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Math Help - AB = I_n can we obtain BA =I_n

  1. #1
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    AB = I_n can we obtain BA =I_n

    Let A,B are square matrices n x n; and I_n is identity matrix.
    If AB =I_n, can we obtain BA =I_n?
    Thanks for your help.
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  2. #2
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    Quote Originally Posted by mahefo View Post
    Let A,B are square matrices n x n; and I_n is identity matrix.
    If AB =I_n, can we obtain BA =I_n?
    Thanks for your help.
    Yes, since A and B are inverses of each other.
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    I know the definition of inverse matrix.
    Can you show me the way from AB=I_n to get BA=I_n?
    Thanks a lot.
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    Quote Originally Posted by mahefo View Post
    I know the definition of inverse matrix.
    Can you show me the way from AB=I_n to get BA=I_n?
    Thanks a lot.

    Ok, then you hopefully know that the product of two matrices is invertible iff every one of them is invertible, so in our case both matrices are invertible, but then:

    BA=A^{-1}(AB)A=A^{-1}\cdot I\cdot A = A^{-1}A=I

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post

    BA=A^{-1}(AB)A
    well, that is not quite right because what we have is that A has a right inverse. we don't know yet if the left inverse of A exists or not.

    Quote Originally Posted by mahefo
    Let A,B are square matrices n x n; and I_n is identity matrix.
    If AB =I_n, can we obtain BA =I_n?
    Thanks for your help.
    see my second post in this thread: http://www.mathhelpforum.com/math-he...-fun-16-a.html
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