Let $A, B$ be $ntimes n(nge 2)$ nonsingular matrices with real entries such that $A^{-1} + B^{-1}...












2












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Whole question looks like-
Let $A, B$ be $ntimes n(nge 2)$ nonsingular matrices with real entries such that $A^{-1} + B^{-1 }=(A+B)^{-1}$, then prove that $operatorname{det}(A)=operatorname{det}(B)$.

Also show that this result is not valid for complex matrices.

I'm not getting any way out to solve this. Can anybody solve the problem?

Thanks for assistance in advance.






linear-algebra matrices matrix-equations







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    2












    $begingroup$


    Whole question looks like-
    Let $A, B$ be $ntimes n(nge 2)$ nonsingular matrices with real entries such that $A^{-1} + B^{-1 }=(A+B)^{-1}$, then prove that $operatorname{det}(A)=operatorname{det}(B)$.

    Also show that this result is not valid for complex matrices.

    I'm not getting any way out to solve this. Can anybody solve the problem?

    Thanks for assistance in advance.






    linear-algebra matrices matrix-equations







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Whole question looks like-
      Let $A, B$ be $ntimes n(nge 2)$ nonsingular matrices with real entries such that $A^{-1} + B^{-1 }=(A+B)^{-1}$, then prove that $operatorname{det}(A)=operatorname{det}(B)$.

      Also show that this result is not valid for complex matrices.

      I'm not getting any way out to solve this. Can anybody solve the problem?

      Thanks for assistance in advance.






      linear-algebra matrices matrix-equations







      share|cite|improve this question









      $endgroup$




      Whole question looks like-
      Let $A, B$ be $ntimes n(nge 2)$ nonsingular matrices with real entries such that $A^{-1} + B^{-1 }=(A+B)^{-1}$, then prove that $operatorname{det}(A)=operatorname{det}(B)$.

      Also show that this result is not valid for complex matrices.

      I'm not getting any way out to solve this. Can anybody solve the problem?

      Thanks for assistance in advance.






      linear-algebra matrices matrix-equations




      linear-algebra matrices matrix-equations






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 5 at 17:56









      Biswarup SahaBiswarup Saha

      557110




      557110






















          1 Answer
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          $begingroup$

          By multiplying both sides in $A$ we obtain $$I+AB^{-1}=A(A+B)^{-1}=(I+BA^{-1})^{-1}$$by defining $U=AB^{-1}$ we have $$I+U=(I+U^{-1})^{-1}$$therefore $$(I+U)(I+U^{-1})=I$$which leads to $$I+U+U^{-1}=0$$by multiplying both sides in $U(U-I)$ we obtain $$(I+U+U^{-1})U(U-I)=(U^2+U+I)(U-I)=U^3-I=0$$therefore $$U^3=I$$ which means that $$det(U)=1$$or equivalently$$det(A)=det(B)$$




          Counter example on $Bbb C$



          Let $A=I_2$ and $b=kI_2$ with $k={-1+isqrt 3over 2}$. Then we have $$A^{-1}+B^{-1}=I+{1over k}I\(A+B)^{-1}={1over 1+k}I$$since $$k^2+k+1=0$$ we have $$A^{-1}+B^{-1}=(A+B)^{-1}$$but $$det(B)=k^2=-1-k={-1-isqrt 3over 2}ne 1=det(A)$$




          Comment



          Even on $Bbb C$, from $U^3=I$ we can conclude that$$left|det(A)right|=left|det(B)right|$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
            $endgroup$
            – loup blanc
            Jan 5 at 18:44










          • $begingroup$
            You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
            $endgroup$
            – Mostafa Ayaz
            Jan 5 at 18:48










          • $begingroup$
            The OP asked a second question about the complex case. It's true that he could work (that would not hurt him) in order to clarify this point.
            $endgroup$
            – loup blanc
            Jan 5 at 19:21










          • $begingroup$
            Thank you for pointing that out. I will add a counter example on $Bbb C$
            $endgroup$
            – Mostafa Ayaz
            Jan 5 at 19:34











