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

Powers and More Powers

David S. Watkins

Department of Mathematics Washington State University

University of Calgary, March 17, 2017

David S. Watkins Powers and More Powers

A bit of my history

David S. Watkins Powers and More Powers

A bit of my history

Growing up in California . . .

David S. Watkins Powers and More Powers

A bit of my history

Growing up in California . . . I want to be

David S. Watkins Powers and More Powers

A bit of my history

Growing up in California . . . I want to be a baseball star,

David S. Watkins Powers and More Powers

A bit of my history

Growing up in California . . . I want to be a baseball star, a teacher,

David S. Watkins Powers and More Powers

A bit of my history

Growing up in California . . . I want to be a baseball star, a teacher, a functional analyst.

David S. Watkins Powers and More Powers

A bit of my history

Growing up in California . . . I want to be a baseball star, a teacher, a functional analyst. I graduate from UCSB in 1970.

David S. Watkins Powers and More Powers

A bit of my history

Growing up in California . . . I want to be a baseball star, a teacher, a functional analyst. I graduate from UCSB in 1970. I move to Canada.

David S. Watkins Powers and More Powers

A bit of my history

Growing up in California . . . I want to be a baseball star, a teacher, a functional analyst. I graduate from UCSB in 1970. I move to Canada. First stop: University of Toronto

David S. Watkins Powers and More Powers

A bit of my history

Growing up in California . . . I want to be a baseball star, a teacher, a functional analyst. I graduate from UCSB in 1970. I move to Canada. First stop: University of Toronto

David S. Watkins Powers and More Powers

A bit of my history

Chandler Davis and I have some things in common,

David S. Watkins Powers and More Powers

A bit of my history

Chandler Davis and I have some things in common, but I hasten to add . . .

David S. Watkins Powers and More Powers

A bit of my history

Chandler Davis and I have some things in common, but I hasten to add . . .

David S. Watkins Powers and More Powers

A bit of my history

David S. Watkins Powers and More Powers

A bit of my history

I take functional analysis from Chandler Davis.

David S. Watkins Powers and More Powers

A bit of my history

I take functional analysis from Chandler Davis. M.Sc. from University of Toronto, 1971

David S. Watkins Powers and More Powers

A bit of my history

I take functional analysis from Chandler Davis. M.Sc. from University of Toronto, 1971 I decide to move back out west.

David S. Watkins Powers and More Powers

A bit of my history

I take functional analysis from Chandler Davis. M.Sc. from University of Toronto, 1971 I decide to move back out west. I ask Chandler Davis for advice.

David S. Watkins Powers and More Powers

A bit of my history

David S. Watkins Powers and More Powers

A bit of my history

Chandler:

David S. Watkins Powers and More Powers

A bit of my history

Chandler: It’s okay to move to Calgary . . .

David S. Watkins Powers and More Powers

A bit of my history

Chandler: It’s okay to move to Calgary . . . . . . if you work with this guy named

David S. Watkins Powers and More Powers

A bit of my history

Chandler: It’s okay to move to Calgary . . . . . . if you work with this guy named Peter Lancaster.

David S. Watkins Powers and More Powers

A bit of my history

Chandler: It’s okay to move to Calgary . . . . . . if you work with this guy named Peter Lancaster. So I came to Calgary.

David S. Watkins Powers and More Powers

A bit of my history

I think about the job market.

David S. Watkins Powers and More Powers

A bit of my history

I think about the job market. I migrate toward numerical analysis.

David S. Watkins Powers and More Powers

A bit of my history

I think about the job market. I migrate toward numerical analysis. Finite Elements

David S. Watkins Powers and More Powers

A bit of my history

I think about the job market. I migrate toward numerical analysis. Finite Elements Ph.D. University of Calgary, 1974.

David S. Watkins Powers and More Powers

A bit of my history

I think about the job market. I migrate toward numerical analysis. Finite Elements Ph.D. University of Calgary, 1974.

• Postdoc. 1974–1975.

