Question

3 (The UL factorization.) Show how to compute the factorization A = UL where U is upper triangular with ls along the diagonal and L is lower triangular. Show how this relates to a way of solving Ax = b by transforming the system into an equivalent system with a lower triangular matrix. (In other words, show that what we did for the LU factorization also works for a UL factorization.) Note: For the purposes of this exercise you may assume that no pivoting is required. This is of course unrealistic but pivoting would only distract from the point of this exercise (which is that conceptually there is no difference between an LU and a UL factorization).

Answer 1

Ax=B - 221 I as 422 423 asz a 33 J Coa run 42 Uig 0 دين : ALU LEO L'si 132 1 Them 0 becomes Lux=B torting Uxrv, - & becomes tv=B which is equivalent to the equation w=5 lai vitle = bz, 23, N2 + d 31 N₂ + 3 = 63 Solving for these for V, vz, vz . We know n then from comes , 4+U12 Q2 + Uiga;=v, U22² 2 + 423 72 V2, 37-13 from which no, I, and on can be found. by Back Sub Stilution to compule the matrices Land u 953/00 . Co on can ch is 2.1 a Tan 0.03 - U2 033 42 422 423 Lis, l123 Lagi 952 433 Matheply the matces on the left and Ouating Correspon ding elements from both sides. We oblom. 4) Un =an z = 0,2 g = ais . " 210g = 02 (21= aula 131 w as 1313 131 te) 121 Viz+ U22 = 922 U22 - 92 - - C₂, Wrz + 4z = arz od Uzg = Az - Quais. i (316)2 + tez uzz = Aszol 62 = [ 432 - 2. Quz] v) ds, Us +133 Uzz + U33=ags of U33. = ag3-13, Cz-133 023- ... This method is superior to Gauss elimination ndis olen used the solution of linear System and for finding the inverse of a matry Among direct methods factonization method is also I preferred as the Software for Computers had and for food fact allew are

of AE c mxm has an Lo. factorization. A=L0 and has the upper band width q and lower bandwidth p. thew o has an upper band cotho q and L has a lower band widthe P . Now let PA= LO be the factorization Computed by Gaussian Elimination with partial pivoting , P = Pne... p, a product of permulation matrices Pp = [es, es ... esn] where si, sz, ... Sn u a permulation of 1,2,...n . . Of Six itp . then at follows that the leading ixi principal Submatrix of PA is sngular , Since (PA) ij = as j f j = 1,2... Se-p-1 and Si-P-1 i This implies that w and a are singular. Thus, we reach a contradiction Therefore Sic itp for i=1,2,...n Thus, PA has an upperband width prp=2p Thus, by the above mentioned fact we see that whas an upper band wordle zp. Note that pivoting destroy the band structure in a Senge that o ends up having a higher bandwidth while nothing at all can be said about the band Structure of L infach L = P (Lepna .. 4P) The only thing we can say is that L Contains at must pri non quo Clements per Column In particular ljil spa ay of a be an NAN banded matrix.