6.2 Problems 89
The second stage is similar: x consists of the last n — 2 entries in the second
column, z is the first unit coordinate vector of matching length, and H 2 is of
order n — 2:
Uo =
100 0
010 0
00
0 0 H 2
00
= U2^1 , and U^iU^AUiM 0 * * * *
0 0***
00***
Following a similar approach, one may operate on the upper right corner
of A simultaneously to generate a tridiagonal matrix at the end. This process
is the main motivation of the QR algorithm.
Problems
6.1. Show that for orthogonal matrices ||Q|| = c(Q) — 1. Orthogonal matrices
and their multipliers (aQ) are only perfect condition matrices.
6.2. Apply the QR algorithm for
A =
0.5000 -1.1180 0 0 0
-1.1180 91.2000 -80.0697 0 0
0 -80.0697 81.0789 4.1906 0
0 0 4.1906 2.5913 0.2242
0 0 0 0.2242 0.1257 -
0 0 0 0 -0.0100
6.3. Let A{n) 6 Rnxn, A(n) = (a^-
, wnere aij — i,j_^.
(a) Take A{2).
- Let 6/ =
1.0
0.5
and bu =
1.5
1.0
. Calculate the relative
0
0
0
0
-0.0100
0.0041
error.
- Find a good upper bound for the relative error obtained after perturbing
the right hand side. - Find the relative error of perturbing A(2) by AA( 2 ). Take
h
1.0
0.5
as the right hand side.
- Find a good upper bound for the relative error obtained after perturbing
.4(2).
(b) Take ^4(3)TA(3) and find its condition number and compare with the
condition number of A(3).
(c) Take A(4) and calculate its condition number after finding the eigen values
using the QR algorithm.