David S. Watkins Powers and More Powers

A bit of my history

I think about the job market. I migrate toward numerical analysis. Finite Elements Ph.D. University of Calgary, 1974.

• Postdoc. 1974–1975.

. . . and I am on my way!

David S. Watkins Powers and More Powers

My trajectory

Finite Elements (PDE)

David S. Watkins Powers and More Powers

My trajectory

Finite Elements (PDE) ODE, stability, stiffness

David S. Watkins Powers and More Powers

My trajectory

Finite Elements (PDE) ODE, stability, stiffness Eigenvalues

David S. Watkins Powers and More Powers

My trajectory

Finite Elements (PDE) ODE, stability, stiffness Eigenvalues Computing eigenvalues of matrices

David S. Watkins Powers and More Powers

My trajectory

Finite Elements (PDE) ODE, stability, stiffness Eigenvalues Computing eigenvalues of matrices I get stuck.

David S. Watkins Powers and More Powers

Today’s talk . . .

David S. Watkins Powers and More Powers

Today’s talk . . .

But this talk is supposed to be about analysis.

David S. Watkins Powers and More Powers

Today’s talk . . .

But this talk is supposed to be about analysis. I’ll start out talking about eigenvalue computation.

David S. Watkins Powers and More Powers

Today’s talk . . .

But this talk is supposed to be about analysis. I’ll start out talking about eigenvalue computation. But we’ll soon drift into analysis.

David S. Watkins Powers and More Powers

Today’s talk . . .

But this talk is supposed to be about analysis. I’ll start out talking about eigenvalue computation. But we’ll soon drift into analysis. This material is not new,

David S. Watkins Powers and More Powers

Today’s talk . . .

But this talk is supposed to be about analysis. I’ll start out talking about eigenvalue computation. But we’ll soon drift into analysis. This material is not new, but I think it’s interesting,

David S. Watkins Powers and More Powers

Today’s talk . . .

But this talk is supposed to be about analysis. I’ll start out talking about eigenvalue computation. But we’ll soon drift into analysis. This material is not new, but I think it’s interesting, and I hope you will too.

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

A ∈ Cn×n

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

A ∈ Cn×n Compute some eigenvalues, eigenvectors, invariant subspaces, . . .

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

A ∈ Cn×n Compute some eigenvalues, eigenvectors, invariant subspaces, . . . What we don’t do:

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

A ∈ Cn×n Compute some eigenvalues, eigenvectors, invariant subspaces, . . . What we don’t do: We don’t ever form the characteristic polynomial.

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

So what do we do?

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

So what do we do? The simplest idea

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

So what do we do? The simplest idea Power Method

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

So what do we do? The simplest idea Power Method v, Av, A2v, A3v, . . .

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

So what do we do? The simplest idea Power Method v, Av, A2v, A3v, . . . This method is forgetful.

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

Krylov subspace methods

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

Krylov subspace methods are not forgetful.

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

Krylov subspace methods are not forgetful. span

• v, Av, A2v, . . . , Ak−1v
• David S. Watkins

Powers and More Powers

Solving Eigenvalue Problems

Krylov subspace methods are not forgetful. span

• v, Av, A2v, . . . , Ak−1v
• rthonormalize!

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

Krylov subspace methods are not forgetful. span

• v, Av, A2v, . . . , Ak−1v
• rthonormalize!

Arnoldi process

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

Arnoldi Process (Variant of Gram-Schmidt)

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

Arnoldi Process (Variant of Gram-Schmidt) kth step (We already have o.n. q1, . . . , qk)

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

Arnoldi Process (Variant of Gram-Schmidt) kth step (We already have o.n. q1, . . . , qk) ˆ qk+1 = Aqk −

k

• j=1

qjhjk

David S. Watkins Powers and More Powers

Solving Eigenvalue Problems

Arnoldi Process (Variant of Gram-Schmidt) kth step (We already have o.n. q1, . . . , qk) ˆ qk+1 = Aqk −

k

• j=1

qjhjk Normalize ˆ qk+1 to get qk+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 ˆ qk+1 = Aqk − qkαk − qk−1βk−1

David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950)

Now specialize to A = A∗ Symmetric Lanczos Process ˆ qk+1 = Aqk − qkαk − qk−1βk−1 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 ˆ qk+1 = Aqk − qkαk − qk−1βk−1 Almost all of the coefficients are zero! Now what do we do?

