Interlacing Families II: Mixed Characteristic Polynomials and the - - PowerPoint PPT Presentation

Interlacing Families II: Mixed Characteristic Polynomials and the Kadison-Singer Problem

Nikhil Srivastava (Microsoft Research) Adam Marcus (Yale), Daniel Spielman (Yale)

Discrepancy Theory

“How well can you approximate a discrete

• bject by a continuous one.”

e.g. Given a set family 𝑇1, … , 𝑇𝑛 ⊂ [𝑜], find a red-blue coloring 𝑜 = 𝑆 ∪ 𝐶 such that every set is half red and half blue.

Discrepancy Theory

“How well can you approximate a discrete

• bject by a continuous one.”

e.g. Given a set family 𝑇1, … , 𝑇𝑛 ⊂ [𝑜], find a red-blue coloring 𝑜 = 𝑆 ∪ 𝐶 such that every set is half red and half blue.

Discrepancy Theory

“How well can you approximate a discrete

• bject by a continuous one.”

e.g. Given a set family 𝑇1, … , 𝑇𝑛 ⊂ [𝑜], find a red-blue coloring 𝑜 = 𝑆 ∪ 𝐶 such that every set is half red and half blue.

Discrepancy Theory

“How well can you approximate a discrete

• bject by a continuous one.”

e.g. Given a set family 𝑇1, … , 𝑇𝑛 ⊂ [𝑜], find a red-blue coloring 𝑜 = 𝑆 ∪ 𝐶 such that every set is half red and half blue.

Discrepancy Theory

“How well can you approximate a discrete

• bject by a continuous one.”

e.g. Given a set family 𝑇1, … , 𝑇𝑛 ⊂ [𝑜], find a red-blue coloring 𝑜 = 𝑆 ∪ 𝐶 such that every set is half red and half blue.

Discrepancy Theory

“How well can you approximate a discrete

• bject by a continuous one.”

e.g. Given a set family 𝑇1, … , 𝑇𝑛 ⊂ [𝑜], find a red-blue coloring 𝑜 = 𝑆 ∪ 𝐶 such that every set is half red and half blue.

𝑒𝑗𝑡𝑑 = 1

Discrepancy Theory

“How well can you approximate a discrete

• bject by a continuous one.”

e.g. Given a set family 𝑇1, … , 𝑇𝑛 ⊂ [𝑜], find a red-blue coloring 𝑜 = 𝑆 ∪ 𝐶 such that every set is half red and half blue. How well can you do in general?

Discrepancy Theory

“How well can you approximate a discrete

• bject by a continuous one.”

e.g. Given a set family 𝑇1, … , 𝑇𝑛 ⊂ [𝑜], find a red-blue coloring 𝑜 = 𝑆 ∪ 𝐶 such that every set is half red and half blue. In general, a random coloring gives 𝑒𝑗𝑡𝑑: = max

𝑗

|𝑇𝑗 ∩ 𝑆| − |𝑇𝑗| 2 ≤ 𝑃 𝑜𝑚𝑝𝑕 𝑛 .

Discrepancy Theory

“How well can you approximate a discrete

• bject by a continuous one.”

e.g. Given a set family 𝑇1, … , 𝑇𝑛 ⊂ [𝑜], find a red-blue coloring 𝑜 = 𝑆 ∪ 𝐶 such that every set is half red and half blue. Spencer: There exists a coloring with 𝑒𝑗𝑡𝑑: = max

𝑗

|𝑇𝑗 ∩ 𝑆| − |𝑇𝑗| 2 ≤ 3 nlog(𝑛 𝑜 ).

Discrepancy Theory

“How well can you approximate a discrete

• bject by a continuous one.”

e.g. Given a set family 𝑇1, … , 𝑇𝑜 ⊂ [𝑜], find a red-blue coloring 𝑜 = 𝑆 ∪ 𝐶 such that every set is half red and half blue. Spencer: There exists a coloring with 𝑒𝑗𝑡𝑑: = max

𝑗

|𝑇𝑗 ∩ 𝑆| − |𝑇𝑗| 2 ≤ 3 𝑜.

A Spectral Discrepancy Theorem

Quadratic Forms and Energy

Given vectors 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜, their energy in a test direction 𝑣 ∈ 𝑺𝑜, ||𝑣|| = 1, is the quadratic form 𝑅 𝑣 ≔

𝑗

𝑣, 𝑤𝑗 2

Quadratic Forms and Energy

Given vectors 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜, their energy in a test direction 𝑣 ∈ 𝑺𝑜, ||𝑣|| = 1, is the quadratic form 𝑅 𝑣 ≔

𝑗

𝑣, 𝑤𝑗 2 Example.

𝑤1, 𝑤2, 𝑤3, 𝑤4 ∈ 𝑺2

Quadratic Forms and Energy

Given vectors 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜, their energy in a test direction 𝑣 ∈ 𝑺𝑜, ||𝑣|| = 1, is the quadratic form 𝑅 𝑣 ≔

𝑗

𝑣, 𝑤𝑗 2 Example.

𝑣 = (1,0)

𝑣, 𝑤1 2 + 𝑣, 𝑤2 2 + 𝑣, 𝑤3 2 + 𝑣, 𝑤4 2 =1 + 0 + ¼ + ¼ =1.5

Given vectors 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜, their energy in a test direction 𝑣 ∈ 𝑺𝑜, ||𝑣|| = 1, is the quadratic form 𝑅 𝑣 ≔

𝑗

𝑣, 𝑤𝑗 2 Example.

Quadratic Forms and Energy

𝑣 = (1,0)

Quadratic Forms and Energy

Given vectors 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜, their energy in a test direction 𝑣 ∈ 𝑺𝑜, ||𝑣|| = 1, is the quadratic form 𝑅 𝑣 ≔

𝑗

𝑣, 𝑤𝑗 2 Example.

𝑣 = (0,1)

𝑣, 𝑤1 2 + 𝑣, 𝑤2 2 + 𝑣, 𝑤3 2 + 𝑣, 𝑤4 2 =0 + 1 + ¾ + ¾ =2.5

Quadratic Forms and Energy

Given vectors 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜, their energy in a test direction 𝑣 ∈ 𝑺𝑜, ||𝑣|| = 1, is the quadratic form 𝑅 𝑣 ≔

𝑗

𝑣, 𝑤𝑗 2 Example.

𝑣 = (1/ 2, 1/ 2)

𝑣, 𝑤1 2 + 𝑣, 𝑤2 2 + 𝑣, 𝑤3 2 + 𝑣, 𝑤4 2 ½ + ½ + ¾ + ¼ =2

Quadratic Forms and Energy

Given vectors 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜, their energy in a test direction 𝑣 ∈ 𝑺𝑜, ||𝑣|| = 1, is the quadratic form 𝑅 𝑣 ≔

𝑗

𝑣, 𝑤𝑗 2 Example.