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          4












          $begingroup$

          By multiplying both sides in $A$ we obtain $$I+AB^{-1}=A(A+B)^{-1}=(I+BA^{-1})^{-1}$$by defining $U=AB^{-1}$ we have $$I+U=(I+U^{-1})^{-1}$$therefore $$(I+U)(I+U^{-1})=I$$which leads to $$I+U+U^{-1}=0$$by multiplying both sides in $U(U-I)$ we obtain $$(I+U+U^{-1})U(U-I)=(U^2+U+I)(U-I)=U^3-I=0$$therefore $$U^3=I$$ which means that $$det(U)=1$$or equivalently$$det(A)=det(B)$$




          Counter example on $Bbb C$



          Let $A=I_2$ and $b=kI_2$ with $k={-1+isqrt 3over 2}$. Then we have $$A^{-1}+B^{-1}=I+{1over k}I\(A+B)^{-1}={1over 1+k}I$$since $$k^2+k+1=0$$ we have $$A^{-1}+B^{-1}=(A+B)^{-1}$$but $$det(B)=k^2=-1-k={-1-isqrt 3over 2}ne 1=det(A)$$




          Comment



          Even on $Bbb C$, from $U^3=I$ we can conclude that$$left|det(A)right|=left|det(B)right|$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
            $endgroup$
            – loup blanc
            Jan 5 at 18:44










          • $begingroup$
            You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
            $endgroup$
            – Mostafa Ayaz
            Jan 5 at 18:48










          • $begingroup$
            The OP asked a second question about the complex case. It's true that he could work (that would not hurt him) in order to clarify this point.
            $endgroup$
            – loup blanc
            Jan 5 at 19:21










          • $begingroup$
            Thank you for pointing that out. I will add a counter example on $Bbb C$
            $endgroup$
            – Mostafa Ayaz
            Jan 5 at 19:34
















          4












          $begingroup$

          By multiplying both sides in $A$ we obtain $$I+AB^{-1}=A(A+B)^{-1}=(I+BA^{-1})^{-1}$$by defining $U=AB^{-1}$ we have $$I+U=(I+U^{-1})^{-1}$$therefore $$(I+U)(I+U^{-1})=I$$which leads to $$I+U+U^{-1}=0$$by multiplying both sides in $U(U-I)$ we obtain $$(I+U+U^{-1})U(U-I)=(U^2+U+I)(U-I)=U^3-I=0$$therefore $$U^3=I$$ which means that $$det(U)=1$$or equivalently$$det(A)=det(B)$$




          Counter example on $Bbb C$



          Let $A=I_2$ and $b=kI_2$ with $k={-1+isqrt 3over 2}$. Then we have $$A^{-1}+B^{-1}=I+{1over k}I\(A+B)^{-1}={1over 1+k}I$$since $$k^2+k+1=0$$ we have $$A^{-1}+B^{-1}=(A+B)^{-1}$$but $$det(B)=k^2=-1-k={-1-isqrt 3over 2}ne 1=det(A)$$




          Comment



          Even on $Bbb C$, from $U^3=I$ we can conclude that$$left|det(A)right|=left|det(B)right|$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
            $endgroup$
            – loup blanc
            Jan 5 at 18:44










          • $begingroup$
            You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
            $endgroup$
            – Mostafa Ayaz
            Jan 5 at 18:48










          • $begingroup$
            The OP asked a second question about the complex case. It's true that he could work (that would not hurt him) in order to clarify this point.
            $endgroup$
            – loup blanc
            Jan 5 at 19:21










          • $begingroup$
            Thank you for pointing that out. I will add a counter example on $Bbb C$
            $endgroup$
            – Mostafa Ayaz
            Jan 5 at 19:34














          4












          4








          4





          $begingroup$

          By multiplying both sides in $A$ we obtain $$I+AB^{-1}=A(A+B)^{-1}=(I+BA^{-1})^{-1}$$by defining $U=AB^{-1}$ we have $$I+U=(I+U^{-1})^{-1}$$therefore $$(I+U)(I+U^{-1})=I$$which leads to $$I+U+U^{-1}=0$$by multiplying both sides in $U(U-I)$ we obtain $$(I+U+U^{-1})U(U-I)=(U^2+U+I)(U-I)=U^3-I=0$$therefore $$U^3=I$$ which means that $$det(U)=1$$or equivalently$$det(A)=det(B)$$




