Finding k largest eigenvalues

[mjadidi]

Is there any method or future to implement algorithms like lanczos algorithms to find the k largest eigen values? Is there any solution for now?

[john]

These exist for symmetric and hermitian matrices in the syevx/heevx routines.

[mjadidi]

yes, I just realized that I can find eigenvalues in a specific range or specific indices.But what would be the corresponding eigenvectors if I set rang to "I"

[john]

Assuming you set JOBZ to 'V' then you'll get the appropriate number of vectors in Z.

[mjadidi]

Yes, it works now.Thanks.

but can you tell me y when I set m= iu-il+1 I get kind of wired answer? Because when I set m=n it makes more sense :

for example for a matrix of 8x8 dimension I get following result for top 4 eigenvectors:
eigenvalues:
[ 8.00000000e-04 8.00000000e-04 8.00000000e-04 8.00000000e-04
2.45898573e+00 3.11167768e+00 9.95596484e+00 2.56381985e+01]

eigenvectors:
scipy.linalg.eigh :
[[-0.14611719 0.78394033 -0.31431961 -0.25958747]
[-0.00860211 -0.23798928 0.39718488 -0.46557788]
[ 0.18952489 0.0831798 -0.09898343 -0.16464493]
[ 0.28435409 -0.21043358 -0.61696779 -0.44713126]
[-0.04668508 -0.27840988 -0.45123004 -0.04021424]
[ 0.5469773 -0.07060557 0.24148952 -0.39681434]
[ 0.17376323 0.41502762 0.29308223 -0.25266402]
[ 0.72817684 0.15116997 -0.07195456 0.5156543 ]]

with m=n:

[[-0.14611719 0.78394033 -0.31431961 -0.25958747 0. 0. 0.
0. ]
[-0.00860211 -0.23798928 0.39718488 -0.46557788 0. 0. 0.
0. ]
[ 0.18952489 0.0831798 -0.09898343 -0.16464493 0. 0. 0.
0. ]
[ 0.28435409 -0.21043358 -0.61696779 -0.44713126 0. 0. 0.
0. ]
[-0.04668508 -0.27840988 -0.45123004 -0.04021424 0. 0. 0.
0. ]
[ 0.5469773 -0.07060557 0.24148952 -0.39681434 0. 0. 0.
0. ]
[ 0.17376323 0.41502762 0.29308223 -0.25266402 0. 0. 0.
0. ]
[ 0.72817684 0.15116997 -0.07195456 0.5156543 0. 0. 0.
0. ]]

with m= iu-il+1

[[-0.14611719 -0.04668508 0.78394033 -0.27840988 -0.31431961 -0.45123004
-0.25958747 -0.04021424]
[-0.00860211 0.5469773 -0.23798928 -0.07060557 0.39718488 0.24148952
-0.46557788 -0.39681434]
[ 0.18952489 0.17376323 0.0831798 0.41502762 -0.09898343 0.29308223
-0.16464493 -0.25266402]
[ 0.28435409 0.72817684 -0.21043358 0.15116997 -0.61696779 -0.07195456
-0.44713126 0.5156543 ]]

[john]

I'm sorry, I don't understand the question - M is an output from this routine.

[mjadidi]

I'm sorry, I don't understand the question - M is an output from this routine.

sorry for bad explanation.
Yes, but the question is what would be the value of M . Base on the documents when I set RANGE to "I" the value of M should be iu-il+1 . in this case the eigenvectors (Z) has some confusing format. but when I set to m=n the resulted eigenvectors have different indexes.


for comparison look at the first two rows of eigenvectors:

in case m=n :
[[-0.14611719 0.78394033 -0.31431961 -0.25958747 0. 0. 0.
0. ]
[-0.00860211 -0.23798928 0.39718488 -0.46557788 0. 0. 0.
0. ]

but for m=iu-il+1

[[-0.14611719 -0.04668508 0.78394033 -0.27840988 -0.31431961 -0.45123004
-0.25958747 -0.04021424]
[-0.00860211 0.5469773 -0.23798928 -0.07060557 0.39718488 0.24148952
-0.46557788 -0.39681434]


you see the latter case I have some other values between my eigenvectors.

Sorry If don't explain well the problem.

[kyle]

Like John said, M is an output value that when using the integer range option should be equal to IU-IL+1.

Are you setting IL and IU? For example, if you want the top 2 values it should be IL=1 and UL=2 (not the Fortran 1-base indexing).

From the docs:

Code: Select all

If RANGE='I', the indices (in ascending order) of the smallest and largest eigenvalues to be returned.

1 <= IL <= IU <= N, if N > 0; IL = 1 and IU = 0 if N = 0.

[mjadidi]

Yes I know this about M but when I set to to IU-IL+1. the answer is not obvious for me. Did you check my result on previews post about different values of M?

Are you setting IL and IU? For example, if you want the top 2 values it should be IL=1 and UL=2 (not the Fortran 1-base indexing).

But by setting IL=1 and UL=2 I am getting the 2 smallest eigenvalues not largest. for a 8x8 matrix for getting top 2 I should set IL=6 and UL=8.

[kyle]

You are correct about the ordering, this routine does indeed order the eigenvalues from lowest to smallest.

I just ran a similar problem.

Range = I; N = 8; IL = 6; IU = 8

Values:

Code: Select all

[0]   0.60990590602159556   
[1]   0.70335476519539919   


Vectors:

Code: Select all

[0]   0.30120404264911910   double
[1]   0.34704179211947256   double
[2]   0.22719182935148297   double
[3]   0.42993973500585869   double
[4]   0.33341002307228246   double
[5]   0.43737133822625307   double
[6]   0.33032203696194246   double
[7]   0.37524023913192073   double

[8]   0.076775878678010015   double
[9]   -0.41740538801404498   double
[10]   -0.25349112551809067   double
[11]   0.56032238991829530   double
[12]   -0.60328068615833719   double
[13]   0.037490411343624956   double
[14]   0.25657054787550626   double
[15]   0.10236069498767375   double


And with

Range = A; N = 8

Values:

Code: Select all

...
[6]   0.60990590602159567
[7]   0.70335476519539941


Vectors:

Code: Select all

...
[48]   0.30120404264911949   double
[49]   0.34704179211947273   double
[50]   0.22719182935148283   double
[51]   0.42993973500585947   double
[52]   0.33341002307228307   double
[53]   0.43737133822625301   double
[54]   0.33032203696194229   double
[55]   0.37524023913192039   double

[56]   0.076775878678010057   double
[57]   -0.41740538801404492   double
[58]   -0.25349112551809044   double
[59]   0.56032238991829464   double
[60]   -0.60328068615833819   double
[61]   0.037490411343625317   double
[62]   0.25657054787550626   double
[63]   0.10236069498767436   double


My results were as expected, the last 2 vectors and values from the 'all' version matched the results from the 'range' version.

Perhaps you have a data layout issue? CULA is column major which might differ from your reference implementation.

[mjadidi]

I have to investigate more about it then. Thanks