Splitting A Quadratic Form in Half

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1

“isotropic”

Splitting A Quadratic Form in Half

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1

“isotropic”

Splitting A Quadratic Form in Half

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1

“isotropic”

Splitting A Quadratic Form in Half

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1

“isotropic”

Splitting A Quadratic Form in Half

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1

Splitting A Quadratic Form in Half

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1 Then there is a partition 𝑈

1 ∪ 𝑈2 such that each

part has energy close to half in each direction: ∀||𝑣|| = 1

𝑗∈𝑈𝑘

𝑣, 𝑤𝑗 2 = 1 2 ± 5𝜗

Splitting A Quadratic Form in Half

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1 Then there is a partition 𝑈

1 ∪ 𝑈2 such that each

part has energy close to half in each direction: ∀||𝑣|| = 1

𝑗∈𝑈𝑘

𝑣, 𝑤𝑗 2 = 1 2 ± 5𝜗

∪

Splitting A Quadratic Form in Half

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1 Then there is a partition 𝑈

1 ∪ 𝑈2 such that each

part has energy close to half in each direction: ∀||𝑣|| = 1

𝑗∈𝑈𝑘

𝑣, 𝑤𝑗 2 = 1 2 ± 5𝜗

∪

“Each part approximates the whole.”

Many Possible Partitions

Many Possible Partitions

Many Possible Partitions

Many Possible Partitions

Theorem: Good partition always exists.

The norm condition

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1 Then there is a partition 𝑈

1 ∪ 𝑈2 such that each

part has energy close to half in each direction: ∀||𝑣|| = 1

𝑗∈𝑈𝑘

𝑣, 𝑤𝑗 2 = 1 2 ± 5𝜗

The norm condition

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1 Then there is a partition 𝑈

1 ∪ 𝑈2 such that each

part has energy close to half in each direction: ∀||𝑣|| = 1

𝑗∈𝑈𝑘

𝑣, 𝑤𝑗 2 = 1 2 ± 5𝜗

In One Dimension

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺1 are numbers with |𝑤𝑗| ≤ 𝜗 and energy one

𝑗

𝑤𝑗

2 = 1.

Then there is a partition 𝑈

1 ∪ 𝑈2 such that

each part has energy close to half

𝑗∈𝑈𝑘

𝑤𝑗

2 = 1

2 ± 5𝜗

In One Dimension

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺1 are numbers with |𝑤𝑗|2 ≤ 𝜗2 and energy one

𝑗

𝑤𝑗

2 = 1.

Then there is a partition 𝑈

1 ∪ 𝑈2 such that

each part has energy close to half

𝑗∈𝑈𝑘

𝑤𝑗

2 = 1

2 ± 5𝜗

In One Dimension

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺1 are numbers with |𝑤𝑗|2 ≤ 𝜗2 and energy one

𝑗

𝑤𝑗

2 = 1.

Then there is a partition 𝑈

1 ∪ 𝑈2 such that

each part has energy close to half

𝑗∈𝑈𝑘

𝑤𝑗

2 = 1

2 ± 𝜗2 2

greedy bin packing

In One Dimension

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺1 are numbers with |𝑤𝑗|2 ≤ 𝜗2 and energy one

𝑗

𝑤𝑗

2 = 1.

Then there is a partition 𝑈

1 ∪ 𝑈2 such that

each part has energy close to half

𝑗∈𝑈𝑘

𝑤𝑗

2 = 1

2 ± 5𝜗

In Higher Dimensions

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1 Then there is a partition 𝑈

1 ∪ 𝑈2 such that each

part has energy close to half in each direction: ∀||𝑣|| = 1

𝑗∈𝑈𝑘

𝑣, 𝑤𝑗 2 = 1 2 ± 5𝜗

In Higher Dimensions

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and energy one in each direction: ∀||𝑣|| = 1

𝑗

𝑣, 𝑤𝑗 2 = 1 Then there is a partition 𝑈

1 ∪ 𝑈2 such that each

part has energy close to half in each direction: ∀||𝑣|| = 1

𝑗∈𝑈𝑘

𝑣, 𝑤𝑗 2 = 1 2 ± 5𝜗

Optimal in high dim

Matrix Notation

Given vectors 𝑤1, … , 𝑤𝑛 write quadratic form as 𝑅 𝑣 =

𝑗

𝑤𝑗, 𝑣 2 = 𝑣𝑈

𝑗

𝑤𝑗𝑤𝑗

𝑈

𝑣

Matrix Notation

Given vectors 𝑤1, … , 𝑤𝑛 write quadratic form as 𝑅 𝑣 =

𝑗

𝑤𝑗, 𝑣 2 = 𝑣𝑈

𝑗

𝑤𝑗𝑤𝑗

𝑈

𝑣 Isotropy:

𝑗

𝑤𝑗𝑤𝑗

𝑈 = 𝐽𝑜

Matrix Notation

Given vectors 𝑤1, … , 𝑤𝑛 write quadratic form as 𝑅 𝑣 =

𝑗

𝑤𝑗, 𝑣 2 = 𝑣𝑈

𝑗

𝑤𝑗𝑤𝑗

𝑈

𝑣 Isotropy: Comparision: 𝐵 ≼ 𝐶 ⟺ 𝑣𝑈𝐵𝑣 ≤ 𝑣𝑈𝐶 𝑣 ∀𝑣

𝑗

𝑤𝑗𝑤𝑗

𝑈 = 𝐽𝑜

Matrix Notation

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗|| ≤ 𝜗 and Then there is a partition 𝑈

1 ∪ 𝑈2 such that

1 2 − 5𝜗 𝐽 ≼

𝑗∈𝑈𝑘

𝑤𝑗𝑤𝑗

𝑈 ≼

1 2 + 5𝜗 𝐽

𝑗

𝑤𝑗𝑤𝑗

𝑈 = 𝐽𝑜

Matrix Notation

Main Theorem. Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗||2 ≤ 𝜗 and Then there is a partition 𝑈

1 ∪ 𝑈2 such that

1 2 − 5 𝜗 𝐽 ≼

𝑗∈𝑈𝑘

𝑤𝑗𝑤𝑗

𝑈 ≼

1 2 + 5 𝜗 𝐽

𝑗

𝑤𝑗𝑤𝑗

𝑈 = 𝐽𝑜

Unnormalized Version

Suppose I get some vectors 𝑥1, … , 𝑥𝑛which are not isotropic:

𝑗

𝑥𝑗𝑥𝑗

𝑈 = 𝑋 ≽ 0

Consider 𝑤𝑗 ≔ 𝑋−1

2𝑥𝑗 and apply theorem to 𝑤𝑗.

Normalized vectors have ||𝑤𝑗||2 = ||𝑋−1

2𝑥𝑗||2 = 𝜗

1 2 − 5 𝜗 𝐽 ≼

𝑗∈𝑈𝑘

𝑤𝑗𝑤𝑗

𝑈 ≼

1 2 + 5 𝜗 𝐽

• Thm. gives

Unnormalized Version

Suppose I get some vectors 𝑥1, … , 𝑥𝑛which are not isotropic:

𝑗

𝑥𝑗𝑥𝑗

𝑈 = 𝑋 ≽ 0

Consider 𝑤𝑗 ≔ 𝑋−1

2𝑥𝑗 and apply theorem to 𝑤𝑗.