          Counter example on $Bbb C$



          Let $A=I_2$ and $b=kI_2$ with $k={-1+isqrt 3over 2}$. Then we have $$A^{-1}+B^{-1}=I+{1over k}I\(A+B)^{-1}={1over 1+k}I$$since $$k^2+k+1=0$$ we have $$A^{-1}+B^{-1}=(A+B)^{-1}$$but $$det(B)=k^2=-1-k={-1-isqrt 3over 2}ne 1=det(A)$$




          Comment



          Even on $Bbb C$, from $U^3=I$ we can conclude that$$left|det(A)right|=left|det(B)right|$$






          share|cite|improve this answer











          $endgroup$



          By multiplying both sides in $A$ we obtain $$I+AB^{-1}=A(A+B)^{-1}=(I+BA^{-1})^{-1}$$by defining $U=AB^{-1}$ we have $$I+U=(I+U^{-1})^{-1}$$therefore $$(I+U)(I+U^{-1})=I$$which leads to $$I+U+U^{-1}=0$$by multiplying both sides in $U(U-I)$ we obtain $$(I+U+U^{-1})U(U-I)=(U^2+U+I)(U-I)=U^3-I=0$$therefore $$U^3=I$$ which means that $$det(U)=1$$or equivalently$$det(A)=det(B)$$




          Counter example on $Bbb C$



          Let $A=I_2$ and $b=kI_2$ with $k={-1+isqrt 3over 2}$. Then we have $$A^{-1}+B^{-1}=I+{1over k}I\(A+B)^{-1}={1over 1+k}I$$since $$k^2+k+1=0$$ we have $$A^{-1}+B^{-1}=(A+B)^{-1}$$but $$det(B)=k^2=-1-k={-1-isqrt 3over 2}ne 1=det(A)$$




          Comment



          Even on $Bbb C$, from $U^3=I$ we can conclude that$$left|det(A)right|=left|det(B)right|$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 5 at 19:41

























          answered Jan 5 at 18:13









          Mostafa AyazMostafa Ayaz

          15.5k3939




          15.5k3939












          • $begingroup$
            Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
            $endgroup$
            – loup blanc
            Jan 5 at 18:44










          • $begingroup$
            You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
            $endgroup$
            – Mostafa Ayaz
            Jan 5 at 18:48










          • $begingroup$
            The OP asked a second question about the complex case. It's true that he could work (that would not hurt him) in order to clarify this point.
            $endgroup$
            – loup blanc
            Jan 5 at 19:21










          • $begingroup$
            Thank you for pointing that out. I will add a counter example on $Bbb C$
            $endgroup$
            – Mostafa Ayaz
            Jan 5 at 19:34


















          • $begingroup$
            Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
            $endgroup$
            – loup blanc
            Jan 5 at 18:44










          • $begingroup$
            You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
            $endgroup$
            – Mostafa Ayaz
            Jan 5 at 18:48










          • $begingroup$
            The OP asked a second question about the complex case. It's true that he could work (that would not hurt him) in order to clarify this point.
            $endgroup$
            – loup blanc
            Jan 5 at 19:21










          • $begingroup$
            Thank you for pointing that out. I will add a counter example on $Bbb C$
            $endgroup$
            – Mostafa Ayaz
            Jan 5 at 19:34
















          $begingroup$
          Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
          $endgroup$
          – loup blanc
          Jan 5 at 18:44




          $begingroup$
          Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
          $endgroup$
          – loup blanc
          Jan 5 at 18:44












          $begingroup$
          You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 18:48




          $begingroup$
          You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 18:48












          $begingroup$
          The OP asked a second question about the complex case. It's true that he could work (that would not hurt him) in order to clarify this point.
          $endgroup$
          – loup blanc
          Jan 5 at 19:21




          $begingroup$
          The OP asked a second question about the complex case. It's true that he could work (that would not hurt him) in order to clarify this point.
          $endgroup$
          – loup blanc
          Jan 5 at 19:21












          $begingroup$
          Thank you for pointing that out. I will add a counter example on $Bbb C$
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 19:34




          $begingroup$
          Thank you for pointing that out. I will add a counter example on $Bbb C$
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 19:34


















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