Page 158 - Numerical methods for chemical engineering

144 3 Matrix eigenvalue analysis

and write Ay in the form of (3.238),
 [1] 
σ 1 v · y  (0) v [1] · y 

.
 . .   . . 
   . 
σ r v · y  [r] 
 [r] 
   (0) v
 = W0 = 0
Ay = W  [r+1]  = W  · y  (3.242)
 σ r+1 v
· y 
 σ r+1 (0) 
  .
.  
.  . 
  .
 . 

σ N v [N] · y σ N (0)
Thus, the right singular vectors for the zero singular values form an orthonormal basis for
the null space (kernel) of A,
[1] [r]
K A = span v ,..., v σ 1 = ... = σ r = 0 dim(K A ) = r (3.243)
We obtain a basis for the range of A by continuing the calculation of Ax,
 [1] 
σ 1 v · x
 
| |  [2] 
 σ 2 v · x 
Ax =  w [1] ... w [N]   . 
 . 
| |  . 
[N]
σ N v · x
w σ 1 v · x +· · · + w σ N v · x
 [1] [1] [N] [N] 
1 1
[1] [N] 
 w σ 1 v [1] · x +· · · + w σ N v [N]

2 2 · x 
=  .  (3.244)
.
 
 . 
[1]

w σ 1 v [1] · x +· · · + w [N] σ N v [N] · x
N N
to obtain
N

[ j] [ j]
Ax = σ j v · x w (3.245)
j=1
If σ 1 =· · · = σ r = 0 and σ j > 0 for j = r + 1,..., N, then
N
[ j] [ j]
Ax = σ j v · x w (3.246)
j=r+1
σ j >0
N
Thus, any Ax ∈ can be written as linear combination of the left singular vec-
tors {w [r+1] ,..., w [N] }, so that the range of A is
[r+1] [N]
R A = span w ,..., w σ j∈ [r+1,N] > 0 dim(K A ) = N − r (3.247)
The right and left singular vectors thus provide orthonormal basis sets for the kernel and
range respectively,

K A = span v [ j] σ j = 0 R A = span w [ j] σ j > 0 (3.248)

For Ax = b,if A is singular, we can check easily for the existence of solutions using SVD.
If b ∈ A , there exist an infinite number of solutions
N
[ j]
x = z + c j v c j ∈ (3.249)
j=1
σ j =0