Normalized vectors have ||𝑤𝑗||2 = ||𝑋−1

2𝑥𝑗||2 = 𝜗

1 2 − 5 𝜗 𝐽 ≼

𝑗∈𝑈𝑘

𝑋−1

2𝑥𝑗𝑥𝑗 𝑈𝑋−1 2 ≼

1 2 + 5 𝜗 𝐽

Unnormalized Version

Suppose I get some vectors 𝑥1, … , 𝑥𝑛which are not isotropic:

𝑗

𝑥𝑗𝑥𝑗

𝑈 = 𝑋 ≽ 0

Consider 𝑤𝑗 ≔ 𝑋−1

2𝑥𝑗 and apply theorem to 𝑤𝑗.

Normalized vectors have ||𝑤𝑗||2 = ||𝑋−1

2𝑥𝑗||2 = 𝜗

1 2 − 5 𝜗 𝐽 ≼ 𝑋−1

2 𝑗∈𝑈𝑘

𝑥𝑗𝑥𝑗

𝑈

𝑋−1

2 ≼

1 2 + 5 𝜗 𝐽

Unnormalized Version

Suppose I get some vectors 𝑥1, … , 𝑥𝑛which are not isotropic:

𝑗

𝑥𝑗𝑥𝑗

𝑈 = 𝑁 ≽ 0

Consider 𝑤𝑗 ≔ 𝑋−1

2𝑥𝑗 and apply theorem to 𝑤𝑗.

Normalized vectors have ||𝑤𝑗||2 = ||𝑋−1

2𝑥𝑗||2 = 𝜗

1 2 − 5 𝜗 𝐽 ≼ 𝑋−1

2 𝑗∈𝑈𝑘

𝑥𝑗𝑥𝑗

𝑈

𝑋−1

2 ≼

1 2 + 5 𝜗 𝐽

Fact: 𝐵 ≼ 𝐶 ⟺ 𝐷𝐵𝐷 ≼ 𝐷𝐶𝐷 for invertible 𝐷

Unnormalized Version

Suppose I get some vectors 𝑥1, … , 𝑥𝑛which are not isotropic:

𝑗

𝑥𝑗𝑥𝑗

𝑈 = 𝑋 ≽ 0

Consider 𝑤𝑗 ≔ 𝑋−1

2𝑥𝑗 and apply theorem to 𝑤𝑗.

Normalized vectors have ||𝑤𝑗||2 = ||𝑋−1

2𝑥𝑗||2 = 𝜗

1 2 − 5 𝜗 𝑋 ≼

𝑗∈𝑈𝑘

𝑥𝑗𝑥𝑗

𝑈

≼ 1 2 + 5 𝜗 𝑋

Unnormalized Theorem

Given arbitrary vectors 𝑥1, … , 𝑥𝑛 ∈ ℝ𝑜 there is a partition 𝑛 = 𝑈

1 ∪ 𝑈2 with

1 2 − 𝜗

𝑗

𝑥𝑗𝑥𝑗

𝑈

≼

𝑗∈𝑈𝑘

𝑥𝑗𝑥𝑗

𝑈 ≼ 1

2 − 𝜗

𝑗

𝑥𝑗𝑥𝑗

𝑈

Where 𝜗 ≔ max

𝑗

|| 𝑋−1

2𝑥𝑗||2

Unnormalized Theorem

Given arbitrary vectors 𝑥1, … , 𝑥𝑛 ∈ ℝ𝑜 there is a partition 𝑛 = 𝑈

1 ∪ 𝑈2 with

1 2 − 𝜗

𝑗

𝑥𝑗𝑥𝑗

𝑈

≼

𝑗∈𝑈𝑘

𝑥𝑗𝑥𝑗

𝑈 ≼ 1

2 − 𝜗

𝑗

𝑥𝑗𝑥𝑗

𝑈

Where 𝜗 ≔ max

𝑗

|| 𝑋−1

2𝑥𝑗||2

Any quadratic form in which no vector has too much influence can be split into two representative pieces.

Applications

• 1. Graph Theory

Given an undirected graph 𝐻 = (𝑊, 𝐹), consider its Laplacian matrix: 𝑀𝐻 =

𝑗𝑘∈𝐹

𝜀𝑗 − 𝜀

𝑘)(𝜀𝑗 − 𝜀 𝑘 𝑈

• 1. Graph Theory

Given an undirected graph 𝐻 = (𝑊, 𝐹), consider its Laplacian matrix: 𝑀𝐻 =

𝑗𝑘∈𝐹

𝜀𝑗 − 𝜀

𝑘)(𝜀𝑗 − 𝜀 𝑘 𝑈

i j i j

• 1. Graph Theory

Given an undirected graph 𝐻 = (𝑊, 𝐹), consider its Laplacian matrix: 𝑀𝐻 =

𝑗𝑘∈𝐹

𝜀𝑗 − 𝜀

𝑘)(𝜀𝑗 − 𝜀 𝑘 𝑈 = 𝐸 − 𝐵

• 1. Graph Theory

Given an undirected graph 𝐻 = (𝑊, 𝐹), consider its Laplacian matrix: 𝑀𝐻 =

𝑗𝑘∈𝐹

𝜀𝑗 − 𝜀

𝑘)(𝜀𝑗 − 𝜀 𝑘 𝑈 = 𝐸 − 𝐵

Quadratic form: 𝑦𝑈𝑀𝑦 =

𝑗𝑘∈𝐹

𝑦𝑗 − 𝑦𝑘

2 𝑔𝑝𝑠 𝑦 ∈ 𝑺𝑜

The Laplacian Quadratic Form

An example:

• 1

+2 +1 +1 +3

An example: xTLx=  i,j 2 E (x(i)-x(j))2

• 1

+2 +1 +1 +3 12 32 22 22 32

The Laplacian Quadratic Form

12

An example: xTLx=  i,j 2 E (x(i)-x(j))2 = 28

• 1

+2 +1 +1 +3 12 32 22 12 22 32

The Laplacian Quadratic Form

The Laplacian Quadratic Form

Another example:

+1 +1

The Laplacian Quadratic Form

Another example: xT LGx = 3

+1 +1 02 02 12 12 12 02

Cuts and the Quadratic Form

For characteristic vector

Cuts and the Quadratic Form

For characteristic vector

The Laplacian Quadratic form encodes the entire cut structure of the graph.

