Quantitative Diagonalizability Part I: Three Measures of - - PowerPoint PPT Presentation
Quantitative Diagonalizability Part I: Three Measures of - - PowerPoint PPT Presentation
Quantitative Diagonalizability Part I: Three Measures of Nonnormality pseudospectrum || 1 || 1 pseudospectrum || 1
π βpseudospectrum
Ξπ π β π¨ β β βΆ || π¨ β π β1|| β₯ πβ1
π βpseudospectrum
Ξπ π β π¨ β β βΆ || π¨ β π β1|| β₯ πβ1 = {π¨ β β βΆ ππ π¨ β π β€ π } = {π¨ β β: π¨ β π‘πππ π΅ + πΉ , ||πΉ|| β€ π}
π βpseudospectrum
Ξπ π β π¨ β β βΆ || π¨ β π β1|| β₯ πβ1
For normal matrices, Ξπ π = Ξ0 π + πΈ 0, π
Pseudospectrum of Toeplitz Example
π βpseudospectrum
Ξπ π β π¨ β β βΆ || π¨ β π β1|| β₯ πβ1
e.g. discretization of pde from acoustics:
Ξπ π β π¨ β β βΆ || π¨ β π β1|| β₯ πβ1 = {π¨ β β βΆ ππ π¨ β π β€ π } = π¨ β β: π¨ β π‘πππ π΅ + πΉ , ||πΉ|| β€ π [Bauer-Fike]: Ξπ π β Ξ0 π + ππ π πΈ 0, π For distinct eigs Ξπ π = Ξ0 π +βͺπ πΈ(ππ, π ππ π) + π(π)
π βpseudospectrum
Part II: Daviesβ Conjecture
(with Jess Banks, Archit Kulkarni, Satyaki Mukherjee)
Diagonalization
π΅ β βπΓπ is diagonalizable if π΅ = ππΈπβ1 for invertible π, diagonal πΈ. Every matrix is a limit of diagonalizable matrices. Let ππ π΅ β ||π|| β ||πβ1|| be the eigenvector condition number of π΅. Question: Given a matrix π΅ and π > 0 , what is min{ππ π΅ + πΉ : ||πΉ|| β€ π}?
ππ = β ππ βͺ β
Diagonalization
π΅ β βπΓπ is diagonalizable if π΅ = ππΈπβ1 for invertible π, diagonal πΈ. Every matrix is a limit of diagonalizable matrices. Let ππ π΅ β ||π|| β ||πβ1|| be the eigenvector condition number of π΅. Question: Given a matrix π΅ and π > 0 , what is min{ππ π΅ + πΉ : ||πΉ|| β€ π}?
ππ = β ππ βͺ β
ππ π΅ = 1 for normal, β for nondiagonalizable
Diagonalization
π΅ β βπΓπ is diagonalizable if π΅ = ππΈπβ1 for invertible π, diagonal πΈ. Every matrix is a limit of diagonalizable matrices. Let ππ π΅ β ||π|| β ||πβ1|| be the eigenvector condition number of π΅. Question: Given a matrix π΅ and π > 0 , what is min{ππ π΅ + πΉ : ||πΉ|| β€ π}?
ππ = β ππ βͺ β
- Problem. Compute π(π΅) for analytic function π, e.g. π π¨ = ππ¨, π¨π.
NaΓ―ve Approach. π π΅ = ππ πΈ πβ1. Highly unstable if ππ(π΅) is big.
e.g. π Γ π Toeplitz:
Empirically: π΅ is close to a matrix with much better ππ β¦
Motivation: Computing Matrix Functions
ππ π΅ = 2πβ1 β 1030
ππ(π΅ + πΉ) β₯ πΉ β₯
π = 100 πΉ~Gaussian
- Problem. Compute π(π΅) for analytic function π, e.g. π π¨ = ππ¨, π¨π.
NaΓ―ve Approach. π π΅ = ππ πΈ πβ1. Highly unstable if ππ(π΅) is big.
e.g. π Γ π Toeplitz:
Empirically: π΅ is close to a matrix with much better ππ β¦
Motivation: Computing Matrix Functions
ππ π΅ = 2πβ1 β 1030
ππ(π΅ + πΉ) β₯ πΉ β₯
π = 100 πΉ~Gaussian
- Problem. Compute π(π΅) for analytic function π, e.g. π π¨ = ππ¨, π¨π.