David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950)

ˆ qk+1 = Aqk − qkαk − qk−1βk−1

David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950)

ˆ qk+1 = Aqk − qkαk − qk−1βk−1 Collect the coefficients.

David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950)

ˆ qk+1 = Aqk − qkαk − qk−1βk−1 Collect the coefficients. Tk =       α1 β1 β1 α2 ... ... ... βk−1 βk−1 αk      

David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950)

ˆ qk+1 = Aqk − qkαk − qk−1βk−1 Collect the coefficients. Tk =       α1 β1 β1 α2 ... ... ... βk−1 βk−1 αk       Compute the eigenvalues of Tk.

David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950)

ˆ qk+1 = Aqk − qkαk − qk−1βk−1 Collect the coefficients. Tk =       α1 β1 β1 α2 ... ... ... βk−1 βk−1 αk       Compute the eigenvalues of Tk. 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 : H → H, A = A∗

David S. Watkins Powers and More Powers

Symmetric Lanczos Process (1950)

How A is used q → Aq A could be an operator! A : H → H, A = A∗ 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 : H → H, A = A∗ 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 . . . I(f ) = b

a

f (x)w(x)dx.

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Abruptly changing the subject . . . I(f ) = b

a

f (x)w(x)dx. (or integrate with respect to a measure µ)

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Abruptly changing the subject . . . I(f ) = b

a

f (x)w(x)dx. (or integrate with respect to a measure µ) Approximate by Q(f ) =

k

• j=1

wjf (xj).

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Abruptly changing the subject . . . I(f ) = b

a

f (x)w(x)dx. (or integrate with respect to a measure µ) Approximate by Q(f ) =

k

• j=1

wjf (xj). How to choose sample points and weights?

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Abruptly changing the subject . . . I(f ) = b

a

f (x)w(x)dx. (or integrate with respect to a measure µ) Approximate by Q(f ) =

k

• j=1

wjf (xj). 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? Inner product: f , g = b

a

f (x)g(x)w(x)dx

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

How to maximize degree? Inner product: f , g = b

a

f (x)g(x)w(x)dx Generate orthogonal polynomials: p0(x), p1(x), p2(x), . . .

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

How to maximize degree? Inner product: f , g = b

a

f (x)g(x)w(x)dx Generate orthogonal polynomials: p0(x), p1(x), p2(x), . . . Optimal sample points are the zeros of pk. (weights dealt with later)

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

How to maximize degree? Inner product: f , g = b

a

f (x)g(x)w(x)dx Generate orthogonal polynomials: p0(x), p1(x), p2(x), . . . Optimal sample points are the zeros of pk. (weights dealt with later) How to generate pk 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: ˆ pk+1(x) = x pk(x) − αkpk(x) − βk−1pk−1(x).

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Stieltjes Procedure: ˆ pk+1(x) = x pk(x) − αkpk(x) − βk−1pk−1(x). Obtain pk+1 from ˆ pk+1 by normalization.

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Stieltjes Procedure: ˆ pk+1(x) = x pk(x) − αkpk(x) − βk−1pk−1(x). Obtain pk+1 from ˆ pk+1 by normalization. and

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Stieltjes Procedure: ˆ pk+1(x) = x pk(x) − αkpk(x) − βk−1pk−1(x). Obtain pk+1 from ˆ pk+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: ˆ pk+1(x) = x pk(x) − αkpk(x) − βk−1pk−1(x). Obtain pk+1 from ˆ pk+1 by normalization. and this is an instance of the symmetric Lanczos process! Af (x) = x f (x) (A : L2 → L2, A = A∗)

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Stieltjes Procedure: ˆ pk+1(x) = x pk(x) − αkpk(x) − βk−1pk−1(x). Obtain pk+1 from ˆ pk+1 by normalization. and this is an instance of the symmetric Lanczos process! Af (x) = x f (x) (A : L2 → L2, A = A∗) Now what?

