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**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
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Chapter 7
Eigenvalues and Eigenvectors
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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...

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