Re: please solve

*To*: mathgroup at smc.vnet.net*Subject*: [mg52942] Re: please solve*From*: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>*Date*: Fri, 17 Dec 2004 05:19:25 -0500 (EST)*References*: <cprk0m$r87$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Your matrix x'x is singular because x is rank 1 (all rows the same) so it can't be uniquely inverted. You could try replacing it by x'x + eps I where eps is a small positive number and I is the identity matrix. This has the effect of regularising the inverse matrix. x = Table[t^i, {j, 3}, {i, 0, 2}] gives {{1, t, t^2}, {1, t, t^2}, {1, t, t^2}} and Simplify[Inverse[x.Transpose[x]+\[Epsilon] IdentityMatrix[3]]] gives {{(2 + 2*t^2 + 2*t^4 + \[Epsilon])/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon])), -((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon]))), -((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon])))}, {-((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon]))), (2 + 2*t^2 + 2*t^4 + \[Epsilon])/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon])), -((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon])))}, {-((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon]))), -((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon]))), (2 + 2*t^2 + 2*t^4 + \[Epsilon])/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon]))}} and Simplify[Inverse[x.Transpose[x]+\[Epsilon] IdentityMatrix[3]].x] gives {{1/(3 + 3*t^2 + 3*t^4 + \[Epsilon]), t/(3 + 3*t^2 + 3*t^4 + \[Epsilon]), t^2/(3 + 3*t^2 + 3*t^4 + \[Epsilon])}, {1/(3 + 3*t^2 + 3*t^4 + \[Epsilon]), t/(3 + 3*t^2 + 3*t^4 + \[Epsilon]), t^2/(3 + 3*t^2 + 3*t^4 + \[Epsilon])}, {1/(3 + 3*t^2 + 3*t^4 + \[Epsilon]), t/(3 + 3*t^2 + 3*t^4 + \[Epsilon]), t^2/(3 + 3*t^2 + 3*t^4 + \[Epsilon])}} Steve Luttrell <rcmcll at yahoo.com> wrote in message news:cprk0m$r87$1 at smc.vnet.net... > Greetings: > > Could someone please solve this symbolically? > This is just the ols formula for beta-hat but I need a symbolic > solution for this special case. > > b = inv(x'x)x'y > > where > > x = 1 t t^2 > 1 t t^2 > 1 t t^2 > > > and simplify simplify simplify!! > > Thanks, > > Bob >