Application to Graphs

G

Application to Graphs

G

𝑀𝐻 =

𝑗𝑘∈𝐹

𝜀𝑗 − 𝜀

𝑘)(𝜀𝑗 − 𝜀 𝑘 𝑈

Application to Graphs

G

Theorem

Application to Graphs

G

Theorem

H1 H2

𝑀𝐼1 𝑀𝐼2

Application to Graphs

G

Theorem

H1 H2

𝑦𝑈𝑀𝐼1𝑦 ≈ 1 2 𝑦𝑈𝑀𝐻𝑦

Recursive Application Gives:

• 1. Graph Sparsification Theorem [Batson-

Spielman-S’09]: Every graph G has a weighted O(1)-cut approximation H with O(n) edges. G H 𝑃 𝑜2 edges 𝑃 𝑜 edges

Unweighted

Weighted

Approximating One Graph by Another

Cut Approximation [Benczur-Karger’96] For every cut, weight of edges in G ≈ weight of edges in H

G H

Approximating One Graph by Another

Cut Approximation [Benczur-Karger’96] For every cut, weight of edges in G ≈ weight of edges in H

G H

Approximating One Graph by Another

Cut Approximation [Benczur-Karger’96] For every cut, weight of edges in G ≈ weight of edges in H

G H

Approximating One Graph by Another

Cut Approximation [Benczur-Karger’96] For every cut, weight of edges in G ≈ weight of edges in H

G H

Approximating One Graph by Another

Cut Approximation [Benczur-Karger’96] For every cut, weight of edges in G ≈ weight of edges in H

G H

Approximating One Graph by Another

Cut Approximation [Benczur-Karger’96] For every cut, weight of edges in G ≈ weight of edges in H

G H

G and H have same cuts. Equivalent for min cut, max cut, sparsest cut…

Recursive Application Gives:

• 2. Unweighted Graph Sparsification Every

transitive graph G has an unweighted O(1)-cut approximation H with O(n) edges.

𝐿𝑜 𝐼 Expander graph

Recursive Application Gives:

• 2. Unweighted Graph Sparsification Every

transitive graph G can be partitioned into O(1)-cut approximations with O(n) edges.

𝐿𝑜 𝐼1 … 𝐼𝑜 Expander graphs

Recursive Application Gives:

• 2. Unweighted Graph Sparsification Every

transitive graph G can be partitioned into O(1)-cut approximations with O(n) edges.

𝐿𝑜 𝐼1 … 𝐼𝑜 Expander graphs

Generalizes [Frieze-Molloy]

Recursive Application Gives:

• 2. Unweighted Graph Sparsification Every

transitive graph G can be partitioned into O(1)-cut approximations with O(n) edges.

Same cut structure

G H1 H2

• 2. Uncertainty Principles

Signal 𝑦 ∈ ℂ𝑜. Discrete Fourier Transform 𝑦 a = 〈𝑦, exp(−

𝑏2𝜌𝑗𝑙 𝑜

)

𝑙≤𝑜〉

• 2. Uncertainty Principles

Signal 𝑦 ∈ ℂ𝑜. Discrete Fourier Transform 𝑦 a = 〈𝑦, exp(−

𝑏2𝜌𝑗𝑙 𝑜

)

𝑙≤𝑜〉

Uncertainty Principle: 𝑦 and 𝑦 cannot be simultaneously localized. 𝑡𝑣𝑞𝑞 𝑦 × 𝑡𝑣𝑞𝑞 𝑦 ≥ 𝑜

• 2. Uncertainty Principles

Signal 𝑦 ∈ ℂ𝑜. Discrete Fourier Transform 𝑦 a = 〈𝑦, exp(−

𝑏2𝜌𝑗𝑙 𝑜

)

𝑙≤𝑜〉

Uncertainty Principle: 𝑦 and 𝑦 cannot be simultaneously localized. 𝑡𝑣𝑞𝑞 𝑦 × 𝑡𝑣𝑞𝑞 𝑦 ≥ 𝑜 If 𝑦 is supported on 𝑇 = √𝑜 coordinates, 𝑡𝑣𝑞𝑞 𝑦 ≥ 𝑜

• 2. Uncertainty Principles

Signal 𝑦 ∈ ℂ𝑜. Discrete Fourier Transform 𝑦 a = 〈𝑦, exp(−

𝑏2𝜌𝑗𝑙 𝑜

)

𝑙≤𝑜〉

Stronger Uncertainty Principle: For every subset 𝑇 = 𝑜 , there is a partition 𝑜 = 𝑈

1 ∪ ⋯ 𝑈 𝑜

||𝑦|𝑇||2 ≈

1 𝑜 ||

𝑦|𝑈𝑗||2 for all 𝑦 and 𝑈𝑗

• 2. Uncertainty Principles

Signal 𝑦 ∈ ℂ𝑜. Discrete Fourier Transform 𝑦 a = 〈𝑦, exp(−

𝑏2𝜌𝑗𝑙 𝑜

)

𝑙≤𝑜〉

Stronger Uncertainty Principle: For every subset 𝑇 = 𝑜 , there is a partition 𝑜 = 𝑈

1 ∪ ⋯ 𝑈 𝑜

||𝑦|𝑇||2 ≈

1 𝑜 ||

𝑦|𝑈𝑗||2 for all 𝑦 and 𝑈𝑗

• 2. Uncertainty Principles

Proof. Let 𝑔

𝑙 = exp(− 𝑏2𝜌𝑗𝑙 𝑜

)

𝑙≤𝑜 be the Fourier basis.

Fix a subset 𝑇 ⊂ 𝑜 of √𝑜 coords. The restricted norm is: ||𝑦|𝑇||2 = 𝑙 𝑦|𝑇, 𝑔

𝑙 2

a quadratic form in 𝑜 dimensions. Apply the theorem.

• 2. Uncertainty Principles

Proof. Let 𝑔

𝑙 = exp(− 𝑏2𝜌𝑗𝑙 𝑜

)

𝑙≤𝑜 be the Fourier basis.

Fix a subset 𝑇 ⊂ 𝑜 of √𝑜 coords. The restricted norm is: ||𝑦|𝑇||2 = 𝑙 𝑦|𝑇, 𝑔

𝑙 2

a quadratic form in 𝑜 dimensions. Apply the theorem.

Applications in analytic number theory, harmonic analysis.

• 3. The Kadison-Singer Problem

Dirac 1930’s: Bra-Ket formalism for quantum states. 𝜔 , 𝑞 …

• 3. The Kadison-Singer Problem

Dirac 1930’s: Bra-Ket formalism for quantum states. 𝜔 , 𝑞 … What are Bras and Kets? NOT vectors.

• 3. The Kadison-Singer Problem

Dirac 1930’s: Bra-Ket formalism for quantum states. 𝜔 , 𝑞 … What are Bras and Kets? NOT vectors. Von Neumann 1936: Theory of 𝐷∗algebras.

• 3. The Kadison-Singer Problem

Dirac 1930’s: Bra-Ket formalism for quantum states. 𝜔 , 𝑞 … What are Bras and Kets? NOT vectors. Von Neumann 1936: Theory of 𝐷∗algebras. Kadison-Singer 1959: Does this lead to a satisfactory notion of measurement? Conjecture: about ∞ matrices

• 3. The Kadison-Singer Problem

Kadison-Singer 1959: Does this lead to a satisfactory notion of measurement? Conjecture: about ∞ matrices

• 3. The Kadison-Singer Problem

Kadison-Singer 1959: Does this lead to a satisfactory notion of measurement? Conjecture: about ∞ matrices Anderson 1979: Reduced to a question about finite matrices. “Paving Conjecture”

• 3. The Kadison-Singer Problem

Kadison-Singer 1959: Does this lead to a satisfactory notion of measurement? Conjecture: about ∞ matrices Anderson 1979: Reduced to a question about finite matrices. Akemann-Anderson 1991: Reduced to a question about finite projection matrices.

