Optimal pivoting strategy in LU factorization
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I'm currently reading the book Numerical Linear Algebra by Lloyd N. Trefethen and David Bau, III, working my way through the required lectures for my Numerical Analysis class. The current subject is LU factorization, and while reading Lecture 21 about Pivoting (in Gaussian Elimination), I came across the statement that in practice, partial pivoting is equally as good as complete pivoting, while requiring inspection of a much smaller number of entries.
I'm pretty sure I understand correctly what partial and complete pivoting mean, namely with complete pivoting the entire $A_{k:m, k:m}$ submatrix is examined while with partial pivoting only the $A_{k:m, k}$ column is examined for an optimal pivot (the largest possible value, that is (right?)).
This seems very wrong to me and I'm surely missing some trivial argument for why this must be true. How can we be sure that part of the $k$th column of A always contains the biggest value in the entire submatrix? Am I right in assuming that the author meant to say that the increased gain from choosing the optimal value in the entire submatrix would be vastly overshadowed by the sheer amount of time needed to find it, and that only considering the first column of the submatrix yields good results?
numerical-methods numerical-linear-algebra gaussian-elimination lu-decomposition
asked Nov 17 at 20:14
Peiffap
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I'm currently reading the book Numerical Linear Algebra by Lloyd N. Trefethen and David Bau, III, working my way through the required lectures for my Numerical Analysis class. The current subject is LU factorization, and while reading Lecture 21 about Pivoting (in Gaussian Elimination), I came across the statement that in practice, partial pivoting is equally as good as complete pivoting, while requiring inspection of a much smaller number of entries.
I'm pretty sure I understand correctly what partial and complete pivoting mean, namely with complete pivoting the entire $A_{k:m, k:m}$ submatrix is examined while with partial pivoting only the $A_{k:m, k}$ column is examined for an optimal pivot (the largest possible value, that is (right?)).
This seems very wrong to me and I'm surely missing some trivial argument for why this must be true. How can we be sure that part of the $k$th column of A always contains the biggest value in the entire submatrix? Am I right in assuming that the author meant to say that the increased gain from choosing the optimal value in the entire submatrix would be vastly overshadowed by the sheer amount of time needed to find it, and that only considering the first column of the submatrix yields good results?
numerical-methods numerical-linear-algebra gaussian-elimination lu-decomposition
asked Nov 17 at 20:14
Peiffap
367
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm currently reading the book Numerical Linear Algebra by Lloyd N. Trefethen and David Bau, III, working my way through the required lectures for my Numerical Analysis class. The current subject is LU factorization, and while reading Lecture 21 about Pivoting (in Gaussian Elimination), I came across the statement that in practice, partial pivoting is equally as good as complete pivoting, while requiring inspection of a much smaller number of entries.
I'm pretty sure I understand correctly what partial and complete pivoting mean, namely with complete pivoting the entire $A_{k:m, k:m}$ submatrix is examined while with partial pivoting only the $A_{k:m, k}$ column is examined for an optimal pivot (the largest possible value, that is (right?)).
This seems very wrong to me and I'm surely missing some trivial argument for why this must be true. How can we be sure that part of the $k$th column of A always contains the biggest value in the entire submatrix? Am I right in assuming that the author meant to say that the increased gain from choosing the optimal value in the entire submatrix would be vastly overshadowed by the sheer amount of time needed to find it, and that only considering the first column of the submatrix yields good results?
numerical-methods numerical-linear-algebra gaussian-elimination lu-decomposition
asked Nov 17 at 20:14
Peiffap
367
I'm currently reading the book Numerical Linear Algebra by Lloyd N. Trefethen and David Bau, III, working my way through the required lectures for my Numerical Analysis class. The current subject is LU factorization, and while reading Lecture 21 about Pivoting (in Gaussian Elimination), I came across the statement that in practice, partial pivoting is equally as good as complete pivoting, while requiring inspection of a much smaller number of entries.
I'm pretty sure I understand correctly what partial and complete pivoting mean, namely with complete pivoting the entire $A_{k:m, k:m}$ submatrix is examined while with partial pivoting only the $A_{k:m, k}$ column is examined for an optimal pivot (the largest possible value, that is (right?)).
