B C$ for any conformable matrix $C.$



              Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x ge 0$ for any conformable vector $x$ so that $C^T(B-A)C ge 0.$



              Lemma 2: $I preceq B Rightarrow$ $B$ is invertible and $B^{-1} preceq I.$



              Proof.




              • Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G le G^T B G = D.$ Thus all eigenvalues of $B$ are $ge 1 >0,$ hence $B$ is invertible.

              • Now write $B^{-1} = B^{-1/2}IB^{-1/2} le B^{-1/2}BB^{-1/2} = I,$ where $B^alpha := GD^{alpha}G^T.$


              Proposition: $0 < A le B Rightarrow$ $B$ invertible and $B^{-1} le A^{-1}.$



              Proof.
              begin{align*}
              Ale B &Rightarrow B-A ge 0 \
              &Rightarrow A^{-1/2}(B-A)A^{-1/2} ge 0\
              &Rightarrow A^{-1/2}BA^{-1/2} ge I\
              &Rightarrow A^{1/2}B^{-1}A^{1/2} le I,\
              end{align*}
              hence
              begin{align*}
              B^{-1} &= A^{-1/2}A^{1/2}BA^{1/2}A^{-1/2}\
              &= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\
              &le A^{-1/2}IA^{-1/2}\
              &= A^{-1}.
              end{align*}






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                For ease of notation I'll use $ge$ instead of $succeq$ and so on.



                Lemma 1: $A le B Rightarrow C^T A C le C^T B C$ for any conformable matrix $C.$



                Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x ge 0$ for any conformable vector $x$ so that $C^T(B-A)C ge 0.$



                Lemma 2: $I preceq B Rightarrow$ $B$ is invertible and $B^{-1} preceq I.$



                Proof.




                • Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G le G^T B G = D.$ Thus all eigenvalues of $B$ are $ge 1 >0,$ hence $B$ is invertible.

                • Now write $B^{-1} = B^{-1/2}IB^{-1/2} le B^{-1/2}BB^{-1/2} = I,$ where $B^alpha := GD^{alpha}G^T.$


                Proposition: $0 < A le B Rightarrow$ $B$ invertible and $B^{-1} le A^{-1}.$



                Proof.
                begin{align*}
                Ale B &Rightarrow B-A ge 0 \
                &Rightarrow A^{-1/2}(B-A)A^{-1/2} ge 0\
                &Rightarrow A^{-1/2}BA^{-1/2} ge I\
                &Rightarrow A^{1/2}B^{-1}A^{1/2} le I,\
                end{align*}
                hence
                begin{align*}
                B^{-1} &= A^{-1/2}A^{1/2}BA^{1/2}A^{-1/2}\
                &= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\
                &le A^{-1/2}IA^{-1/2}\
                &= A^{-1}.
                end{align*}






                share|cite|improve this answer









                $endgroup$



                For ease of notation I'll use $ge$ instead of $succeq$ and so on.



                Lemma 1: $A le B Rightarrow C^T A C le C^T B C$ for any conformable matrix $C.$



                Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x ge 0$ for any conformable vector $x$ so that $C^T(B-A)C ge 0.$



                Lemma 2: $I preceq B Rightarrow$ $B$ is invertible and $B^{-1} preceq I.$



                Proof.




                • Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G le G^T B G = D.$ Thus all eigenvalues of $B$ are $ge 1 >0,$ hence $B$ is invertible.

                • Now write $B^{-1} = B^{-1/2}IB^{-1/2} le B^{-1/2}BB^{-1/2} = I,$ where $B^alpha := GD^{alpha}G^T.$


                Proposition: $0 < A le B Rightarrow$ $B$ invertible and $B^{-1} le A^{-1}.$



                Proof.
                begin{align*}
                Ale B &Rightarrow B-A ge 0 \
                &Rightarrow A^{-1/2}(B-A)A^{-1/2} ge 0\
                &Rightarrow A^{-1/2}BA^{-1/2} ge I\
                &Rightarrow A^{1/2}B^{-1}A^{1/2} le I,\
                end{align*}
                hence
                begin{align*}
                B^{-1} &= A^{-1/2}A^{1/2}BA^{1/2}A^{-1/2}\
                &= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\
                &le A^{-1/2}IA^{-1/2}\
                &= A^{-1}.
                end{align*}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 29 '18 at 14:09









                EpiousiosEpiousios

                1,545622




                1,545622






























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