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












2














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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    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

























      2












      2








      2







      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













      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






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      share|cite|improve this question











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      asked 2 days ago









      Biswarup SahaBiswarup Saha

      542110




      542110






















          1 Answer
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          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























          • Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
            – loup blanc
            2 days ago










          • You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
            – Mostafa Ayaz
            2 days ago










          • 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.
            – loup blanc
            2 days ago










          • Thank you for pointing that out. I will add a counter example on $Bbb C$
            – Mostafa Ayaz
            2 days ago











          Your Answer





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          1 Answer
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          active

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          1 Answer
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          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          4














          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























          • Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
            – loup blanc
            2 days ago










          • You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
            – Mostafa Ayaz
            2 days ago










          • 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.
            – loup blanc
            2 days ago










          • Thank you for pointing that out. I will add a counter example on $Bbb C$
            – Mostafa Ayaz
            2 days ago
















          4














          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























          • Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
            – loup blanc
            2 days ago










          • You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
            – Mostafa Ayaz
            2 days ago










          • 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.
            – loup blanc
            2 days ago










          • Thank you for pointing that out. I will add a counter example on $Bbb C$
            – Mostafa Ayaz
            2 days ago














          4












          4








          4






          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














          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 2 days ago

























          answered 2 days ago









          Mostafa AyazMostafa Ayaz

          14.1k3937




          14.1k3937












          • Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
            – loup blanc
            2 days ago










          • You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
            – Mostafa Ayaz
            2 days ago










          • 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.
            – loup blanc
            2 days ago










          • Thank you for pointing that out. I will add a counter example on $Bbb C$
            – Mostafa Ayaz
            2 days ago


















          • Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
            – loup blanc
            2 days ago










          • You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
            – Mostafa Ayaz
            2 days ago










          • 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.
            – loup blanc
            2 days ago










          • Thank you for pointing that out. I will add a counter example on $Bbb C$
            – Mostafa Ayaz
            2 days ago
















          Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
          – loup blanc
          2 days ago




          Over $mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $mathbb{C}$, that does no imply that $det(U)=1$.
          – loup blanc
          2 days ago












          You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
          – Mostafa Ayaz
          2 days ago




          You are right. The statement doesn't hold generally over $Bbb C$ but the question has mentioned with real entries
          – Mostafa Ayaz
          2 days ago












          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.
          – loup blanc
          2 days ago




          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.
          – loup blanc
          2 days ago












          Thank you for pointing that out. I will add a counter example on $Bbb C$
          – Mostafa Ayaz
          2 days ago




          Thank you for pointing that out. I will add a counter example on $Bbb C$
          – Mostafa Ayaz
          2 days ago


















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