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Collect the coefficients. Tk =       α1 β1 β1 α2 ... ... ... βk−1 βk−1 αk      

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Collect the coefficients. Tk =       α1 β1 β1 α2 ... ... ... βk−1 βk−1 αk       Compute the eigenvalues of Tk.

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Collect the coefficients. Tk =       α1 β1 β1 α2 ... ... ... βk−1 βk−1 αk       Compute the eigenvalues of Tk. These are the sample points (zeros of pk).

David S. Watkins Powers and More Powers

Stieltjes Procedure (1884)

Collect the coefficients. Tk =       α1 β1 β1 α2 ... ... ... βk−1 βk−1 αk       Compute the eigenvalues of Tk. These are the sample points (zeros of pk). . . . 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

Spectral Theorem

The Stieltjes procedure is an instance of the symmetric Lanczos process . . . and the converse is true as well! So . . .

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! So . . . symmetric Lanczos = Stieltjes.

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! So . . . symmetric Lanczos = Stieltjes. This is a consequence of the spectral theorem for self-adjoint operators.

David S. Watkins Powers and More Powers

Spectral Theorem

Various formulations

David S. Watkins Powers and More Powers

Spectral Theorem

Various formulations Every Hermitian matrix “is” diagonal.

David S. Watkins Powers and More Powers

Spectral Theorem

Various formulations Every Hermitian matrix “is” diagonal. A =

• σ(A)

λdEλ

David S. Watkins Powers and More Powers

Spectral Theorem

Various formulations Every Hermitian matrix “is” diagonal. A =

• σ(A)

λdEλ Af (x) = x f (x)

David S. Watkins Powers and More Powers

Spectral Theorem

Various formulations Every Hermitian matrix “is” diagonal. A =

• σ(A)

λdEλ Af (x) = x f (x) Popularized by Paul Halmos

David S. Watkins Powers and More Powers

Spectral Theorem

Various formulations Every Hermitian matrix “is” diagonal. A =

• σ(A)

λdEλ Af (x) = x f (x) Popularized by Paul Halmos What does the spectral theorem say?, A. M. Monthly, 1963.

David S. Watkins Powers and More Powers

Spectral Theorem

Various formulations Every Hermitian matrix “is” diagonal. A =

• σ(A)

λdEλ Af (x) = x f (x) Popularized by Paul Halmos What does the spectral theorem say?, A. M. Monthly, 1963. (Thank you, Chandler Davis!)

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof:

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗.

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ)

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ) WLG assume A has a cyclic vector v,

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ) WLG assume A has a cyclic vector v, i.e. span

• v, Av, A2v, A3v, . . .
• is dense in H.

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ) WLG assume A has a cyclic vector v, i.e. span

• v, Av, A2v, A3v, . . .
• is dense in H.

power method! Krylov subspace!

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ) WLG assume A has a cyclic vector v, i.e. span

• v, Av, A2v, A3v, . . .
• is dense in H.

power method! Krylov subspace! span

• v, Av, A2v, A3v, . . .
• = {p(A)v | p ∈ P}

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ) WLG assume A has a cyclic vector v, i.e. span

• v, Av, A2v, A3v, . . .
• is dense in H.

power method! Krylov subspace! span

• v, Av, A2v, A3v, . . .
• = {p(A)v | p ∈ P}

isomorphism: p(A)v → p(x)

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ) WLG assume A has a cyclic vector v, i.e. span

• v, Av, A2v, A3v, . . .
• is dense in H.

power method! Krylov subspace! span

• v, Av, A2v, A3v, . . .
• = {p(A)v | p ∈ P}

isomorphism: p(A)v → p(x) (Find the right µ.)

