52 967
Assignments Done
97,8%
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.

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 &isin; GLn(R), and A = diag(1, . . . , 1, 0, . . . , 0).
First observe that, if u, vare units in any ring S, then an element a &isin; S is von Neumann regular iff uav is. (To see this, it suffices to provethe &ldquo;only if&rdquo; part. Suppose a = asa for some s &isin; S; we call any such s a &ldquo;pseudo-inverse&rdquo;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&minus;1su&minus;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 &isin; GLn(R) such that PBQ = diag(b1, . . . , bn),where bi+1 &isin; biR for each i &lt; n. By the remark above,diag(b1, . . . , bn) is von Neumann regular. Thus, upon interpretingMn(R) as EndR(Rn), im(diag(b1,. . . , bn)) = b1R&oplus;&middot; &middot; &middot;&oplus;bnR &sube; Rn must be a direct summand of Rn. Thisis possible only if each biR is either 0 or R. Sincethe &ldquo;elementary divisors&rdquo; 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&minus;1 and V = Q&minus;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!