This seems very wrong to me and I'm surely missing some trivial argument for why this must be true. How can we be sure that part of the $k$th column of A always contains the biggest value in the entire submatrix? Am I right in assuming that the author meant to say that the increased gain from choosing the optimal value in the entire submatrix would be vastly overshadowed by the sheer amount of time needed to find it, and that only considering the first column of the submatrix yields good results?
numerical-methods numerical-linear-algebra gaussian-elimination lu-decomposition
numerical-methods numerical-linear-algebra gaussian-elimination lu-decomposition
asked Nov 17 at 20:14
Peiffap
367
asked Nov 17 at 20:14
Peiffap
367
asked Nov 17 at 20:14
Peiffap
367
asked Nov 17 at 20:14
Peiffap
367
asked Nov 17 at 20:14
Peiffap
367
367
add a comment |
add a comment |
1 Answer
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The authors do not claim that the partial and complete pivoting are equivalent and that the optimal pivot at each step of the elimination must be in the first column of the working sub-matrix.
What they claim merely is that partial and complete pivoting strategies are equally good in the sense that the growth factor obtained by the partial pivoting strategy is not much worse than that obtained with the complete pivoting for vast majority of matrices arising in practice. There is no way how to prove this as there is no rigorous definition of what a practical matrix is and this claim is simply a result of decades of experience.
There are well known examples showing that the partial pivoting can be much worse than the complete pivoting so this claim is not generally true.
answered Nov 22 at 14:24
Algebraic Pavel
16.2k31839
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The authors do not claim that the partial and complete pivoting are equivalent and that the optimal pivot at each step of the elimination must be in the first column of the working sub-matrix.
What they claim merely is that partial and complete pivoting strategies are equally good in the sense that the growth factor obtained by the partial pivoting strategy is not much worse than that obtained with the complete pivoting for vast majority of matrices arising in practice. There is no way how to prove this as there is no rigorous definition of what a practical matrix is and this claim is simply a result of decades of experience.
There are well known examples showing that the partial pivoting can be much worse than the complete pivoting so this claim is not generally true.
answered Nov 22 at 14:24
Algebraic Pavel
16.2k31839
add a comment |
up vote
1
down vote
accepted
The authors do not claim that the partial and complete pivoting are equivalent and that the optimal pivot at each step of the elimination must be in the first column of the working sub-matrix.
What they claim merely is that partial and complete pivoting strategies are equally good in the sense that the growth factor obtained by the partial pivoting strategy is not much worse than that obtained with the complete pivoting for vast majority of matrices arising in practice. There is no way how to prove this as there is no rigorous definition of what a practical matrix is and this claim is simply a result of decades of experience.
There are well known examples showing that the partial pivoting can be much worse than the complete pivoting so this claim is not generally true.
answered Nov 22 at 14:24
Algebraic Pavel
16.2k31839
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The authors do not claim that the partial and complete pivoting are equivalent and that the optimal pivot at each step of the elimination must be in the first column of the working sub-matrix.
What they claim merely is that partial and complete pivoting strategies are equally good in the sense that the growth factor obtained by the partial pivoting strategy is not much worse than that obtained with the complete pivoting for vast majority of matrices arising in practice. There is no way how to prove this as there is no rigorous definition of what a practical matrix is and this claim is simply a result of decades of experience.
There are well known examples showing that the partial pivoting can be much worse than the complete pivoting so this claim is not generally true.
answered Nov 22 at 14:24
Algebraic Pavel
16.2k31839
The authors do not claim that the partial and complete pivoting are equivalent and that the optimal pivot at each step of the elimination must be in the first column of the working sub-matrix.
What they claim merely is that partial and complete pivoting strategies are equally good in the sense that the growth factor obtained by the partial pivoting strategy is not much worse than that obtained with the complete pivoting for vast majority of matrices arising in practice. There is no way how to prove this as there is no rigorous definition of what a practical matrix is and this claim is simply a result of decades of experience.
There are well known examples showing that the partial pivoting can be much worse than the complete pivoting so this claim is not generally true.
answered Nov 22 at 14:24
Algebraic Pavel
16.2k31839
answered Nov 22 at 14:24
Algebraic Pavel
16.2k31839
answered Nov 22 at 14:24
Algebraic Pavel
16.2k31839
answered Nov 22 at 14:24
Algebraic Pavel
16.2k31839
16.2k31839
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