• 3. The Kadison-Singer Problem

Kadison-Singer 1959: Does this lead to a satisfactory notion of measurement? Conjecture: about ∞ matrices Anderson 1979: Reduced to a question about finite matrices. Akemann-Anderson 1991: Reduced to a question about finite projection matrices. Weaver 2002: Discrepancy theoretic formulation

• f the same question.

• 3. The Kadison-Singer Problem

Kadison-Singer 1959: Does this lead to a satisfactory notion of measurement? Conjecture: about ∞ matrices Anderson 1979: Reduced to a question about finite matrices. Akemann-Anderson 1991: Reduced to a question about finite projection matrices. This work: Proof of Weaver’s conjecture.

• 3. The Kadison-Singer Problem

Kadison-Singer 1959: Does this lead to a satisfactory notion of measurement? Conjecture: about ∞ matrices Anderson 1979: Reduced to a question about finite matrices. Akemann-Anderson 1991: Reduced to a question about finite projection matrices. This work: Proof of Weaver’s conjecture.

In General

Anything that can be encoded as a quadratic form can be split into pieces while preserving certain properties. Many different things can be gainfully encoded this way.

Proof

Main Theorem

Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗||2 ≤ 𝜗 and Then there is a partition 𝑈

1 ∪ 𝑈2 such that

1 2 − 5 𝜗 𝐽 ≼

𝑗∈𝑈𝑘

𝑤𝑗𝑤𝑗

𝑈 ≼

1 2 + 5 𝜗 𝐽

𝑗

𝑤𝑗𝑤𝑗

𝑈 = 𝐽𝑜

Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗||2 ≤ 𝜗 and Then there is a partition 𝑈

1 ∪ 𝑈2 such that 𝑗∈𝑈𝑘

𝑤𝑗𝑤𝑗

𝑈 ≼

1 2 + 5 𝜗 𝐽

Equivalent Theorem

𝑗

𝑤𝑗𝑤𝑗

𝑈 = 𝐽𝑜

Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗||2 ≤ 𝜗 and Then there is a partition 𝑈

1 ∪ 𝑈2 such that 𝑗∈𝑈𝑘

𝑤𝑗𝑤𝑗

𝑈 ≼

1 2 + 5 𝜗 𝐽

Equivalent Theorem

𝑗

𝑤𝑗𝑤𝑗

𝑈 = 𝐽𝑜

𝑗∈𝑈

1

𝑤𝑗𝑤𝑗

𝑈 = 𝐽 − 𝑗∈𝑈

2

𝑤𝑗𝑤𝑗

𝑈

Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗||2 ≤ 𝜗 and Then there is a partition 𝑈

1 ∪ 𝑈2 such that 𝑗∈𝑈𝑘

𝑤𝑗𝑤𝑗

𝑈 ≼

1 2 + 5 𝜗 𝐽

Equivalent Theorem

𝑗

𝑤𝑗𝑤𝑗

𝑈 = 𝐽𝑜

Idea 1: Random Partition

Choose 𝑈

1 ∪ 𝑈2 = 𝑛 randomly

Want:

𝑗∈𝑈

1

𝑤𝑗𝑤𝑗

𝑈 ≼

1 2 + 5 𝜗 𝐽

𝑗∈𝑈

2

𝑤𝑗𝑤𝑗

𝑈 ≼

1 2 + 5 𝜗 𝐽

Idea 1: Random Partition

Choose 𝑈

1 ∪ 𝑈2 = 𝑛 randomly

Want:

𝑗∈𝑈

1

𝑤𝑗𝑤𝑗

𝑈

≤ 1 2 + 5 𝜗

𝑗∈𝑈

2

𝑤𝑗𝑤𝑗

𝑈

≤ 1 2 + 5 𝜗

Idea 1: Random Partition

Choose 𝑈

1 ∪ 𝑈2 = 𝑛 randomly

Want:

𝑗∈𝑈

1

𝑤𝑗𝑤𝑗

𝑈

≤ 1 2 + 5 𝜗

𝑗∈𝑈

2

𝑤𝑗𝑤𝑗

𝑈

≤ 1 2 + 5 𝜗

Trick: embed in blocks of 2𝑜 × 2𝑜 matrix

𝑗∈𝑈

1

𝑤𝑗𝑤𝑗

𝑈 𝑗∈𝑈

2

𝑤𝑗𝑤𝑗

𝑈

Idea 1: Random Partition

Choose 𝑈

1 ∪ 𝑈2 = 𝑛 randomly

Want:

𝑗∈𝑈

1

𝑤𝑗𝑤𝑗

𝑈

≤ 1 2 + 5 𝜗

𝑗∈𝑈

2

𝑤𝑗𝑤𝑗

𝑈

≤ 1 2 + 5 𝜗

Trick: embed in blocks of 2𝑜 × 2𝑜 matrix

𝑗∈𝑈

1

𝑤𝑗𝑤𝑗

𝑈 𝑗∈𝑈

2

𝑤𝑗𝑤𝑗

𝑈

= max

j 𝑗∈𝑈𝑘

𝑤𝑗𝑤𝑗

𝑈

Idea 1: Random Partition

Define independent random 𝑤1, … , 𝑤𝑛 ∈ ℝ2𝑜

𝑤𝑗 = 𝑤𝑗 𝑥𝑗𝑢ℎ 𝑞𝑠𝑝𝑐 0.5 𝑤𝑗 𝑥𝑗𝑢ℎ 𝑞𝑠𝑝𝑐. 0.5

Then 𝔽

𝑗

𝑤𝑗𝑤𝑗

𝑈

= 𝔽𝑈 max

𝑘 𝑗∈𝑈𝑘

𝑤𝑗𝑤𝑗

𝑈

The Matrix Chernoff Bound

𝑤1, … , 𝑤𝑛 ∈ ℝ2𝑜 are independent, 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 2 and ||𝑤𝑗||2 ≤ 𝜗

Tropp 2011 𝔽

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 1 2 + 𝑃( 𝜗 log 𝑜) Analogous for the scalar Chernoff bound for sums of independent bdd random numbers.

The Matrix Chernoff Bound

𝑤1, … , 𝑤𝑛 ∈ ℝ2𝑜 are independent, 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 2 and ||𝑤𝑗||2 ≤ 𝜗

Tropp 2011 𝔽

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 1 2 + 𝑃( 𝜗 log 𝑜) Analogous for the scalar Chernoff bound for sums of independent bdd random numbers.

