Now extend s2 s1 b1 b2 to a basis for m0 n l this

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: se vectors as columns in a matrix Qn×n and show that N (Q) = 0. The trick in doing so is to arrange the vectors from the Jb ’s in just the right order. Begin by placing the vectors at the top level in chains emanating from Si as columns in a matrix Xi as depicted in the heuristic diagram in Figure 7.7.3. O C Figure 7.7.3 Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 578 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 The matrix LXi contains all vectors at the second highest level of those chains emanating from Si , while L2 Xi contains all vectors at the third highest level of those chains emanating from Si , and so on. In general, Lj Xi contains all vectors at the (j +1)st highest level of those chains emanating from Si . Proceed by filling in Q = [ Q0 | Q1 | · · · | Qk−1 ] from the bottom up by letting Qj be the matrix whose columns are all vectors at the j th level from the bottom in all chains. For the example illustrated in Figures 7.7.1–7.7.3 with k = 5, It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] Q0 = [ X0 | LX1 | L2 X2 | L3 X3 | L4 X4 ] = vectors at level 0 = basis B for N (L), Q1 = [ X1 | LX2 | L2 X3 | L3 X4 ] = vectors at level 1 (from the bottom), D E Q2 = [ X2 | LX3 | L2 X4 ] = vectors at level 2 (from the bottom), Q3 = [ X3 | LX4 ] = vectors at level 3 (from the bottom), T H Q4 = [ X4 ] = vectors at level 4 (from the bottom). In general, Qj = [ Xj | LXj +1 | L2 Xj +2 | · · · | Lk−1−j Xk−1 ]. Since the columns of Lj Xj are all on the bottom level (level 0), they are part of the basis B for N (L). This means that the columns of Lj Qj are also part of the basis B for N (L), so they are linearly independent, and thus N Lj Qj = 0. Furthermore, since the columns of Lj Qj are in N (L), we have L Lj Qj = 0, and hence Lj +h Qj = 0 for all h ≥ 1. Now use these observations to prove N (Q) = 0. If Qz = 0, then multiplication by Lk−1 yields IG R 0 = Lk−1 Qz = [ Lk−1 Q0 | Lk−1 Q1 | · · · | Lk−1 Qk−1 ] z...
View Full Document

This document was uploaded on 03/06/2014 for the course MA 5623 at City University of Hong Kong.

Ask a homework question - tutors are online