NaΓ―ve Approach. π π΅ = ππ πΈ πβ1. Highly unstable if ππ(π΅) is big.
e.g. π Γ π Toeplitz, n=100:
Motivation: Computing Matrix Functions
ππ π΅ = 2πβ1 β 1030
ππ(π΅ + πΉ) β₯ πΉ β₯
π = 100 πΉ~Gaussian
- Problem. Compute π(π΅) for analytic function π, e.g. π π¨ = ππ¨, π¨π.
NaΓ―ve Approach. π π΅ = ππ πΈ πβ1. Highly unstable if ππ(π΅) is big.
e.g. π Γ π Toeplitz, n=100:
Empirically: π΅ is close to a matrix with much better ππ β¦
Motivation: Computing Matrix Functions
ππ π΅ = 2πβ1 β 1030
ππ(π΅ + πΉ) β₯ πΉ β₯
πΉ~Gaussian
experiment by M. Embree
- Problem. Compute π(π΅) for analytic function π, e.g. π π¨ = ππ¨, π¨π.
NaΓ―ve Approach. π π΅ = ππ πΈ πβ1. Highly unstable if ππ(π΅) is big.
e.g. π Γ π Toeplitz, n=100:
Empirically: π΅ is close to a matrix with much better ππ.
Motivation: Computing Matrix Functions
ππ π΅ = 2πβ1 β 1030
ππ(π΅ + πΉ) β₯ πΉ β₯
πΉ~Gaussian
- Idea. Approximate π(π΅) by π π΅ + πΉ for some small πΉ.
e.g.π π΅ = π΅ E = randn(n)*delta [V,D]=eig(A+E) S = V*D.^(1/2)*inv(V)
π π
- Idea. Approximate π(π΅) by π π΅ + πΉ for some small πΉ.
e.g.π π΅ = π΅ E = randn(n)*delta [V,D]=eig(A+E) S = V*D.^(1/2)*inv(V)
π π
- Idea. Approximate π(π΅) by π π΅ + πΉ for some small πΉ.
e.g.π π΅ = π΅ E = randn(n)*delta [V,D]=eig(A+E) S = V*D.^(1/2)*inv(V)
π π experiment by M. Embree
Approximate Diagonalization
- Theorem. [Daviesβ06] For every π΅ β βπΓπ with ||π΅|| β€ 1 and π β (0,1)
there is a perturbation πΉ such that ππ π΅ + πΉ β€ π· π π
πβ1
- Conjecture. For every π΅ β βπΓπ with ||π΅|| β€ 1 and π β (0,1) there is a
perturbation πΉ such that ππ π΅ + πΉ β€ π·π π [Daviesβ06]: true for π = 3 and for special case π΅ = πΎπ, with π·π = 2.
Approximate Diagonalization
- Theorem. [Daviesβ06] For every π΅ β βπΓπ with ||π΅|| β€ 1 and π β (0,1)
there is a perturbation πΉ such that ππ π΅ + πΉ β€ π· π π
πβ1
- Conjecture. For every π΅ β βπΓπ with ||π΅|| β€ 1 and π β (0,1) there is a
perturbation πΉ such that ππ π΅ + πΉ β€ π·π π [Daviesβ06]: true for π = 3 and for special case π΅ = πΎπ, with π·π = 2.
Approximate Diagonalization
- Theorem. [Daviesβ06] For every π΅ β βπΓπ with ||π΅|| β€ 1 and π β (0,1)
there is a perturbation πΉ such that ππ π΅ + πΉ β€ π· π π
πβ1
- Conjecture. For every π΅ β βπΓπ with ||π΅|| β€ 1 and π β (0,1) there is a
perturbation πΉ such that ππ π΅ + πΉ β€ π·π π [Daviesβ06]: true for π = 3 and for special case π΅ = πΎπ, with π·π = 2.
Main Result
Theorem A. For every π΅ β βπΓπ with ||π΅|| β€ 1 and π β (0,1) there is a perturbation πΉ such that ππ π΅ + πΉ β€ 4π3/2 π Implies every matrix has a 1/ππππ§(π) perturbation with ππ β€ ππππ§(π) Implied by a stronger probabilistic result on condition number of eigenvalues.
Main Result
Theorem A. For every π΅ β βπΓπ with ||π΅|| β€ 1 and π β (0,1) there is a perturbation πΉ such that ππ π΅ + πΉ β€ 4π3/2 π Implies every matrix has a 1/ππππ§(π) perturbation with ππ β€ ππππ§(π) Implied by a stronger probabilistic result on condition number of eigenvalues.