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ) WLG assume A has a cyclic vector v, i.e. span

• v, Av, A2v, A3v, . . .
• is dense in H.

power method! Krylov subspace! span

• v, Av, A2v, A3v, . . .
• = {p(A)v | p ∈ P}

isomorphism: p(A)v → p(x) (Find the right µ.) p(A)v → A p(A)v

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ) WLG assume A has a cyclic vector v, i.e. span

• v, Av, A2v, A3v, . . .
• is dense in H.

power method! Krylov subspace! span

• v, Av, A2v, A3v, . . .
• = {p(A)v | p ∈ P}

isomorphism: p(A)v → p(x) (Find the right µ.) p(A)v → A p(A)v maps to

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ) WLG assume A has a cyclic vector v, i.e. span

• v, Av, A2v, A3v, . . .
• is dense in H.

power method! Krylov subspace! span

• v, Av, A2v, A3v, . . .
• = {p(A)v | p ∈ P}

isomorphism: p(A)v → p(x) (Find the right µ.) p(A)v → A p(A)v maps to p(x) → x p(x)

David S. Watkins Powers and More Powers

Spectral Theorem

Sketch of proof: A : H → H, A = A∗. Establish isomorphism: H ≈ L2(µ) WLG assume A has a cyclic vector v, i.e. span

• v, Av, A2v, A3v, . . .
• is dense in H.

power method! Krylov subspace! span

• v, Av, A2v, A3v, . . .
• = {p(A)v | p ∈ P}

isomorphism: p(A)v → p(x) (Find the right µ.) p(A)v → A p(A)v maps to p(x) → x p(x) Thus the action of A maps to multiplication by x.

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process:

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process: qk+1 = Aqk − αkqk − βk−1qk−1

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process: qk+1 = Aqk − αkqk − βk−1qk−1 q0 = v

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process: qk+1 = Aqk − αkqk − βk−1qk−1 q0 = v q1 = Av − α0v = (A − α0I)v = p1(A)v

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process: qk+1 = Aqk − αkqk − βk−1qk−1 q0 = v q1 = Av − α0v = (A − α0I)v = p1(A)v qk = pk(A)v

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process: qk+1 = Aqk − αkqk − βk−1qk−1 q0 = v q1 = Av − α0v = (A − α0I)v = p1(A)v qk = pk(A)v pk+1(A)v = A pk(A)v − αkpk(A)v − βk−1pk−1(A)v

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process: qk+1 = Aqk − αkqk − βk−1qk−1 q0 = v q1 = Av − α0v = (A − α0I)v = p1(A)v qk = pk(A)v pk+1(A)v = A pk(A)v − αkpk(A)v − βk−1pk−1(A)v Now apply the isomorphism p(A)v → p(x)

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process: qk+1 = Aqk − αkqk − βk−1qk−1 q0 = v q1 = Av − α0v = (A − α0I)v = p1(A)v qk = pk(A)v pk+1(A)v = A pk(A)v − αkpk(A)v − βk−1pk−1(A)v Now apply the isomorphism p(A)v → p(x) pk+1(x) = x pk(x) − αkpk(x) − βk−1pk−1(x)

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process: qk+1 = Aqk − αkqk − βk−1qk−1 q0 = v q1 = Av − α0v = (A − α0I)v = p1(A)v qk = pk(A)v pk+1(A)v = A pk(A)v − αkpk(A)v − βk−1pk−1(A)v Now apply the isomorphism p(A)v → p(x) pk+1(x) = x pk(x) − αkpk(x) − βk−1pk−1(x) This is exactly the Stieltjes procedure!

