Jacobi Eigen
Jacobi Eigen
Jacobi Eigen
c s 0
Q0
Q 1 = s c 0 =
0 1
0 0 1
1 0 0
Q 2 = 0 c s obtained from Q1 using a permutation matrix
0 s c
Basic ideas from earlier.
If A is symmetric then Q TAQ is also.
If Q is orthogonal, then Q TAQ and A have the same eigenvalues.
5 2 0
A = 2 5 0
0 0 3
1/ 2 1/ 2 0
c s 0
and Q 1 = s c 0 = 1/ 2 1/ 2 0 is a Givens rotation
0 0 1
0
01
designed to zero out the (2,1) element.
300
A1 =
= 070
003
The columns of Q are eigenvectors and the eigenvalues are on the
diagonal.
Q T1 AQ 1
Rotations that zero out a specified element for this note are called
Givens rotations after the scientist at Oak Ridge who pioneerred
their use more than fifty years ago.
Related orthogonal transformations that zero out a portion of a
column are called Householder transformations.
312
Start with A = 1 2 1 and construct a Givens rotation Q 1that
214
zeros out the (2,1) element in then sense that the (2,1) element of
A 1 = Q T1 AQ 1 is zero.
Then find a Givens rotation Q 2 that zeros out the (3,1) element of
A1 in the same manner A 2 = Q T2 A 1Q 2 . Proceed with another
rotation Q 3 that zeros out the (3,2) element of A2 and set
A p = Q Tp A p1Q p or A p = Q TAQ
for the orthogonal matrix Q = Q 1Q 2
Q p1 Q p .
1
2
1
2
1
4
Q
0.8507 -0.5257
0.5257
0.8507
0.0000
0.0000
0.0000
0.0000
1.0000
*** ***
|
|
|
|
current product
of Givens rotations
|
\|/
Q
0.6267 -0.5257 0.5752
0.3873 0.8507 0.3555
-0.6762
0.0000
0.7367
|
|
|
|
Q
0.6267 -0.5072
0.3873 0.8615
-0.6762 0.0233
0.5916
0.3284
0.7363
|
|
|
|
|
|
|
|
Q
0.3275 -0.7367
0.7367 0.5911
-0.5917 0.3283
Q
0.3280 -0.7367
0.7370 0.5911
-0.5910 0.3283
0.5916
0.3284
0.7363
0.5913
0.3277
0.7368
|
|
|
|
Q
0.3280 -0.7370
0.7370 0.5910
-0.5910 0.3280
0.5910
0.3280
0.7370
|
|
|
|
Q
0.3280 -0.7370
0.7370 0.5910
-0.5910
0.3280
0.5910
0.3280
0.7370
|
|
|
|
Q
0.3280 -0.7370
0.7370 0.5910
-0.5910
0.3280
0.5910
0.3280
0.7370
|
|
|
|
Q
0.3280 -0.7370
0.7370 0.5910
-0.5910 0.3280
0.5910
0.3280
0.7370
0.0000
1.3080
0
-0.0000
0
6.0489
-0.7370
0.5910
0.3280
0.5910
0.3280
0.7370
*****************************************************
*** *** ***
Summary Information
*** *** ***
*****************************************************
At termination, 3 sweeps were completed.
D should be diagonal and it is
1.6431
0.0000
-0.0000
0.0000
1.3080
0
-0.0000
0
6.0489
Orthogonal Q is
0.3280
-0.7370
0.7370
0.5910
-0.5910
0.3280
0.5910
0.3280
0.7370
0.5910
0.3280
0.7370
disp(D)
1.3080
0
0
0
0
6.0489
0
1.6431
0