# A simple Demonstration...

• Sep 1st 2013, 10:11 AM
Leviathantheesper
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...
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Sorry, Problem solved...It was truly a very simple (Trivial) demonstration...
• Sep 4th 2013, 10:14 PM
Drexel28
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}. :)
• Sep 5th 2013, 05:50 PM
Leviathantheesper
Re: A simple Demonstration...
Quote:

Originally Posted by Drexel28
To spoil it for anyone--take B to be the identity matrix.

Yes, that's it...I solved it seeing that...