Page 227 - MATLAB an introduction with applications

212 ——— MATLAB: An Introduction with Applications

If any vector {x} is multiplied by [R.(p, q, θ)] such that y = [R]{x}, then
y = r x + r x = cos xθ p + sin xθ q
p
pq q
p
pp
and Y = r x + r x = − sin xθ p + cos xθ q
qp
p
qq q
q
i.e., y = x if i ≠ p or q. Therefore it is necessary to select θ such that y or y is equal to zero. It should also
p
i
i
q
be noted that if x = x = 0 then y = y = 0. If we write
p
q
q
p
n
[] ( x x= (1) (2) ...x ( ) )
A
n
( , ,q θ
then   Rp ) [] ( y y  A = (1) (2) ...y ( ) )
An element in either pth row or qth row is reduced by a proper choice of θ. Thus, all the elements in the
first column except the first element can be made zero by suitable values of θ, and they remain zero even
after multiplying with [R]. Similarly for all other columns elements are made zero for the elements below the
diagonal elements by proper choice of θ values. Thus, we obtain an upper triangular matrix by forming the
product
[] u = R ( ,n n − 1,θ , nn− 1 ) ( ,R n n − 2,θ , nn− 2 ) (3,1,R θ 3,1 ) (2,1,R θ 2,1 )[] A
...
 n− 1 n 
( ,,j θ
) [] A
= ∏∏ Ri ij 
 j= 1 i j+ 1 
=
Since [R] is an orthogonal matrix, the products of [R] are also orthogonal. Hence, we have [U] = [S][A]
–1
T
where [S] is orthogonal an d[U] is upper triangular. Since [S] is orthogonal, we have [A] = [S] [U] = [S] [U].
If [Q] = [S] , then [A] = [Q][U].
T
We can construct a sequence of matrices [A ], [A ][A ]… where
1
0
2
A =
A =
A =
A
Q
[ ] [] [ ][ ] [ ][ ] [Q k − 1 ][ k− 1 ]
, Q
A
k
0
0
0
k
for all values of k ≥ 1. That is, we start with [A ] = [A] and put it in the form [Q ][U ] to obtain [Q ] and [U ]
0
0
0
0
0
and then obtain [A ] which is equal to the product of [U ][Q ] in reverse order. Repeating this procedure,
0
0
1
we can obtain any number of sequence.
[A]= [A ] = [Q ][U ] determines [Q ] and [U ]
0
0
0
0
0
[A ]= [U ][Q ] determines [A ]
0
1
1
0
=[Q ][U ] determines [Q ] and [U ]
1
1
1
1
[A ]= [U ][Q ] determines [A ]
2
1
2
1
=[Q ][U ] determines [Q ] and [U ]
2
2
2
2
[A ]= [U ][Q ] determines [A ]
k
k–1
k
k–1
=[Q ][U ] determines [Q ] and [U ]
k
k
k
k
From the above sequence, we can show that the product Q Q Q …Q converges as k → ∞. Then [A ]
1
k
2
k
0
converges to an upper triangular matrix with the eigenvalues of [A] as its diagonal elements. This can be
proved as follows. We have
=
A =
Q
U
[ ] [ ][ ] [U k− 1 ][Q k− 1 ] ...(4.20)
k
k
k
=
A ] [Q ][U ] ...(4.21)
[ k − 1 k − 1 k − 1