Proofs on Idempotent Matrices
Date: 05/09/2000 at 02:07:07 From: Laura Jones Subject: Abstract Algebra: Idempotents Hello. My final exam in abstract algebra is tomorrow and this problem appeared on an exam early in the semester, but I'm a little confused about it. Can you explain and help me work through it? An element e of a ring is said to be IDEMPOTENT if e^2 = e. (a) Determine all idempotent 2x2 matrices over R (b) Prove that in an integral domain, the only idempotents are 0 and 1. I think I've got part (b): e^2 = e e^2 - e = 0 e(e-1) = 0 so e = 0 or e = 1 This is what I've done for part (a): [ a b ] [a b ] = [ a b ] [ c d ] [c d ] [ c d ] so [ a^2 + bc ab + bd ] = [ a b ] [ ac + ad bd + d^2 ] [ c d ] so a^2 + bc = a ac + ad = c a(c+d) = c ab + bd = b b(a+d) = b a+d = 1 a = 1-d bd + d^2 = d then (1-d)(1-d) + bc = (1-d) But now what? [1 0] [0 0] I know that [0 1] and [0 0] are idempotents, but there are more. What are they? Thanks for any help you may be able to offer. I'm sure the exam won't have anything to do with this problem but I would still like to understand it! Laura
Date: 05/09/2000 at 13:53:35 From: Doctor Rob Subject: Re: Abstract Algebra: Idempotents Thanks for writing to Ask Dr. Math, Laura. You have started on a path which can get you the right answer, but you've made a couple of errors. [ a b ] [ a b ] = [ a b ] [ c d ] [ c d ] [ c d ] so [ a^2 + b*c a*b + b*d ] = [ a b ] [ a*c + c*d b*c + d^2 ] [ c d ] (notice two changes from your work above), so a^2 + b*c = a a*c + c*d = c c*(a+d) = c c = 0 or a + d = 1 a*b + b*d = b b*(a+d) = b b = 0 or a + d = 1 b*c + d^2 = d If a + d = 1 is false, then you can see that b = 0 and c = 0, so a^2 = a and d^2 = d. The solutions here are the two you gave above, namely I and 0. If a + d = 1 is true, then you get the equations d = 1 - a b*c = a - a^2 = a*(1-a) = a*d = (1-d)*d = d - d^2. Pick any value for a you want, make d = 1 - a, and then pick b and c so that b*c = a*d. That will give you an idempotent matrix. Furthermore, every idempotent matrix will either be I, 0, or one formed in this way. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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