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Thread: AB=A+B;Prove A and B commute

  1. #1
    Senior Member pankaj's Avatar
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    AB=A+B;Prove A and B commute

    If $\displaystyle A$ and $\displaystyle B$ are two square matrices of the same order such that $\displaystyle AB=A+B$ then prove that A and B commute i.e. AB=BA
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  2. #2
    Super Member Anonymous1's Avatar
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    Umm working on it.

    $\displaystyle AB = A+B$

    $\displaystyle A(BB^{-1}) = (A + B)B^{-1}$

    $\displaystyle A = (AB^{-1} + I)$

    $\displaystyle BA = BAB^{-1} + B$


    ...
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  3. #3
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    Quote Originally Posted by pankaj View Post
    If $\displaystyle A$ and $\displaystyle B$ are two square matrices of the same order such that $\displaystyle AB=A+B$ then prove that A and B commute i.e. AB=BA
    If $\displaystyle AB=A+B$ then $\displaystyle (A-I)(B-I)=I$ (I is the identity matrix). So $\displaystyle B-I$ is the inverse of $\displaystyle A-I$ and therefore commutes with it. Thus $\displaystyle (B-I)(A-I) = I$, from which $\displaystyle BA = A+B = AB$.
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    Senior Member pankaj's Avatar
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    So if $\displaystyle CD=I$ does it always mean that $\displaystyle C$ and $\displaystyle D$ are inverse of each other.What if any one of $\displaystyle C$ or $\displaystyle D$ is singular matrix
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  5. #5
    MHF Contributor
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    Quote Originally Posted by pankaj View Post
    So if $\displaystyle CD=I$ does it always mean that $\displaystyle C$ and $\displaystyle D$ are inverse of each other.What if any one of $\displaystyle C$ or $\displaystyle D$ is singular matrix
    If $\displaystyle CD = I$ then C and D cannot be singular. One way to see this is to notice that the determinant is multiplicative: $\displaystyle \det(C)\det(D) = \det(I) = 1$, so that $\displaystyle \det(C)$ and $\displaystyle \det(D)$ cannot be 0.
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