SLIDE 1
Determinants of Perturbations of Finite Toeplitz Matrices
Estelle Basor American Institute of Mathematics October 2010
SLIDE 2 One of the purposes of this talk is to describe a Helton
- bservation that fundamentally made the computing the
asymptotics of determinants of finite matrices quite simple. We begin with the classical Szegö Limit Theorem and Toeplitz matrices.
SLIDE 3
Toeplitz Matrices
The Strong Szegö Limit Theorem states that if the symbol φ defined on the unit circle has a sufficiently well-behaved logarithm then the determinant of the Toeplitz matrix TN(φ) = (φj−k)j,k=0,··· ,N−1 where φk = 1 2π 2π φ(eiθ)e−ikθ dθ has the asymptotic behavior DN(φ) = det TN(φ) ∼ G(φ)N E(φ) as N → ∞.
SLIDE 4 Here the constants are G(φ) = e(log φ)0 E(φ) = exp ∞
k (log φ)k (log φ)−k
The last constant E(φ) can also be described by det
T(φ) = (φj−k) 0 ≤ j, k < ∞ is the Toeplitz Operator defined on the Hardy space.
SLIDE 5
This constant makes sense because if φ is sufficiently well-behaved, then the operator T(φ)T(φ−1) − I is trace class. Here is a sketch of the proof. The finite Toeplitz matrix TN(φ) can be thought of as the upper left hand corner of the matrix representation of the operator T(φ). We can think of it then as PNT(φ)PN where PN : {xn}∞
n=0 ∈ ℓ2 → {yn}∞ n=0 ∈ ℓ2,
yn = xn if n < N if n ≥ N .
SLIDE 6
Now suppose that U is an operator whose matrix representation has an upper triangular form. Then PNUPN = UPn. If L is an operator whose matrix representation has an lower triangular form. Then PNLPN = PnL. So if we had an operator of the form LU, then PNLUPN = PNLPNUPN and the corresponding determinants would be easy to compute.
SLIDE 7 What happens for Toeplitz operators is the opposite. If the symbol is sufficiently nice then T(φ) = UL so we need to do something else. We write PNT(φ)PN = PNULPN = PNLL−1ULU−1UPN = PNLPNL−1ULU−1PNUPN Now what else makes this works is that it turns out that the
- perator L−1ULU−1 is actually T(φ)T(φ−1) and we know that
this is I plus a trace class operator and thus has a well defined infinite determinant.
SLIDE 8
So putting this all together we have that det PNLPN × det PNUPN = G(φ)N and that lim
N−>∞ PNL−1ULU−1PN = det T(φ)T(φ−1).
SLIDE 9 Random Matrix Ensembles
There is a fundamental connection between determinants of Toeplitz matrices and random matrix ensembles. For example, one can consider the Circular Unitary Ensemble (CUE) with joint density a constant times
|eiθj − eiθk|2. A linear statistic for this ensemble is a random variable of the form SN =
N
f(eiθj), and it is this quantity which is connected to a Toeplitz determinant.
SLIDE 10 More precisely, if we define g(λ) to be 1 (2π)NN! π
−π
. . . π
−π N
eiλf(eiθj )
j<k
|eiθj − eiθk|2dθ1 . . . dθN then g(λ) is identically equal to det 1 2π π
−π
eiλf(θ)e−i(j−k)θdθ
.
SLIDE 11 The last determinant is a Toeplitz determinant with symbol φ(θ) = eiλf(eiθ). The identity holds because a very old result due to Andréief (1883) says that 1 N!
- · · ·
- det(fj(xk)) det(gj(xk))dx1 · · · dxN
= det
- fj(x)gk(x)dx
- j,k=1,··· ,N
.
SLIDE 12 One is interested in g because it is the inverse Fourier transform of the density of the linear statistic. In the opposite sense, the Toeplitz determinant can be thought
- f as an average or expectation with respect to CUE.
Asymptotics of the determinant gives us information about the linear statistic. This is especially useful when the function f is smooth enough, because we may appeal to the Strong Szegö Limit Theorem to tell us asymptotically the behavior of the density function.
SLIDE 13 To obtain asymptotic information about the linear statistic we apply the Strong Szegö Limit Theorem. This shows g(λ) ∼ G(φ)NE(φ), φ(eiθ) = eiλf(eiθ) where G(φ)N = exp
2π π
−π
f(eiθ)dθ
E(φ) = exp
∞
kfkf−k
SLIDE 14 We see that we can interpret the last formula as saying that asymptotically as N → ∞: For a smooth function f the distribution of SN − Nµ where SN =
N
f(eiθj), µ = 1 2π π
−π
f(eiθ)dθ converges to a Gaussian distribution with mean zero and variance given by σ2 =
∞
kfkf−k =
∞
k|fk|2 (The last equality holds if f is real-valued.)
SLIDE 15 Other examples that arise from RMT
It has also known that for if one considers averages for O+(2N), then the corresponding determinant is of a finite Toeplitz plus Hankel matrix and is of the form det
- aj−k + aj+k
- j,k=0,...,N−1
where subscripts denote Fourier coefficients and the function a is assumed to be even. Hence we are interested in the determinants of a sum of a finite Toeplitz plus a “certain type” of Hankel matrix. To be a little more general we are going to consider a set of
- perators and associated spaces the we will call compatible
pairs.