Main Result
Theorem A. For every π΅ β βπΓπ with ||π΅|| β€ 1 and π β (0,1) there is a perturbation πΉ such that ππ π΅ + πΉ β€ 4π3/2 π Implies every matrix has a 1/ππππ§(π) perturbation with ππ β€ ππππ§(π) Implied by a stronger probabilistic result on eigenvalue condition numbers.
Probabilistic Analysis of ππ
Theorem B. Assume ||π΅|| β€ 1 and let π» have i.i.d. complex standard Gaussian entries. Let π1, β¦ ππ be the eigenvalues of π΅ + πΏπ».
Probabilistic Analysis of ππ
Theorem B. Assume ||π΅|| β€ 1 and let π» have i.i.d. complex standard Gaussian entries. Let π1, β¦ ππ be the eigenvalues of π΅ + πΏπ».
π¨ = π¦ + ππ§ where π¦, π§~π 0,
1 2
Probabilistic Analysis of ππ
Theorem B. Assume ||π΅|| β€ 1 and let π» have i.i.d. complex standard Gaussian entries. Let π1, β¦ ππ be the eigenvalues of π΅ + πΏπ». Then for any open ball πΆ β β: π½ ΰ·
ππβπΆ
π2 ππ β€ π ππΏ2 β π€ππ(πΆ)
π1 π2 π3 π4
Probabilistic Analysis of ππ
Theorem B. Assume ||π΅|| β€ 1 and let π» have i.i.d. complex standard Gaussian entries. Let π1, β¦ ππ be the eigenvalues of π΅ + πΏπ». Then for any open ball πΆ β β: π½ ΰ·
ππβπΆ
π2 ππ β€ π ππΏ2 β π€ππ(πΆ)
- cf. Precise asymptotic results for π΅ = 0 [Chalker-Mehligβ98,β¦Bourgade-Dubachβ18,Fyodorovβ18]
and π΅ =Toeplitz [Davies-Hagerβ08,β¦Basak-Paquette-Zeitouniβ14-18, Sjostrand-Vogelβ18]
π1 π2 π3 π4
Probabilistic Analysis of ππ
Theorem B. Assume ||π΅|| β€ 1 and let π» have i.i.d. complex standard Gaussian entries. Let π1, β¦ ππ be the eigenvalues of π΅ + πΏπ». Then for any open ball πΆ β β: π½ ΰ·
ππβπΆ
π2 ππ β€ π ππΏ2 β π€ππ(πΆ)
- cf. Precise asymptotic results for π΅ = 0 [Chalker-Mehligβ98,β¦Bourgade-Dubachβ18,Fyodorovβ18]
and π΅ =Toeplitz [Davies-Hagerβ08,β¦Basak-Paquette-Zeitouniβ14-18, Sjostrand-Vogelβ18] Remark: Bourgade-Dubach implies that Theorem B is sharp for π΅ = 0
π1 π2 π3 π4
Implication B->A
Theorem B. Assume ||π΅|| β€ 1 and let π» have i.i.d. complex standard Gaussian entries. Let π1, β¦ ππ be the eigenvalues of π΅ + πΏπ». Then for any
- pen ball πΆ β β:
π½ ΰ·
ππβπΆ
π2 ππ β€ π ππΏ2 β π€ππ(πΆ) Proof of Theorem A. Let πΏ < 1/ π. whp ||π΅ + πΏπ»|| β€ 3 so all ππ β πΆ = πΈ(0,3). ππ π΅ + πΏπ» β€ π β ΰ·
ππβπΆ
π2 ππ β€ π π πΏ β€ π π3/2 π
π β πΏ π
π1 π2 π3 π4
Implication B->A
Theorem B. Assume ||π΅|| β€ 1 and let π» have i.i.d. complex standard Gaussian entries. Let π1, β¦ ππ be the eigenvalues of π΅ + πΏπ». Then for any
- pen ball πΆ β β:
π½ ΰ·
ππβπΆ
π2 ππ β€ π ππΏ2 β π€ππ(πΆ) Proof of Theorem A. Let πΏ < 1/ π. whp ||π΅ + πΏπ»|| β€ 3 so all ππ β πΆ = πΈ(0,3). ππ π΅ + πΏπ» β€ π β ΰ·
ππβπΆ
π2 ππ β€ π π πΏ β€ π π3/2 π
π β πΏ π
