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?










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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?










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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?










      share|cite|improve this question













      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






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      asked Nov 17 at 20:14









      Peiffap

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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.






          share|cite|improve this answer





















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            up vote
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            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.






            share|cite|improve this answer

























              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.






              share|cite|improve this answer























                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.






                share|cite|improve this answer












                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 14:24









                Algebraic Pavel

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