Main Theorem

If 𝑤1, … , 𝑤𝑛 ∈ ℝ2𝑜 are independent, 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 2 and ||𝑤𝑗||2 ≤ 𝜗

Then ℙ

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 1 2 + 𝑃( 𝜗) > 0

Main Theorem

If 𝑤1, … , 𝑤𝑛 ∈ ℝ𝑜 are independent, 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 and ||𝑤𝑗||2 ≤ 𝜗

Then ℙ

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 1 + 𝑃( 𝜗) > 0

Main Theorem

If 𝑤1, … , 𝑤𝑛 ∈ ℝ𝑜 are independent, 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 and ||𝑤𝑗||2 ≤ 𝜗

Then ℙ

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 1 + 𝑃( 𝜗) > 0

Main Theorem

If 𝑤1, … , 𝑤𝑛 ∈ ℝ𝑜 are independent, 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 and ||𝑤𝑗||2 ≤ 𝜗

Then ℙ

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 1 + 𝑃( 𝜗) > 0

∃𝑤1, … , 𝑤𝑛 𝑡𝑣𝑑ℎ 𝑢ℎ𝑏𝑢 ||

𝑗

𝑤𝑗𝑤𝑗

𝑈|| ≤ 1 + √𝜗

Idea 2: The Expected Polynomial

Just like yesterday,

𝑗

𝑤𝑗𝑤𝑗

𝑈

= 𝜇𝑛𝑏𝑦 det(𝑦𝐽 −

𝑗

𝑤𝑗𝑤𝑗

𝑈)

Consider 𝜈 𝑦 : = 𝔽det 𝑦𝐽 −

𝑗

𝑤𝑗𝑤𝑗

𝑈

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

𝜓( 𝑗 𝑤𝑗𝑤𝑗

𝑈) 𝑦

𝑤𝑗~𝑤𝑗

is an interlacing family

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

𝔽𝜓( 𝑗 𝑤𝑗𝑤𝑗

𝑈)(𝑦) is real-rooted for all product

distributions on signings. 𝜓( 𝑗 𝑤𝑗𝑤𝑗

𝑈) 𝑦

𝑤𝑗~𝑤𝑗

is an interlacing family

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

𝔽𝜓( 𝑗 𝑤𝑗𝑤𝑗

𝑈)(𝑦) is real-rooted for all product

distributions on signings. 𝜓( 𝑗 𝑤𝑗𝑤𝑗

𝑈) 𝑦

𝑤𝑗~𝑤𝑗

is an interlacing family

𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

=

𝑗=1 𝑛

1 − 𝜖 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=⋯=𝑨𝑛=0

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈 

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

• 2. Calculate

𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈



Central Identity

Suppose 𝑤1, … , 𝑤𝑛 are independent random vectors with 𝐵𝑗 ≔ 𝔽𝑤𝑗𝑤𝑗

𝑈. Then

𝔽det 𝑦𝐽 −

𝑗

𝑤𝑗𝑤𝑗

𝑈

=

𝑗=1 𝑛

1 − 𝜖 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=⋯=𝑨𝑛=0

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

• 2. Calculate

𝔽𝜓 =: 𝜈 𝑦 =

𝑗=1 𝑛

1 − 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=⋯0

 

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

• 2. Calculate
• 3. Bound the largest root 𝜇𝑛𝑏𝑦𝜈 𝑦 ≤ 1 +

𝜗 𝔽𝜓 =: 𝜈 𝑦 =

𝑗=1 𝑛

1 − 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=⋯0

Assuming 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 and ||𝑤𝑗||2 ≤ 𝜗

 

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

• 2. Calculate
• 3. Bound the largest root 𝜇𝑛𝑏𝑦𝜈 𝑦 ≤ 1 +

𝜗 𝔽𝜓 =: 𝜈 𝑦 =

𝑗=1 𝑛

1 − 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=⋯0

Assuming 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 and ||𝑤𝑗||2 ≤ 𝜗

  ?

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

• 2. Calculate
• 3. Bound the largest root 𝜇𝑛𝑏𝑦𝜈 𝑦 ≤ 1 +

𝜗 𝔽𝜓 =: 𝜈 𝑦 =

𝑗=1 𝑛

1 − 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=⋯0

Assuming 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 and ||𝑤𝑗||2 ≤ 𝜗

Bounding the Roots

Need to bound the roots of as a function of the 𝐵𝑗.

𝑗=1 𝑛

1 − 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=…𝑨𝑛=0

Bounding the Roots

Need to bound the roots of as a function of the 𝐵𝑗. Basic Question: What does (1 − 𝜖) do to roots?

𝑗=1 𝑛

1 − 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=…𝑨𝑛=0

Bounding the Roots

Need to bound the roots of as a function of the 𝐵𝑗. Basic Question: What does (1 − 𝜖) do to roots?

𝑗=1 𝑛

1 − 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=…𝑨𝑛=0

Quantitative version of the fact that it preserves real stability.

Bounding the Roots

Need to bound the roots of as a function of the 𝐵𝑗. Basic Question: What does (1 − 𝜖) do to roots?

𝑗=1 𝑛

1 − 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=…𝑨𝑛=0

Quantitative version of the fact that it preserves real stability.

The Univariate Case

Basic Question: What does (1 − 𝜖) do to roots?

The Univariate Case

Basic Question: What does (1 − 𝜖) do to roots? Answer: Interlacing

The Univariate Case

Consider 𝑞 𝑦 = 𝑦 − 𝜇1 … 𝑦 − 𝜇𝑜 distinct 𝑞′ 𝑦 = 𝑘 𝑗≠𝑘 𝑦 − 𝜇𝑗 Then

𝑞′ 𝑦 𝑞 𝑦 = 𝑗 1 𝑦−𝜇𝑗 is a rational function with

n poles.

Basic Question: What does (1 − 𝜖) do to roots? Answer: Interlacing

A Rational Function

𝑞 𝑦 = 𝑦 − 𝜇1 … 𝑦 − 𝜇𝑜 , 𝑞′ 𝑦 𝑞 𝑦 =

𝑗

1 𝑦 − 𝜇𝑗

A Rational Function

𝑞 𝑦 = 𝑦 − 𝜇1 … 𝑦 − 𝜇𝑜 , 𝑞′ 𝑦 𝑞 𝑦 =

𝑗

1 𝑦 − 𝜇𝑗

A Rational Function

𝑞 𝑦 = 𝑦 − 𝜇1 … 𝑦 − 𝜇𝑜 , 𝑞′ 𝑦 𝑞 𝑦 =

𝑗

1 𝑦 − 𝜇𝑗

roots of 𝑞′

A Rational Function

𝑞 𝑦 = 𝑦 − 𝜇1 … 𝑦 − 𝜇𝑜 , 𝑞′ 𝑦 𝑞 𝑦 =

𝑗

1 𝑦 − 𝜇𝑗

A Rational Function

𝑞 𝑦 = 𝑦 − 𝜇1 … 𝑦 − 𝜇𝑜 , 𝑞′ 𝑦 𝑞 𝑦 =

𝑗

1 𝑦 − 𝜇𝑗

𝑞′ 𝑞 = 1

A Rational Function

𝑞 𝑦 = 𝑦 − 𝜇1 … 𝑦 − 𝜇𝑜 , 𝑞′ 𝑦 𝑞 𝑦 =

𝑗

1 𝑦 − 𝜇𝑗

𝑞′ 𝑞 = 1

roots of 𝑞 − 𝑞′!

