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Math Help - A simple Demonstration...

  1. #1
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    A simple Demonstration...

    Hi, I've a problem with a (I know it must be very easy but I don't get it) demonstration about matrices...

    The thing I've to show is...

    Let A be a square matrix such as for all B (With the same order as A) AB=0. Then A=0.

    I don't wanna prove that there are no zero divisors in this ring (Because there are)...just that there isn't a zero divisor A in this ring that AB=0 for ALL B and the only matrix that does that is the zero matrix of that order...this is what i mean...
    --------------------------------------------------------
    Sorry, Problem solved...It was truly a very simple (Trivial) demonstration...
    Last edited by Leviathantheesper; September 1st 2013 at 11:30 AM. Reason: Forget it, it was too easy...
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: A simple Demonstration...

    To spoil it for anyone--take B to be the identity matrix. More generally, if R is a ring, and M is an R-module, then Ann_R(M) denotes the set of all r in R such that rM=0. You're probably realizing that if R is a non-trivial ring, then it's fairly easy to see that Ann_R(R)={0}.
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  3. #3
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    Re: A simple Demonstration...

    Quote Originally Posted by Drexel28 View Post
    To spoil it for anyone--take B to be the identity matrix.
    Yes, that's it...I solved it seeing that...
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