π1 π2 π3 π4
Implication B->A
Theorem B. Assume ||π΅|| β€ 1 and let π» have i.i.d. complex standard Gaussian entries. Let π1, β¦ ππ be the eigenvalues of π΅ + πΏπ». Then for any
- pen ball πΆ β β:
π½ ΰ·
ππβπΆ
π2 ππ β€ π ππΏ2 β π€ππ(πΆ) Proof of Theorem A. Let πΏ < 1/ π. whp ||π΅ + πΏπ»|| β€ 3 so all ππ β πΆ = πΈ(0,3). Then with constant prob. ππ π΅ + πΏπ» β€ π β ΰ·
ππβπΆ
π2 ππ β€ π π πΏ β€ π π3/2 π
π β πΏ π
Implication B->A
Theorem B. Assume ||π΅|| β€ 1 and let π» have i.i.d. complex standard Gaussian entries. Let π1, β¦ ππ be the eigenvalues of π΅ + πΏπ». Then for any
- pen ball πΆ β β:
π½ ΰ·
ππβπΆ
π2 ππ β€ π ππΏ2 β π€ππ(πΆ) Proof of Theorem A. Let πΏ < 1/ π. whp ||π΅ + πΏπ»|| β€ 3 so all ππ β πΆ = πΈ(0,3). Then with constant prob. ππ π΅ + πΏπ» β€ π β ΰ·
ππβπΆ
π2 ππ β€ π π πΏ β€ π π3/2 π
π β πΏ π
Proof of Theorem B
- 1. Area of the pseudospectrum
Lemma 1: If π has distinct eigenvalues then for every open πΆ: π ΰ·
ππβπΆ
π ππ 2 = lim
πβ0
π€ππ Ξπ π β© πΆ π2
<board>
- 1. Area of the pseudospectrum
Lemma 1: If π has distinct eigenvalues then for every open πΆ: π ΰ·
ππβπΆ
π ππ 2 = lim
πβ0 inf π€ππ Ξπ π β© πΆ
π2
<board>
- 2. Real Anticoncentration
Theorem[Sankar-Spielman-Tengβ06]: For any real π Γ π matrix π, and G with i.i.d. real π(0,1) entries: β ππ π + πΏπ» β€ π β€ π· ππ/πΏ Proof Idea: Let π + πΏπ» have columns ππ + πΏππ, Let π = π‘πππ ππ + πΏππ π>2 β πππ‘π’(π1 + πΏπ1, π) β€ π = β π1 + πΏπ1, π₯ β€ π = β π1, π₯ β πΏπ β€ π β€ π/πΏ π1 + πΏπ1 π₯
Orthogonal invariance anticoncentration
<bo
<board>
- 2. Real Anticoncentration
Theorem[Sankar-Spielman-Tengβ06]: For any real π Γ π matrix π, and G with i.i.d. real π(0,1) entries: β ππ π + πΏπ» β€ π β€ π· ππ/πΏ Proof Idea: Let π + πΏπ» have columns ππ + πΏππ, Let π = π‘πππ ππ + πΏππ π>2 β πππ‘π’(π1 + πΏπ1, π) β€ π = β π1 + πΏπ1, π₯ β€ π = β π1, π₯ β πΏπ β€ π β€ π/πΏ π1 + πΏπ1
Orthogonal invariance anticoncentration
π₯
2β. Complex Anticoncentration
Lemma 2. For any complex π Γ π matrix π, and G with i.i.d. complex π(0,1β) entries: β ππ π + πΏπ» β€ π β€ ππ2/πΏ2 Proof Idea: Let π + πΏπ» have columns ππ + πΏππ, Let π = π‘πππ ππ + πΏππ π>2 β πππ‘π’(π1 + πΏπ1, π) β€ π = β π1 + πΏπ1, π₯ β€ π = β π1, π₯ β πΏπ β€ π β€ π2/πΏ2 π1 + πΏπ1 π₯
Unitary invariance anticoncentration
2β. Complex Anticoncentration
Lemma 2. For any complex π Γ π matrix π, and G with i.i.d. complex π(0,1β) entries: β ππ π + πΏπ» β€ π β€ ππ2/πΏ2 Proof Idea: Let π + πΏπ» have columns ππ + πΏππ, Let π = π‘πππ ππ + πΏππ π>2 β πππ‘π’(π1 + πΏπ1, π) β€ π = β π1 + πΏπ1, π₯ β€ π = β π1, π₯ β πΏπ β€ π β€ π2/πΏ2 π1 + πΏπ1 π₯
Unitary invariance anticoncentration