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process: qk+1 = Aqk − αkqk − βk−1qk−1 q0 = v q1 = Av − α0v = (A − α0I)v = p1(A)v qk = pk(A)v pk+1(A)v = A pk(A)v − αkpk(A)v − βk−1pk−1(A)v Now apply the isomorphism p(A)v → p(x) pk+1(x) = x pk(x) − αkpk(x) − βk−1pk−1(x) This is exactly the Stieltjes procedure! Conclusion:

David S. Watkins Powers and More Powers

Equivalence of Lanczos and Stieltjes

Symmetric Lanczos process: qk+1 = Aqk − αkqk − βk−1qk−1 q0 = v q1 = Av − α0v = (A − α0I)v = p1(A)v qk = pk(A)v pk+1(A)v = A pk(A)v − αkpk(A)v − βk−1pk−1(A)v Now apply the isomorphism p(A)v → p(x) pk+1(x) = x pk(x) − αkpk(x) − βk−1pk−1(x) This is exactly the Stieltjes procedure! Conclusion: The symmetric Lanczos process and the Stieltjes procedure are exactly the same thing.

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting.

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role?

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role? Not much!

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role? Not much! Some perspectives on the eigenvalue problem, SIREV, 1993.

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role? Not much! Some perspectives on the eigenvalue problem, SIREV, 1993. Unitary theory

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role? Not much! Some perspectives on the eigenvalue problem, SIREV, 1993. Unitary theory work of Ammar, Gragg, Reichel

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role? Not much! Some perspectives on the eigenvalue problem, SIREV, 1993. Unitary theory work of Ammar, Gragg, Reichel Isometric Arnoldi Process (Gragg 1982)

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role? Not much! Some perspectives on the eigenvalue problem, SIREV, 1993. Unitary theory work of Ammar, Gragg, Reichel Isometric Arnoldi Process (Gragg 1982) (instead of symmetric Lanczos)

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role? Not much! Some perspectives on the eigenvalue problem, SIREV, 1993. Unitary theory work of Ammar, Gragg, Reichel Isometric Arnoldi Process (Gragg 1982) (instead of symmetric Lanczos) Szeg¨

• recursions

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role? Not much! Some perspectives on the eigenvalue problem, SIREV, 1993. Unitary theory work of Ammar, Gragg, Reichel Isometric Arnoldi Process (Gragg 1982) (instead of symmetric Lanczos) Szeg¨

• recursions

(instead of Stieltjes procedure)

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role? Not much! Some perspectives on the eigenvalue problem, SIREV, 1993. Unitary theory work of Ammar, Gragg, Reichel Isometric Arnoldi Process (Gragg 1982) (instead of symmetric Lanczos) Szeg¨

• recursions

(instead of Stieltjes procedure) spectral theorem for unitary operators

David S. Watkins Powers and More Powers

My role in this

I hope you found this interesting. My role? Not much! Some perspectives on the eigenvalue problem, SIREV, 1993. Unitary theory work of Ammar, Gragg, Reichel Isometric Arnoldi Process (Gragg 1982) (instead of symmetric Lanczos) Szeg¨

• recursions

(instead of Stieltjes procedure) spectral theorem for unitary operators citations

David S. Watkins Powers and More Powers

It’s been fun

David S. Watkins Powers and More Powers

It’s been fun

What’s next for this

David S. Watkins Powers and More Powers

It’s been fun

What’s next for this baseball star,

David S. Watkins Powers and More Powers

It’s been fun

What’s next for this baseball star, teacher,

David S. Watkins Powers and More Powers

It’s been fun

What’s next for this baseball star, teacher, numerical analyst?

David S. Watkins Powers and More Powers

It’s been fun

What’s next for this baseball star, teacher, numerical analyst? Retirement in May.

David S. Watkins Powers and More Powers

It’s been fun

What’s next for this baseball star, teacher, numerical analyst? Retirement in May. Then what?

David S. Watkins Powers and More Powers

It’s been fun

What’s next for this baseball star, teacher, numerical analyst? Retirement in May. Then what?

David S. Watkins Powers and More Powers

It’s been fun

What’s next for this baseball star, teacher, numerical analyst? Retirement in May. Then what? Perhaps I should move back to Canada!

David S. Watkins Powers and More Powers

It’s been fun

David S. Watkins Powers and More Powers

It’s been fun

Thank you for your attention.

David S. Watkins Powers and More Powers