The Barrier Function

Define: Φ𝑞 𝑦 = 𝑞′ 𝑦 𝑞 𝑦

The Barrier Function

Define: Φ𝑞 𝑦 = 𝑞′ 𝑦 𝑞 𝑦 To bound roots of 𝑞 − 𝑞′, find a point above the roots of p with Φ𝑞(𝑐) < 1.

The Barrier Function

Define: Φ𝑞 𝑦 = 𝑞′ 𝑦 𝑞 𝑦 To bound roots of 𝑞 − 𝑞′, find a point above the roots of p with Φ𝑞(𝑐) < 1. Level sets {Φ𝑞 < 1} contain no zeros of 𝑞 − 𝑞′

The Barrier Method [BSS’09]

• Theorem. If b is above the roots of p and

Φ𝑞 𝑐 ≤ 1 − 1/𝜀 then Φ𝑞−𝑞′ 𝑐 + 𝜀 ≤ 1 − 1/𝜀

The Barrier Method [BSS’09]

• Theorem. If b is above the roots of p and

Φ𝑞 𝑐 ≤ 1 − 1/𝜀 then Φ𝑞−𝑞′ 𝑐 + 𝜀 ≤ 1 − 1/𝜀 Relates level sets of Φ𝑞 to level sets of Φ 1−𝜖 𝑞

The Barrier Method [BSS’09]

• Theorem. If b is above the roots of p and

Φ𝑞 𝑐 ≤ 1 − 1/𝜀 then Φ𝑞−𝑞′ 𝑐 + 𝜀 ≤ 1 − 1/𝜀 Relates level sets of Φ𝑞 to level sets of Φ 1−𝜖 𝑞

Robust version of {Φ𝑞 < 1} argument – can be iterated.

Evolution of level sets of Φ𝑞

Evolution of level sets of Φ𝑞

Evolution of level sets of Φ𝑞

Evolution of level sets of Φ𝑞

Evolution of level sets of Φ𝑞

Evolution of level sets of Φ𝑞

Gives bounds on 𝜇𝑛 1 − 𝜖 𝑙𝑞(𝑦)

The Bivariate Case

Given 𝑞 𝑦, 𝑧 , I want to bound the roots of 1 − 𝜖𝑦 1 − 𝜖𝑧 𝑞(𝑦, 𝑧)

The Bivariate Case

Example: roots of 𝑞 𝑦, 𝑧 real stable.

The Bivariate Case

Example: roots of 1 − 𝜖𝑧 𝑞 𝑦, 𝑧

The Bivariate Case

Example: roots of 1 − 𝜖𝑧 𝑞 𝑦, 𝑧

How to track these?

The Bivariate Case

Given 𝑞 𝑦, 𝑧 , I want to bound the roots of 1 − 𝜖𝑦 1 − 𝜖𝑧 𝑞(𝑦, 𝑧) Define bivariate barrier function Φ𝑞 𝑦, 𝑧 = 𝜖𝑦𝑞 𝑦, 𝑧 𝑞 𝑦, 𝑧 , 𝜖𝑧𝑞 𝑦, 𝑧 𝑞 𝑦, 𝑧

Evolution of Bivariate Φ Level Sets

𝑞 𝑦, 𝑧 real stable.

Evolution of Bivariate Φ Level Sets

𝑞 𝑦, 𝑧 real stable.

The Bivariate Case

1 − 𝜖𝑧 𝑞 𝑦, 𝑧

The Bivariate Case

1 − 𝜖𝑧 𝑞 𝑦, 𝑧

Several Perturbations

Several Perturbations

Several Perturbations

Several Perturbations

Several Perturbations

Several Perturbations

Several Perturbations

Key Ingredient

Helton-Vinnikov’92: All bivariate real stable polynomials are determinants: 𝑞 𝑦, 𝑧 = det(𝑦𝐵 + 𝑧𝐶 + 𝐷) with 𝐵 ≽ 0, 𝐶 ≽ 0 This implies that the bivariate barrier has the same properties as the univariate one, and the old proof goes through.

Basic Principle

Can track the evolution of multivariate zeros under 1 − 𝜖𝑨 operators by studying related rational functions. A quantitative version of the stability preserving property.

End Result

If 𝑈𝑠 𝐵𝑗 ≤ 𝜗 and 𝑗 𝐵𝑗 = 𝐽 then

𝑗=1 𝑛

1 − 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=…𝑨𝑛=0

Has roots bounded by 1 + 𝜗 2

3-Step Plan

• 1. Show that there exist 𝑤1, … , 𝑤𝑛 with

𝜇𝑛𝑏𝑦 𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 𝜇𝑛𝑏𝑦𝔽𝜓

𝑗

𝑤𝑗𝑤𝑗

𝑈

• 2. Calculate
• 3. Bound the largest root 𝜇𝑛𝑏𝑦𝜈 𝑦 ≤ 1 +

𝜗 𝔽𝜓 =: 𝜈 𝑦 =

𝑗=1 𝑛

1 − 𝜖𝑨𝑗 det 𝑦𝐽 +

𝑗

𝑨𝑗𝐵𝑗

𝑨1=⋯0

Assuming 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 and ||𝑤𝑗||2 ≤ 𝜗

  

Main Theorem

If 𝑤1, … , 𝑤𝑛 ∈ ℝ𝑜 are independent, 𝔽 𝑗 𝑤𝑗𝑤𝑗

𝑈 = 𝐽 and ||𝑤𝑗||2 ≤ 𝜗

Then ℙ

𝑗

𝑤𝑗𝑤𝑗

𝑈

≤ 1 + 𝑃( 𝜗) > 0



Spectral Discrepancy Theorem

Suppose 𝑤1, … , 𝑤𝑛 ∈ 𝑺𝑜 are vectors ||𝑤𝑗||2 ≤ 𝜗 and Then there is a partition 𝑈

1 ∪ 𝑈2 such that

1 2 − 5 𝜗 𝐽 ≼

𝑗∈𝑈𝑘

𝑤𝑗𝑤𝑗

𝑈 ≼

1 2 + 5 𝜗 𝐽

𝑗

𝑤𝑗𝑤𝑗

𝑈 = 𝐽𝑜



Open Questions

Quantitative analysis of other stability- preserving operators. More applications of discrepancy theorem. Algorithms.

