Problem (C2). Here we are finding a basis for Range(L_S), i.e, a basis for ColSpan(S). We use S-transpose. Recall a previous posting [and classwork]: /--------------------------------------------------------\ /----- Method 2 ------------------------------------------\ An alternative method to compute a basis for ColSpan(S): Put the columns of S as rows of a matrix T; i.e, let T be S-transpose. Let b_1,...,b_{97} denote the rows of T. Define B := RREF(T). Note that even though T is S-transpose, this B is almost-certainly /not/ the transpose of RREF(S). Suppose rows q_7, q_{23}, q_{35}. are the pivot-rows of B. Then (this takes a moment's thought) collection {q_7, q_{23}, q_{35}} is a basis for RowSpan(B). But RowSpan(B) = RowSpan(T) = ColSpan(S). \________________________________________________________/ \________________________________________________________/ on Thu Oct 14 18:58:38 EDT 2010 > 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] So [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] > I3 := eye(3); onerow := <0|0|0> : threecol := ; [1 0 0] [ ] I3 := [0 1 0] [ ] [0 0 1] [1 0 0] [ ] [0 1 0] threecol := [ ] [0 0 1] [ ] [0 0 0] > SWithRsPivots := ; [0 1 2 0 0 2 1 1 0 0] [ ] [0 1 2 1 -1 3 -1 0 1 0] SWithRsPivots := [ ] [0 -1 -2 -1 2 -3 3 0 0 1] [ ] [0 2 4 1 -1 5 0 0 0 0] > RREF(SWithRsPivots) ; [0 1 2 0 0 2 1 0 -1 0] [0 0 0 1 0 1 0 0 3 1] [0 0 0 0 1 0 2 0 1 1] [0 0 0 0 0 0 0 1 1 0] ^ ^ ^ ^ > T := Tpose(S); [0 0 0 0] [1 1 -1 2] [2 2 -2 4] T := [0 1 -1 1] [0 -1 2 -1] [2 3 -3 5] [1 -1 3 0] > RREFTposeS := RREF(T); [1 0 0 1] [0 1 0 1] [0 0 1 0] RREFTposeS := [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] > b1 := Tpose(RREFTposeS[1,1..4]); [1] b1 := [0] [0] [1] > b2 := Tpose(RREFTposeS[2,1..4]); b3 := Tpose(RREFTposeS[3,1..4]); [0] b2 := [1] [0] [1] [0] b3 := [0] [1] [0] > BThenS := ; [1 0 0 0 1 2 0 0 2 1] BThenS = [0 1 0 0 1 2 1 -1 3 -1] [0 0 1 0 -1 -2 -1 2 -3 3] [1 1 0 0 2 4 1 -1 5 0] > RREF(BThenS); [1 0 0 0 1 2 0 0 2 1] [0 1 0 0 1 2 1 -1 3 -1] [0 0 1 0 -1 -2 -1 2 -3 3] [0 0 0 0 0 0 0 0 0 0]