CVXPY Square root of Singular Quadratic
I need to model sqrt(x^T C x) for a singular positive semidefinite matrix C. Here, it is proposed to use norm(Q*x) where Q is obtained from the Cholesky decomposition of C.
How to take the square root of quad_form output in CVXPY?
But, np./scipy.linalg.cholskey does not work for singular matrices.
PS, using SVD or eigenvalue decomposition is too slow for my application.
PS2, this post Numpy Cholesky decomposition LinAlgError does not help as it does not offer a solution. Also, the matrix in the question seems to have negative eigenvalues (rather than being singular).
python square-root cvxpy
asked Nov 14 '18 at 11:51
Behrooz Ns
254
add a comment |
I need to model sqrt(x^T C x) for a singular positive semidefinite matrix C. Here, it is proposed to use norm(Q*x) where Q is obtained from the Cholesky decomposition of C.
How to take the square root of quad_form output in CVXPY?
But, np./scipy.linalg.cholskey does not work for singular matrices.
PS, using SVD or eigenvalue decomposition is too slow for my application.
PS2, this post Numpy Cholesky decomposition LinAlgError does not help as it does not offer a solution. Also, the matrix in the question seems to have negative eigenvalues (rather than being singular).
python square-root cvxpy
asked Nov 14 '18 at 11:51
Behrooz Ns
254
Possible duplicate of Numpy Cholesky decomposition LinAlgError
– sophros
Nov 14 '18 at 12:39
That post does not suggest a solution.
– Behrooz Ns
Nov 14 '18 at 13:59
add a comment |
I need to model sqrt(x^T C x) for a singular positive semidefinite matrix C. Here, it is proposed to use norm(Q*x) where Q is obtained from the Cholesky decomposition of C.
How to take the square root of quad_form output in CVXPY?
But, np./scipy.linalg.cholskey does not work for singular matrices.
PS, using SVD or eigenvalue decomposition is too slow for my application.
PS2, this post Numpy Cholesky decomposition LinAlgError does not help as it does not offer a solution. Also, the matrix in the question seems to have negative eigenvalues (rather than being singular).
python square-root cvxpy
asked Nov 14 '18 at 11:51
Behrooz Ns
254
I need to model sqrt(x^T C x) for a singular positive semidefinite matrix C. Here, it is proposed to use norm(Q*x) where Q is obtained from the Cholesky decomposition of C.
How to take the square root of quad_form output in CVXPY?
But, np./scipy.linalg.cholskey does not work for singular matrices.
PS, using SVD or eigenvalue decomposition is too slow for my application.
PS2, this post Numpy Cholesky decomposition LinAlgError does not help as it does not offer a solution. Also, the matrix in the question seems to have negative eigenvalues (rather than being singular).
python square-root cvxpy
python square-root cvxpy
asked Nov 14 '18 at 11:51
Behrooz Ns
254
asked Nov 14 '18 at 11:51
Behrooz Ns
254
edited Nov 14 '18 at 14:49
asked Nov 14 '18 at 11:51
Behrooz Ns
254
asked Nov 14 '18 at 11:51
Behrooz Ns
254
asked Nov 14 '18 at 11:51
Behrooz Ns
254
254
Possible duplicate of Numpy Cholesky decomposition LinAlgError
– sophros
Nov 14 '18 at 12:39
That post does not suggest a solution.
– Behrooz Ns
Nov 14 '18 at 13:59
add a comment |
Possible duplicate of Numpy Cholesky decomposition LinAlgError
– sophros
Nov 14 '18 at 12:39
That post does not suggest a solution.
– Behrooz Ns
Nov 14 '18 at 13:59
Possible duplicate of Numpy Cholesky decomposition LinAlgError
– sophros
Nov 14 '18 at 12:39
Possible duplicate of Numpy Cholesky decomposition LinAlgError
– sophros
Nov 14 '18 at 12:39
That post does not suggest a solution.
– Behrooz Ns
Nov 14 '18 at 13:59
That post does not suggest a solution.
– Behrooz Ns
Nov 14 '18 at 13:59
add a comment |
1 Answer
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I found a solution using the ldl decomposition
L,d,_ = scipy.linalg.ldl(C)
d = np.diag(d).copy()
inds = d >= d.max()*1e-8
d = d[inds]
d = np.sqrt(d)
d.shape = (-1,1)
Q = d * L.T[inds]
loss = cp.norm(cp.matmul(Q, x))
The ldl decomposition needs scipy >= 1.1 though.
answered Nov 14 '18 at 13:58
Behrooz Ns
254
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I found a solution using the ldl decomposition
L,d,_ = scipy.linalg.ldl(C)
d = np.diag(d).copy()
inds = d >= d.max()*1e-8
d = d[inds]
d = np.sqrt(d)
d.shape = (-1,1)
Q = d * L.T[inds]
loss = cp.norm(cp.matmul(Q, x))
The ldl decomposition needs scipy >= 1.1 though.
answered Nov 14 '18 at 13:58
Behrooz Ns
254
add a comment |
I found a solution using the ldl decomposition
L,d,_ = scipy.linalg.ldl(C)
d = np.diag(d).copy()
inds = d >= d.max()*1e-8
d = d[inds]
d = np.sqrt(d)
d.shape = (-1,1)
Q = d * L.T[inds]
loss = cp.norm(cp.matmul(Q, x))
The ldl decomposition needs scipy >= 1.1 though.
answered Nov 14 '18 at 13:58
Behrooz Ns
254
add a comment |
I found a solution using the ldl decomposition
L,d,_ = scipy.linalg.ldl(C)
d = np.diag(d).copy()
inds = d >= d.max()*1e-8
d = d[inds]
d = np.sqrt(d)
d.shape = (-1,1)
Q = d * L.T[inds]
loss = cp.norm(cp.matmul(Q, x))
The ldl decomposition needs scipy >= 1.1 though.
answered Nov 14 '18 at 13:58
Behrooz Ns
254
I found a solution using the ldl decomposition
L,d,_ = scipy.linalg.ldl(C)
d = np.diag(d).copy()
inds = d >= d.max()*1e-8
d = d[inds]
d = np.sqrt(d)
d.shape = (-1,1)
Q = d * L.T[inds]
loss = cp.norm(cp.matmul(Q, x))
The ldl decomposition needs scipy >= 1.1 though.
answered Nov 14 '18 at 13:58
Behrooz Ns
254
edited Nov 14 '18 at 14:47
answered Nov 14 '18 at 13:58
Behrooz Ns
254
answered Nov 14 '18 at 13:58
Behrooz Ns
254
answered Nov 14 '18 at 13:58
Behrooz Ns
254
254
add a comment |
add a comment |
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Possible duplicate of Numpy Cholesky decomposition LinAlgError
– sophros
Nov 14 '18 at 12:39
That post does not suggest a solution.
– Behrooz Ns
Nov 14 '18 at 13:59