52 967
Assignments Done
Successfully Done
In October 2017
Your physics homework can be a real challenge, and the due date can be really close — feel free to use our assistance and get the desired result.
Be sure that math assignments completed by our experts will be error-free and done according to your instructions specified in the submitted order form.
Our experts will gladly share their knowledge and help you with programming homework. Keep up with the world’s newest programming trends.

Answer on Algebra Question for Mohammad

Question #17363
If R is a commutative principal ideal domain, show that the von Neumann regular elements of Mn(R) are precisely matrices of the form UAV , where U, V ∈ GLn(R), and A = diag(1, . . . , 1, 0, . . . , 0).
Expert's answer
First observe that, if u, vare units in any ring S, then an element a ∈ S is von Neumann regular iff uav is. (To see this, it suffices to provethe “only if” part. Suppose a = asa for some s ∈ S; we call any such s a “pseudo-inverse”of a. To find a pseudo-inverse t for uav, we need to solvethe equation: uav = (uav)t(uav). This amounts to a= a(vtu)a, so it suffices to choose t = v−1su−1,and all pseudo-inverses for uav arise in this way.) Applying this remarkto Mn(R), it follows that, for U, V , A,UAV is von Neumann regular (since A = A2 = A3is). Conversely, let B be any von Neumann regular element in Mn(R).By the Smith Normal Form Theorem, there exist P, Q ∈ GLn(R) such that PBQ = diag(b1, . . . , bn),where bi+1 ∈ biR for each i < n. By the remark above,diag(b1, . . . , bn) is von Neumann regular. Thus, upon interpretingMn(R) as EndR(Rn), im(diag(b1,. . . , bn)) = b1R⊕· · ·⊕bnR ⊆ Rn must be a direct summand of Rn. Thisis possible only if each biR is either 0 or R. Sincethe “elementary divisors” bi are determined up to units anyway, we mayassume that
diag(b1, . . . , bn) =diag(1, . . . , 1, 0, . . . , 0) := A,
and so B = UAV for U= P−1 and V = Q−1in GLn(R).

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


No comments. Be first!

Leave a comment

Ask Your question