Polynomial Kernel not PSD?
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I am currently going through different Machine Learning methods on a low-level basis. Since the polynomial kernel
K(x,z)=(1+x^T*z)^d
is one of the more commonly used kernels, I was under the assumption that this kernel must yield a positive (semi-)definite matrix for a fixed set of data {x1,...,xn}.
However, in my implementation, this seems to not be the case, as the following example shows:
import numpy as np
np.random.seed(93)
x=np.random.uniform(0,1,5)
#Assuming d=1
kernel=(1+x[:,None]@x.T[None,:])
np.linalg.eigvals(kernel)
The output would be
[ 6.9463439e+00 1.6070016e-01 9.5388039e-08 -1.5821310e-08 -3.7724433e-08]
I'm also getting negative eigenvalues for d>2.
Am I totally misunderstanding something here? Or is the polynomia kernel simply not PSD?
EDIT: In a previous version, i used x=np.float32(np.random.uniform(0,1,5)) to reduce computing time, which lead to a greater amount of negative eigenvalues (I believe due to numerical instabilities, as @user2357112 mentioned). I guess this is a good example that precision does matter? Since negative eigenvalues still occur, even for float64 precision, the follow-up question would then be how to avoid such numerical instabilities?
python machine-learning svm kernel-methods
asked Nov 8 at 4:06
emilaz
708
add a comment |
up vote
0
down vote
favorite
I am currently going through different Machine Learning methods on a low-level basis. Since the polynomial kernel
K(x,z)=(1+x^T*z)^d
is one of the more commonly used kernels, I was under the assumption that this kernel must yield a positive (semi-)definite matrix for a fixed set of data {x1,...,xn}.
However, in my implementation, this seems to not be the case, as the following example shows:
import numpy as np
np.random.seed(93)
x=np.random.uniform(0,1,5)
#Assuming d=1
kernel=(1+x[:,None]@x.T[None,:])
np.linalg.eigvals(kernel)
The output would be
[ 6.9463439e+00 1.6070016e-01 9.5388039e-08 -1.5821310e-08 -3.7724433e-08]
I'm also getting negative eigenvalues for d>2.
Am I totally misunderstanding something here? Or is the polynomia kernel simply not PSD?
EDIT: In a previous version, i used x=np.float32(np.random.uniform(0,1,5)) to reduce computing time, which lead to a greater amount of negative eigenvalues (I believe due to numerical instabilities, as @user2357112 mentioned). I guess this is a good example that precision does matter? Since negative eigenvalues still occur, even for float64 precision, the follow-up question would then be how to avoid such numerical instabilities?
python machine-learning svm kernel-methods
asked Nov 8 at 4:06
emilaz
708
3
Those values are 0 up to numerical error.
– user2357112
Nov 8 at 4:13
Also I think you've got your transposes backward.
– user2357112
Nov 8 at 4:16
What would be the way to go to avoid this numerical instability? And no, the transposed should be that way. I want to get a nxn matrix out of a nx1 vector, where the (i,j)th entry is x_i*x_j.
– emilaz
Nov 8 at 4:20
already helped a lot though, thanks.
– emilaz
Nov 8 at 4:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am currently going through different Machine Learning methods on a low-level basis. Since the polynomial kernel
K(x,z)=(1+x^T*z)^d
is one of the more commonly used kernels, I was under the assumption that this kernel must yield a positive (semi-)definite matrix for a fixed set of data {x1,...,xn}.
However, in my implementation, this seems to not be the case, as the following example shows:
import numpy as np
np.random.seed(93)
x=np.random.uniform(0,1,5)
#Assuming d=1
kernel=(1+x[:,None]@x.T[None,:])
np.linalg.eigvals(kernel)
The output would be
[ 6.9463439e+00 1.6070016e-01 9.5388039e-08 -1.5821310e-08 -3.7724433e-08]
I'm also getting negative eigenvalues for d>2.
Am I totally misunderstanding something here? Or is the polynomia kernel simply not PSD?