- Cf. [Edelmanβ88] M=0
- 3. Expected Area of the Pseudospectrum
Lemma 2. For any complex π Γ π matrix π, complex Gaussian π»: β ππ π + πΏπ» β€ π β€ ππ2/πΏ2 Applied to π = π¨ β π΅ β πΏπ» says : β π¨ β Ξπ π΅ + πΏπ» = β[ππ π¨ β π΅ β πΏπ» β€ π] β€ ππ2/πΏ2 So for every fixed ball πΆ, for every π > 0: π½π€ππ Ξπ π΅ + πΏπ» β© πΆ = ΰΆ±
πΆ
β π¨ β Ξπ π΅ + πΏπ» ππ¨ β€ ππ2 πΏ2 β π€ππ(πΆ)
- 3. Expected Area of the Pseudospectrum
Lemma 2. For any complex π Γ π matrix π, complex Gaussian π»: β ππ π + πΏπ» β€ π β€ ππ2/πΏ2 Lemma 3. For every fixed ball πΆ, for every π > 0: π½π€ππ Ξπ π΅ + πΏπ» β© πΆ β€ ππ2 πΏ2 β π€ππ(πΆ)
<board>
- 4. Expected Limiting Area of the Pseudospectrum
Define the function π
π π» β π€ππ(Ξπ π΅ + πΏπ» β© πΆ)/π2
Lemma 2 shows that lim inf
πβ0
π½π
π π» β€ π/πΏ2
By Fatouβs lemma, π½ lim inf
πβ0
π
π π» β€ π/πΏ2
So by Lemma 1: π½ π ΰ·
ππβπΆ
π2 ππ = π½ lim inf
πβ 0 π π(π») β€ π
πΏ2 β π€ππ(πΆ)
- 4. Expected Limiting Area of the Pseudospectrum
Define the function π
π π» β π€ππ(Ξπ π΅ + πΏπ» β© πΆ)/π2
Lemma 2 shows that lim inf
πβ0
π½π
π π» β€ π/πΏ2
By Fatouβs lemma, π½ lim inf
πβ0
π
π π» β€ π/πΏ2
So by Lemma 1: π½ π ΰ·
ππβπΆ
π2 ππ = π½ lim inf
πβ 0 π π(π») β€ π
πΏ2 β π€ππ(πΆ)
- 4. Expected Limiting Area of the Pseudospectrum
Define the function π
π π» β π€ππ(Ξπ π΅ + πΏπ» β© πΆ)/π2
Lemma 3 shows that lim inf
πβ0
π½π
π π» β€ π/πΏ2
By Fatouβs lemma, π½ lim inf
πβ0
π
π π» β€ π/πΏ2
So by Lemma 1: π½ π ΰ·
ππβπΆ
π2 ππ = π½ lim inf
πβ 0 π π(π») β€ π
πΏ2 β π€ππ(πΆ)
- 4. Expected Limiting Area of the Pseudospectrum
Define the function π
π π» β π€ππ(Ξπ π΅ + πΏπ» β© πΆ)/π2
Lemma 3 shows that lim inf
πβ0
π½π
π π» β€ π/πΏ2
By Fatouβs lemma, π½ lim inf
πβ0
π
π π» β€ π/πΏ2
So by Lemma 1: π½ π ΰ·
ππβπΆ
π2 ππ = π½ lim inf
πβ 0 π π(π») β€ π
πΏ2 β π€ππ(πΆ)
Recap of the Proof
Let π = π΅ + πΏπ» and πΆ = πΈ 0,3 . π½ ΰ·
ππβπΆ
π ππ 2 = 1 π β π½ lim inf
πβ0
π€ππ Ξπ π β© πΆ π2 β€
1 π β lim inf πβ0
π½
π€ππ Ξπ π β©πΆ π2
β€
9max
π¨βπΆ β π¨βΞπ π
π2
β€ 9π/πΏ2
Phenomenon behind the result
Summary and Questions
Three related notions of spectral stability (ππ, π(ππ), Ξπ) Can control global quantities by local singular values ππ(π¨ β π) Exploited invariance and anticoncentration of complex Gaussian
Summary and Questions
Three related notions of spectral stability (ππ, π(ππ), Ξπ) Can control global quantities by local singular values ππ(π¨ β π) Exploited invariance and anticoncentration of complex Gaussian
- Does a real Gaussian fail?
- Dimension dependence in Theorem A. Dimension free bound?
- Derandomization of the perturbation
- Non-gaussian perturbations