tensor power method
$begingroup$
Just as we can use the matrix power method to find eigenvalues/eigenvectors of matrices in an iterative way, we can analogously find the eigenvalues/eigenvectors of tensors in a similar way.
My question is about the convergence properties here. For matrices (order 2 tensors), we know the error converges linearly in the eigengap and it is not hard to prove (I'm talking about the basic matrix power method, not e.g. Rayleigh quotient method which I know will actually converge quadratically with eigenvalue gap).
For symmetric orthogonally decomposable (odeco) tensors of order 3, for the basic method (not a more advanced thing like if there is an analogous tensor Rayleigh quotient method) I have heard the rate is now square in the eigengap. Is it true that in general for an order D symmetric odeco tensor, the convergence using the basic tensor power method will be at a rate power of (D-1) of the eigengap?
linear-algebra tensor-products numerical-linear-algebra tensors symmetric-matrices
asked Dec 13 '18 at 20:11
sambajetsonsambajetson
280211
$endgroup$
add a comment |
$begingroup$
Just as we can use the matrix power method to find eigenvalues/eigenvectors of matrices in an iterative way, we can analogously find the eigenvalues/eigenvectors of tensors in a similar way.
My question is about the convergence properties here. For matrices (order 2 tensors), we know the error converges linearly in the eigengap and it is not hard to prove (I'm talking about the basic matrix power method, not e.g. Rayleigh quotient method which I know will actually converge quadratically with eigenvalue gap).
For symmetric orthogonally decomposable (odeco) tensors of order 3, for the basic method (not a more advanced thing like if there is an analogous tensor Rayleigh quotient method) I have heard the rate is now square in the eigengap. Is it true that in general for an order D symmetric odeco tensor, the convergence using the basic tensor power method will be at a rate power of (D-1) of the eigengap?
linear-algebra tensor-products numerical-linear-algebra tensors symmetric-matrices
asked Dec 13 '18 at 20:11
sambajetsonsambajetson
280211
$endgroup$
add a comment |
$begingroup$
Just as we can use the matrix power method to find eigenvalues/eigenvectors of matrices in an iterative way, we can analogously find the eigenvalues/eigenvectors of tensors in a similar way.
My question is about the convergence properties here. For matrices (order 2 tensors), we know the error converges linearly in the eigengap and it is not hard to prove (I'm talking about the basic matrix power method, not e.g. Rayleigh quotient method which I know will actually converge quadratically with eigenvalue gap).
For symmetric orthogonally decomposable (odeco) tensors of order 3, for the basic method (not a more advanced thing like if there is an analogous tensor Rayleigh quotient method) I have heard the rate is now square in the eigengap. Is it true that in general for an order D symmetric odeco tensor, the convergence using the basic tensor power method will be at a rate power of (D-1) of the eigengap?
linear-algebra tensor-products numerical-linear-algebra tensors symmetric-matrices
asked Dec 13 '18 at 20:11
sambajetsonsambajetson
280211
$endgroup$
Just as we can use the matrix power method to find eigenvalues/eigenvectors of matrices in an iterative way, we can analogously find the eigenvalues/eigenvectors of tensors in a similar way.
My question is about the convergence properties here. For matrices (order 2 tensors), we know the error converges linearly in the eigengap and it is not hard to prove (I'm talking about the basic matrix power method, not e.g. Rayleigh quotient method which I know will actually converge quadratically with eigenvalue gap).
For symmetric orthogonally decomposable (odeco) tensors of order 3, for the basic method (not a more advanced thing like if there is an analogous tensor Rayleigh quotient method) I have heard the rate is now square in the eigengap. Is it true that in general for an order D symmetric odeco tensor, the convergence using the basic tensor power method will be at a rate power of (D-1) of the eigengap?
linear-algebra tensor-products numerical-linear-algebra tensors symmetric-matrices
linear-algebra tensor-products numerical-linear-algebra tensors symmetric-matrices
asked Dec 13 '18 at 20:11
sambajetsonsambajetson
280211
asked Dec 13 '18 at 20:11
sambajetsonsambajetson
280211
asked Dec 13 '18 at 20:11
sambajetsonsambajetson
280211
asked Dec 13 '18 at 20:11
sambajetsonsambajetson
280211
asked Dec 13 '18 at 20:11
sambajetsonsambajetson
280211
280211
add a comment |
add a comment |
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