SLIDE 16 Let S stand for a unital Banach algebra of functions on the unit circle continuously embedded into L∞(T) with the property that a ∈ S implies that ˜ a ∈ S and Pa ∈ S. Here ˜ a(eiθ) = a(e−iθ). and P is the Riesz projection defined by P :
∞
akeikθ →
∞
akeikθ. Moreover, define S− =
- a ∈ S : an = 0 for all n > 0
- ,
S0 =
a
SLIDE 17
Assume that M : a ∈ L∞ → M(a) ∈ L(ℓ2) is a continuous linear map such that: (a) If a ∈ S, then M(a) − T(a) ∈ C1(ℓ2) and M(a) − T(a)C1(ℓ2) ≤ C aS. (b) If a ∈ S−, b ∈ S, c ∈ S0, then M(abc) = T(a)M(b)M(c). (c) M(1) = I. Then we say M and S are compatible pairs.
SLIDE 18
Concrete examples
All of the following can be realized as compatible pairs with an appropriate Banach algebra. We define the Hankel operator H(a) with symbol a by its with matrix representation H(a) = (aj+k+1), 0 ≤ j, k < ∞. (I) M(a) = T(a) + H(a), (II) M(a) = T(a) − H(a), (III) M(a) = T(a) − H(t−1a) with t = eiθ, (IV) M(a) = (T(a) + H(ta))R with R = diag(1/2, 1, 1, . . . ). The matrix representations of the operators are of the form aj−k ± aj+k−κ+1 with κ = 0, 1, −1.
SLIDE 19
For each of the previous four examples we can take the Banach algebra to be the Besov class. This is the class of all functions a defined on the unit circle for which π
−π
1 y2 π
−π
|a(eix+iy) + a(eix−iy) − 2a(eix) |dxdy < ∞. A function a is in this class if and only if the Hankel operators H(a) and H(˜ a) are both trace class. Moreover the Riesz projection is bounded on this class and an equivalent norm is given by |a0| + H(a)C1 + H(˜ a)C1.
SLIDE 20 We are interested in the determinants (where the matrices or
- perators are always thought of as acting on the image of the
projection of the appropriate space) of PNM(a)PN.
Theorem
Let M and S be a compatible pair, and let b ∈ S and a = exp(b). Then det PNM(a)PN ∼ G[a]N ˆ E[a] as N → ∞, where ˆ E[a] = exp
2trace H(b)2+trace H(b)H(˜ b)
SLIDE 21 Borodin-Okunkov/Case-Geronimo Identity
We can also produce an exact identity for such matrices. The following is for even functions, but can be made more general. Let M and S be a compatible pair, and let b+ ∈ S+. Put a = a+˜ a+ = exp(b) with a+ = exp(b+), b = b+ + ˜ b+. Then det PNM(a)PN = G[a]N ˆ E[a] det(I + QNKQN), where ˆ E[a] = exp
2trace H(b)2 , and K = M(a−1
+ )T(a+) − I.
SLIDE 22
Other results
An application of the above asymptotics yields an expansion for determinants of finite sections of operators of the form T(a) ± H(atκ). By using the basic identity det PAP = (det A) · (det QA−1Q), where Q = I − P we can reduce these determinants to the previous cases and compute them asymptotically.
SLIDE 23 κ is a negative even integer (κ = −2ℓ, ℓ ≥ 1)
Suppose that a = a−a0, where a0 is even and a− ∈ H2. Then det PN(T(a) ± H(atκ))PN ∼ G[a]N+ℓE1,±[a] det Pℓ(T(a−1
0 ) ± H(a−1 0 ))Pℓ
as N → ∞, where E1,±[a] is given by exp
∞
log a2n+1 − 1 2
∞
n[log a]2
n + ∞
n[log a]−n[log a]n
SLIDE 24 κ = −1 − 2ℓ, ℓ ≥ 1
Then det PN(T(a) − H(atκ))PN ∼ G[a]N+ℓE2[a] det Pℓ(T(a−1
0 ) − H(a−1 0 t−1)Pℓ
as N → ∞, where E2[a] is given by exp
∞
log a2n − 1 2
∞
n[log a]2
n + ∞
n[log a]−n[log a]n
SLIDE 25 κ = 1 − 2ℓ, ℓ ≥ 1
Then det PN(T(a) + H(atκ))PN ∼ G[a]N+ℓE3[a] det Pℓ(T(a−1
0 ) + H(a−1 0 t))Pℓ
as N → ∞, where E3[a] is given by exp
∞
log a2n−1 2
∞
n[log a]2
n+ ∞
n[log a]−n[log a]n
SLIDE 26
κ positive
We have det PN(T(a) + H(atκ))PN = if N ≥ κ ≥ 2, det PN(T(a) − H(atκ))PN = if N ≥ κ ≥ 1.
SLIDE 27
How general can M be?
Let us write K(a) = M(a) − T(a). The main properties for compatible pairs implies the following: Since M(ab) = M(a)M(b) for b even, then K(ab) = K(a)K(b) + T(a)K(b) + K(a)T(b) − H(a)H(b) whenever b is even. Also, T(a)M(b) = M(ab) for a ∈ S−, implies that for a ∈ S− we have K(a) = 0 and T(a)K(b) = K(ab) for any b.
SLIDE 28
Using these algebraic facts one can show that the structure of M is determined by K(t). in fact K(t) = e0xT where e0xT with x ∈ ℓ2 stand for the rank one operator y ∈ ℓ2 → e0y, x ∈ ℓ2. and e0 = (1, 0, 0, . . . ). The question of which x then generate an operator with the proper conditions is still not completely solved.