Two Tools

Matrix-Determinant Lemma: det 𝑁 + 𝑤𝑤𝑈 = det 𝑁 det 𝐽 + 𝑁−1𝑤𝑤𝑈 = det(𝑁)(1 + 𝑤𝑈𝑁−1𝑤)

Two Tools

Matrix-Determinant Lemma: det 𝑁 + 𝑤𝑤𝑈 = det 𝑁 det 𝐽 + 𝑁−1𝑤𝑤𝑈 = det(𝑁)(1 + 𝑤𝑈𝑁−1𝑤) Jacobi’s Formula: 𝜖𝑢 det 𝑁 + 𝑢𝐵 = det 𝑁 𝜖𝑢 I + M−1A = det 𝑁 𝑈𝑠(𝑁−1𝐵)

(1 − 𝜖𝑨) operators = rank-1 updates

𝔽 det 𝑁 − 𝑤𝑤𝑈 = 𝔽det(M)(1 − 𝑤𝑈𝑁−1𝑤) = det(M)(1 − 𝔽𝑈𝑠(𝑁−1𝑤𝑤𝑈)) = det(𝑁)(1 − 𝑈𝑠(𝑁−1𝔽𝑤𝑤𝑈)) = det 𝑁 − det 𝑁 𝑈𝑠(𝑁−1 𝔽𝑤𝑤𝑈) = (1 − 𝜖𝑢)det 𝑁 + 𝑢𝔽𝑤𝑤𝑈 |𝑢=0

(1 − 𝜖𝑨) operators = rank-1 updates

𝔽 det 𝑁 − 𝑤𝑤𝑈 = 𝔽det(M)(1 − 𝑤𝑈𝑁−1𝑤) = det(M)(1 − 𝔽𝑈𝑠(𝑁−1𝑤𝑤𝑈)) = det(𝑁)(1 − 𝑈𝑠(𝑁−1𝔽𝑤𝑤𝑈)) = det 𝑁 − det 𝑁 𝑈𝑠(𝑁−1 𝔽𝑤𝑤𝑈) = (1 − 𝜖𝑢)det 𝑁 + 𝑢𝔽𝑤𝑤𝑈 |𝑢=0

(1 − 𝜖𝑨) operators = rank-1 updates

𝔽 det 𝑁 − 𝑤𝑤𝑈 = 𝔽det(M)(1 − 𝑤𝑈𝑁−1𝑤) = det(M)(1 − 𝔽𝑈𝑠(𝑁−1𝑤𝑤𝑈)) = det(𝑁)(1 − 𝑈𝑠(𝑁−1𝔽𝑤𝑤𝑈)) = det 𝑁 − det 𝑁 𝑈𝑠(𝑁−1 𝔽𝑤𝑤𝑈) = (1 − 𝜖𝑢)det 𝑁 + 𝑢𝔽𝑤𝑤𝑈 |𝑢=0

(1 − 𝜖𝑨) operators = rank-1 updates

𝔽 det 𝑁 − 𝑤𝑤𝑈 = 𝔽det(M)(1 − 𝑤𝑈𝑁−1𝑤) = det(M)(1 − 𝔽𝑈𝑠(𝑁−1𝑤𝑤𝑈)) = det(𝑁)(1 − 𝑈𝑠(𝑁−1𝔽𝑤𝑤𝑈)) = det 𝑁 − det 𝑁 𝑈𝑠(𝑁−1 𝔽𝑤𝑤𝑈) = (1 − 𝜖𝑢)det 𝑁 + 𝑢𝔽𝑤𝑤𝑈 |𝑢=0

(1 − 𝜖𝑨) operators = rank-1 updates

𝔽 det 𝑁 − 𝑤𝑤𝑈 = 𝔽det(M)(1 − 𝑤𝑈𝑁−1𝑤) = det(M)(1 − 𝔽𝑈𝑠(𝑁−1𝑤𝑤𝑈)) = det(𝑁)(1 − 𝑈𝑠(𝑁−1𝔽𝑤𝑤𝑈)) = det 𝑁 − det 𝑁 𝑈𝑠(𝑁−1 𝔽𝑤𝑤𝑈) = (1 − 𝜖𝑢)det 𝑁 + 𝑢𝔽𝑤𝑤𝑈 |𝑢=0

Proof of Central Identity

𝔽 det 𝑁 − 𝑤𝑤𝑈 = 1 − 𝜖𝑨 det 𝑁 + 𝑨𝔽𝑤𝑤𝑈

𝑨=0

𝔽 det 𝑦𝐽 − 𝑤1𝑤1

𝑈 − 𝑤2𝑤2 𝑈

= 1 − 𝜖𝑨1 𝔽 det(𝑦𝐽 − 𝑤2𝑤2

𝑈 + 𝑨1𝐵1)|𝑨1=0

= det(𝑦𝐽 + 𝑨2𝐵2 + 𝑨1𝐵1)|𝑨1=𝑨2=0

Proof of Central Identity

𝔽 det 𝑁 − 𝑤𝑤𝑈 = 1 − 𝜖𝑨 det 𝑁 + 𝑨𝔽𝑤𝑤𝑈

𝑨=0

𝔽 det 𝑦𝐽 − 𝑤1𝑤1

𝑈 − 𝑤2𝑤2 𝑈

= 1 − 𝜖𝑨1 𝔽 det(𝑦𝐽 − 𝑤2𝑤2

𝑈 + 𝑨1𝐵1)|𝑨1=0

= det(𝑦𝐽 + 𝑨2𝐵2 + 𝑨1𝐵1)|𝑨1=𝑨2=0

Proof of Central Identity

𝔽 det 𝑁 − 𝑤𝑤𝑈 = 1 − 𝜖𝑨 det 𝑁 + 𝑨𝔽𝑤𝑤𝑈

𝑨=0

𝔽 det 𝑦𝐽 − 𝑤1𝑤1

𝑈 − 𝑤2𝑤2 𝑈

= 1 − 𝜖𝑨1 𝔽 det(𝑦𝐽 − 𝑤2𝑤2

𝑈 + 𝑨1𝐵1)|𝑨1=0

= det(𝑦𝐽 + 𝑨2𝐵2 + 𝑨1𝐵1)|𝑨1=𝑨2=0

Proof of Central Identity

𝔽 det 𝑁 − 𝑤𝑤𝑈 = 1 − 𝜖𝑨 det 𝑁 + 𝑨𝔽𝑤𝑤𝑈

𝑨=0

𝔽 det 𝑦𝐽 − 𝑤1𝑤1

𝑈 − 𝑤2𝑤2 𝑈

= 1 − 𝜖𝑨1 𝔽 det(𝑦𝐽 − 𝑤2𝑤2

𝑈 + 𝑨1𝐵1)|𝑨1=0

= det(𝑦𝐽 + 𝑨2𝐵2 + 𝑨1𝐵1)|𝑨1=𝑨2=0

Proof of Central Identity

𝔽 det 𝑁 − 𝑤𝑤𝑈 = 1 − 𝜖𝑨 det 𝑁 + 𝑨𝔽𝑤𝑤𝑈

𝑨=0

𝔽 det 𝑦𝐽 − 𝑤1𝑤1

𝑈 − 𝑤2𝑤2 𝑈

= 1 − 𝜖𝑨1 𝔽 det(𝑦𝐽 − 𝑤2𝑤2

𝑈 + 𝑨1𝐵1)|𝑨1=0

= det(𝑦𝐽 + 𝑨2𝐵2 + 𝑨1𝐵1)|𝑨1=𝑨2=0