Powers and More Powers David S. Watkins Department of Mathematics - PowerPoint PPT Presentation

Solving Eigenvalue Problems Arnoldi Process (Variant of Gram-Schmidt) k th step (We already have o.n. q 1 , . . . , q k ) k � q k +1 = Aq k − ˆ q j h jk j =1 Normalize ˆ q k +1 to get q k +1 . David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) Now specialize to A = A ∗ David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) Now specialize to A = A ∗ Symmetric Lanczos Process David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) Now specialize to A = A ∗ Symmetric Lanczos Process q k +1 = Aq k − q k α k − q k − 1 β k − 1 ˆ David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) Now specialize to A = A ∗ Symmetric Lanczos Process q k +1 = Aq k − q k α k − q k − 1 β k − 1 ˆ Almost all of the coefficients are zero! David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) Now specialize to A = A ∗ Symmetric Lanczos Process q k +1 = Aq k − q k α k − q k − 1 β k − 1 ˆ Almost all of the coefficients are zero! Now what do we do? David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) q k +1 = Aq k − q k α k − q k − 1 β k − 1 ˆ David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) q k +1 = Aq k − q k α k − q k − 1 β k − 1 ˆ Collect the coefficients. David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) q k +1 = Aq k − q k α k − q k − 1 β k − 1 ˆ Collect the coefficients.   α 1 β 1 ...   β 1 α 2   T k =   ... ...   β k − 1   β k − 1 α k David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) q k +1 = Aq k − q k α k − q k − 1 β k − 1 ˆ Collect the coefficients.   α 1 β 1 ...   β 1 α 2   T k =   ... ...   β k − 1   β k − 1 α k Compute the eigenvalues of T k . David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) q k +1 = Aq k − q k α k − q k − 1 β k − 1 ˆ Collect the coefficients.   α 1 β 1 ...   β 1 α 2   T k =   ... ...   β k − 1   β k − 1 α k Compute the eigenvalues of T k . Some of these approximate peripheral eigenvalues of A . David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) How A is used David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) How A is used q → Aq David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) How A is used q → Aq A could be an operator! A = A ∗ A : H → H , David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) How A is used q → Aq A could be an operator! A = A ∗ A : H → H , We can run the Lanczos process, David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950) How A is used q → Aq A could be an operator! A = A ∗ A : H → H , We can run the Lanczos process, potentially forever. David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Abruptly changing the subject . . . David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Abruptly changing the subject . . . � b I ( f ) = f ( x ) w ( x ) dx . a David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Abruptly changing the subject . . . � b I ( f ) = f ( x ) w ( x ) dx . a (or integrate with respect to a measure µ ) David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Abruptly changing the subject . . . � b I ( f ) = f ( x ) w ( x ) dx . a (or integrate with respect to a measure µ ) k � Approximate by Q ( f ) = w j f ( x j ). j =1 David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Abruptly changing the subject . . . � b I ( f ) = f ( x ) w ( x ) dx . a (or integrate with respect to a measure µ ) k � Approximate by Q ( f ) = w j f ( x j ). j =1 How to choose sample points and weights? David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Abruptly changing the subject . . . � b I ( f ) = f ( x ) w ( x ) dx . a (or integrate with respect to a measure µ ) k � Approximate by Q ( f ) = w j f ( x j ). j =1 How to choose sample points and weights? One answer: maximize degree. David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) How to maximize degree? David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) How to maximize degree? � b Inner product: � f , g � = f ( x ) g ( x ) w ( x ) dx a David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) How to maximize degree? � b Inner product: � f , g � = f ( x ) g ( x ) w ( x ) dx a Generate orthogonal polynomials: p 0 ( x ), p 1 ( x ), p 2 ( x ), . . . David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) How to maximize degree? � b Inner product: � f , g � = f ( x ) g ( x ) w ( x ) dx a Generate orthogonal polynomials: p 0 ( x ), p 1 ( x ), p 2 ( x ), . . . Optimal sample points are the zeros of p k . (weights dealt with later) David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) How to maximize degree? � b Inner product: � f , g � = f ( x ) g ( x ) w ( x ) dx a Generate orthogonal polynomials: p 0 ( x ), p 1 ( x ), p 2 ( x ), . . . Optimal sample points are the zeros of p k . (weights dealt with later) How to generate p k efficiently? David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Stieltjes Procedure: David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Stieltjes Procedure: p k +1 ( x ) = x p k ( x ) − α k p k ( x ) − β k − 1 p k − 1 ( x ). ˆ David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Stieltjes Procedure: p k +1 ( x ) = x p k ( x ) − α k p k ( x ) − β k − 1 p k − 1 ( x ). ˆ Obtain p k +1 from ˆ p k +1 by normalization. David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Stieltjes Procedure: p k +1 ( x ) = x p k ( x ) − α k p k ( x ) − β k − 1 p k − 1 ( x ). ˆ Obtain p k +1 from ˆ p k +1 by normalization. and David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Stieltjes Procedure: p k +1 ( x ) = x p k ( x ) − α k p k ( x ) − β k − 1 p k − 1 ( x ). ˆ Obtain p k +1 from ˆ p k +1 by normalization. and this is an instance of the symmetric Lanczos process! David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Stieltjes Procedure: p k +1 ( x ) = x p k ( x ) − α k p k ( x ) − β k − 1 p k − 1 ( x ). ˆ Obtain p k +1 from ˆ p k +1 by normalization. and this is an instance of the symmetric Lanczos process! Af ( x ) = x f ( x ) ( A : L 2 → L 2 , A = A ∗ ) David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Stieltjes Procedure: p k +1 ( x ) = x p k ( x ) − α k p k ( x ) − β k − 1 p k − 1 ( x ). ˆ Obtain p k +1 from ˆ p k +1 by normalization. and this is an instance of the symmetric Lanczos process! Af ( x ) = x f ( x ) ( A : L 2 → L 2 , A = A ∗ ) Now what? David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Collect the coefficients.   α 1 β 1 ...   β 1 α 2   T k =  ... ...    β k − 1   β k − 1 α k David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Collect the coefficients.   α 1 β 1 ...   β 1 α 2   T k =  ... ...    β k − 1   β k − 1 α k Compute the eigenvalues of T k . David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Collect the coefficients.   α 1 β 1 ...   β 1 α 2   T k =  ... ...    β k − 1   β k − 1 α k Compute the eigenvalues of T k . These are the sample points (zeros of p k ). David S. Watkins Powers and More Powers

Stieltjes Procedure (1884) Collect the coefficients.   α 1 β 1 ...   β 1 α 2   T k =  ... ...    β k − 1   β k − 1 α k Compute the eigenvalues of T k . These are the sample points (zeros of p k ). . . . and the weights are . . . David S. Watkins Powers and More Powers

Spectral Theorem The Stieltjes procedure is an instance of the symmetric Lanczos process . . . David S. Watkins Powers and More Powers

Spectral Theorem The Stieltjes procedure is an instance of the symmetric Lanczos process . . . and the converse is true as well! David S. Watkins Powers and More Powers