Problem (C2). Here we are finding a basis for Range(L_S), i.e, a basis for ColSpan(S). Recall a previous posting [and classwork]: /--------------------------------------------------------\ Let c_1, ..., c_{97} denote the columns of S. Let R := RREF(S). Suppose the pivot-cols of R have indices 2, 5 and 19. /----- Method 1 ------------------------------------------\ To compute a basis for ColSpan(S), use the same col-numbers as give a basis for R. Thus collection {c_2, c_5, c_{19}} is a basis for ColSpan(S). After all, row-equiv doesn't change lin-relations among cols. \________________________________________________________/ \________________________________________________________/ date @poincare.math.ufl.edu: START of interactive part of "~/Top/jk.cshrc" unsetenv DISPLAY on Tue Oct 19 12:10:32 EDT 2010 "jk: END of 0.mapleinit" > S := <<0|1|2|0|0|2|1>,<0|1|2|1|-1|3|-1>,<0|-1|-2|-1|2|-3|3>,<0|2|4|1|-1|5|0>>; R := RREF(S); [0 1 2 0 0 2 1] [ ] [0 1 2 1 -1 3 -1] S := [ ] [0 -1 -2 -1 2 -3 3] [ ] [0 2 4 1 -1 5 0] [0 1 2 0 0 2 1] [ ] [0 0 0 1 0 1 0] R := [ ] [0 0 0 0 1 0 2] [ ] [0 0 0 0 0 0 0] %% Matrix R (which is in RREF) has pivot column-indices 2,4,5. %% Hence the corresponding columns in S form a %% basis for the column-span of S. > s2 := S[1..4,2]; s4 := S[1..4,4]; s5 := S[1..4,5]; [ 1] [ ] [ 1] s2 := [ ] [-1] [ ] [ 2] [ 0] [ ] [ 1] s4 := [ ] [-1] [ ] [ 1] [ 0] [ ] [-1] s5 := [ ] [ 2] [ ] [-1] > ScolsThenS := ; [ 1 0 0 0 1 2 0 0 2 1] [ ] [ 1 1 -1 0 1 2 1 -1 3 -1] ScolsThenS := [ ] [-1 -1 2 0 -1 -2 -1 2 -3 3] [ ] [ 2 1 -1 0 2 4 1 -1 5 0] %% We expect that there will be /no/ pivot-cols in the S-copy, since %% we've made the first 3 cols to be a basis for the colspan of S. > RREF(ScolsThenS); [1 0 0 0 1 2 0 0 2 1] [ ] [0 1 0 0 0 0 1 0 1 0] [ ] [0 0 1 0 0 0 0 1 0 2] [ ] [0 0 0 0 0 0 0 0 0 0] %% As expected... > SThenScols := ; [0 1 2 0 0 2 1 1 0 0] [ ] [0 1 2 1 -1 3 -1 1 1 -1] SThenScols := [ ] [0 -1 -2 -1 2 -3 3 -1 -1 2] [ ] [0 2 4 1 -1 5 0 2 1 -1] > RREF(SThenScols); [0 1 2 0 0 2 1 1 0 0] [ ] [0 0 0 1 0 1 0 0 1 0] [ ] [0 0 0 0 1 0 2 0 0 1] [ ] [0 0 0 0 0 0 0 0 0 0] :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: > s7 := S[1..4,7]; [ 1] [ ] [-1] s7 := [ ] [ 3] [ ] [ 0] > ScolsThenS := ; RREF(ScolsThenS); [ 1 1 0 0 1 2 0 0 2 1] [ ] [-1 1 1 0 1 2 1 -1 3 -1] ScolsThenS := [ ] [ 3 -1 -1 0 -1 -2 -1 2 -3 3] [ ] [ 0 2 1 0 2 4 1 -1 5 0] [1 0 0 0 0 0 0 1/2 0 1] [ ] [0 1 0 0 1 2 0 -1/2 2 0] [ ] [0 0 1 0 0 0 1 0 1 0] [ ] [0 0 0 0 0 0 0 0 0 0] > SThenScols := ; RREF(SThenScols) ; [0 1 2 0 0 2 1 0 1 1] [ ] [0 1 2 1 -1 3 -1 1 -1 1] SThenScols := [ ] [0 -1 -2 -1 2 -3 3 -1 3 -1] [ ] [0 2 4 1 -1 5 0 1 0 2] [0 1 2 0 0 2 1 0 1 1] [ ] [0 0 0 1 0 1 0 1 0 0] RREF(SThenScols) =[ ] [0 0 0 0 1 0 2 0 2 0] [ ] [0 0 0 0 0 0 0 0 0 0]