EDIT: In a previous version, i used x=np.float32(np.random.uniform(0,1,5)) to reduce computing time, which lead to a greater amount of negative eigenvalues (I believe due to numerical instabilities, as @user2357112 mentioned). I guess this is a good example that precision does matter? Since negative eigenvalues still occur, even for float64 precision, the follow-up question would then be how to avoid such numerical instabilities?
python machine-learning svm kernel-methods
asked Nov 8 at 4:06
emilaz
708
I am currently going through different Machine Learning methods on a low-level basis. Since the polynomial kernel
K(x,z)=(1+x^T*z)^d
is one of the more commonly used kernels, I was under the assumption that this kernel must yield a positive (semi-)definite matrix for a fixed set of data {x1,...,xn}.
However, in my implementation, this seems to not be the case, as the following example shows:
import numpy as np
np.random.seed(93)
x=np.random.uniform(0,1,5)
#Assuming d=1
kernel=(1+x[:,None]@x.T[None,:])
np.linalg.eigvals(kernel)
The output would be
[ 6.9463439e+00 1.6070016e-01 9.5388039e-08 -1.5821310e-08 -3.7724433e-08]
I'm also getting negative eigenvalues for d>2.
Am I totally misunderstanding something here? Or is the polynomia kernel simply not PSD?
EDIT: In a previous version, i used x=np.float32(np.random.uniform(0,1,5)) to reduce computing time, which lead to a greater amount of negative eigenvalues (I believe due to numerical instabilities, as @user2357112 mentioned). I guess this is a good example that precision does matter? Since negative eigenvalues still occur, even for float64 precision, the follow-up question would then be how to avoid such numerical instabilities?
python machine-learning svm kernel-methods
python machine-learning svm kernel-methods
asked Nov 8 at 4:06
emilaz
708
asked Nov 8 at 4:06
emilaz
708
edited Nov 8 at 4:55
asked Nov 8 at 4:06
emilaz
708
asked Nov 8 at 4:06
emilaz
708
asked Nov 8 at 4:06
emilaz
708
708
3
Those values are 0 up to numerical error.
– user2357112
Nov 8 at 4:13
Also I think you've got your transposes backward.
– user2357112
Nov 8 at 4:16
What would be the way to go to avoid this numerical instability? And no, the transposed should be that way. I want to get a nxn matrix out of a nx1 vector, where the (i,j)th entry is x_i*x_j.
– emilaz
Nov 8 at 4:20
already helped a lot though, thanks.
– emilaz
Nov 8 at 4:41
add a comment |
3
Those values are 0 up to numerical error.
– user2357112
Nov 8 at 4:13
Also I think you've got your transposes backward.
– user2357112
Nov 8 at 4:16
What would be the way to go to avoid this numerical instability? And no, the transposed should be that way. I want to get a nxn matrix out of a nx1 vector, where the (i,j)th entry is x_i*x_j.
– emilaz
Nov 8 at 4:20
already helped a lot though, thanks.
– emilaz
Nov 8 at 4:41
3
3
Those values are 0 up to numerical error.
– user2357112
Nov 8 at 4:13
Those values are 0 up to numerical error.
– user2357112
Nov 8 at 4:13
Also I think you've got your transposes backward.
– user2357112
Nov 8 at 4:16
Also I think you've got your transposes backward.
– user2357112
Nov 8 at 4:16
What would be the way to go to avoid this numerical instability? And no, the transposed should be that way. I want to get a nxn matrix out of a nx1 vector, where the (i,j)th entry is x_i*x_j.
– emilaz
Nov 8 at 4:20
What would be the way to go to avoid this numerical instability? And no, the transposed should be that way. I want to get a nxn matrix out of a nx1 vector, where the (i,j)th entry is x_i*x_j.
– emilaz
Nov 8 at 4:20
already helped a lot though, thanks.
– emilaz
Nov 8 at 4:41
already helped a lot though, thanks.
– emilaz
Nov 8 at 4:41
add a comment |
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3
Those values are 0 up to numerical error.
– user2357112
Nov 8 at 4:13
Also I think you've got your transposes backward.
– user2357112
Nov 8 at 4:16
What would be the way to go to avoid this numerical instability? And no, the transposed should be that way. I want to get a nxn matrix out of a nx1 vector, where the (i,j)th entry is x_i*x_j.
– emilaz
Nov 8 at 4:20
already helped a lot though, thanks.
– emilaz
Nov 8 at 4:41