Spectral Radius Theorem

6. Spectral Radius Theorem

Let A be a Banach algebra with unit element e and make the additional assumption that ||e|| = 1.

Definition. The spectrum σ(x) of an element x ∈ A is the set of all complex numbers λ such that x - λe is not invertible.

In the finite dimensional case, this means that λ is an eigenvalue of x when A is treated as a subalgebra of a matrix algebra (the so called regular representation). In the infinite dimensional case, eigenvectors for such a λ might not exist.

Formula for Inverse. If x ∈ A and ||x|| < 1 then e + x is invertible and

(e + x)^-1 = ∑_{n = 1, ... , ∞} (-1)ⁿxⁿ

Proof.Since ||ab|| ≤ ||a|| ||b||, it follows that ||xⁿ|| ≤ ||x||ⁿ. Thus if

s_N = e - x + x² - ... + (-1)^Nx^N

then {x_N} is a Cauchy sequence. Moreover

(e + x) s_N = e + (-1)^Nx^N+1 = s_N (e + x)

and, taking limits as N → ∞, (e + x)y = e = y(e + x) where y = lim_{N → ∞} s_N.

Corollary 1. The set G of invertible elements of A is open, and the mapping x |→ x^-1 is a homeomorphism of G onto G.

Proof. If x ∈ G then ||(x + h)^-1 - x^-1|| ≤ ||x^-1|| ||(e + x^-1h)^-1 - e|| for any h ∈ G, and as h → 0, ||x^-1h|| ≤ ||x^-1|| ||h|| → 0, so the above formula shows first that G ccontains an open neighbourhood of x, next that ||(x + h)^-1 - x^-1|| → 0, establishing the continuity of the inverse.

Corollary 2. For every x ∈ A, σ(x) is compact, and |λ| ≤ ||x|| if λ ∈ σ(x).

Proof. If |λ| > ||x||, then e - λ^-1x ∈ G as the formula for inverse converges. The same is true of x - λe = λ(e - λ^-1x); hence λ ∉ σ(x). Now λ ∈ σ(x) iff x - λe ∉ G; since the complement of G is closed, so is its inverse image under the continuous mapping λ |→ x - λe. Thus σ(x) is closed and bounded; hence compact.

The Resolvent Equation. The resolvent set of x ∈ A, denoted by ρ(x), is the complement of σ(x). This is clearly open and contains {z: |z| > ||x||}. The resolvent of x is the function defined by

x(λ) = (x - λe)^-1

x(λ) is a continuous function of λ; and since x(λ) = λ^-1(x/λ - 1)^-1, x(λ) → 0 as λ → ∞. For λ, μ ∈ ρ(x)

x(λ) = x(λ) [x - μe] x(μ)

= x(λ) [x - λe + (λ - μ)e] x(μ)

= [e + (λ - μ) x(λ)] x(μ)

= x(μ) + (λ - μ) x(λ) x(μ)

x(λ) - x(μ) = (λ - μ) x(λ) x(μ)

This relation is called the resolvent equation.

Theorem. σ(x) is non-empty.

Remark. We have already proved that σ(x) is compact.

Proof. Let f be a continuous linear functional on A, and define F(λ) by F(λ) = f(x(λ)). F is continuous on ρ(x) and, by the resolvent equation

[F(λ) - F(μ)]/(λ - μ) = f(x(λ) x(μ)))

hence

lim_{λ → μ} [F(λ) - F(μ)]/(λ - μ) = f(x(μ)²)

so F(λ) is differentiable throughout ρ(x). Further

|F(λ)| ≤ ||f|| ||x(λ)||

so F(λ) → 0 as λ → ∞. If σ(x) is empty then ρ(x) is the entire complex plane. By Liouville's theorem, F(λ) = 0 for all λ. Since f is an arbitrary functional on A, the Hahn-Banach theorem implies that x(λ) implies that x(λ) = 0 for all λ. [We can always find a functional f such that f(x) ≠ 0 for a given x ≠ 0.] This is impossible, so σ(x) is non-empty.

Definition. The number

r(x) = sup {|λ|: λ ∈ σ(x)}

is called the spectral radius of x. Clearly 0 ≤ r(x) ≤ ||x||.

Spectral Radius Theorem. r(x) = lim ||xⁿ||^1/n

Proof. (i) We first show that r(x) ≤ lim inf ||xⁿ||^1/n.

For x ∈ A, n ∈ Z⁺, λ ∈ C, we have

(xⁿ - λⁿe) = (x - λe)(x^{n - 1} + λx^{n - 2} + ... + λ^{n - 1}e)

Multiplying both sides of the above by (xⁿ - λⁿe)^-1, we see that x - λe is invertible, hence λ ∉ σ(x).

So if λ ∈ σ(x) then λⁿ ∈ σ(xⁿ) for all n. Thus |λⁿ| ≤ ||xⁿ|| and therefore |λ| ≤ ||xⁿ||^1/n, proving the assertion.

(ii) Finally we have to show that lim sup ||xⁿ||^1/n ≤ r(x).

If |λ| > ||x||, we have

(λe - x) ∑_{n = 0, ... , ∞} λ^{- n - 1}xⁿ = e

that is, - (x - λe)^-1 = ∑_{n = 0, ... , ∞} λ^{- n - 1}xⁿ. Let f be a bounded linear functional and define F as in the previous theorem. Then we have

F(λ) = - ∑_{n = 0. ... , ∞} f(xⁿ) λ^{- n - 1}

where the series converges uniformly for |λ| > ||x|| ≥ r(x). Thus

sup_n |f(λⁿxⁿ)| < ∞ (|λ| > r(x))

for every bounded linear functional f on A.

Now, by the Corollary of the Hahn-Banach Theorem, the norm of any element of A is the same as its norm as a linear functional on A^*, and, applying the uniform boundedness theorem, we conclude that to each λ with |λ| > r(x) there corresponds a real number C(λ) such that

||λ^{- n}xⁿ|| ≤ C(λ)

for all n. Multiplying by |λ|ⁿ and taking n-th roots, we get

||xⁿ||^1/n ≤ |λ| [C(λ)]^1/n

for |λ| > r(x), so

lim sup ||xⁿ||^1/n ≤ r(x)

proving the theorem.

Remark. In the case of finite dimensional Banach algebras, it is possible to give several elementary proofs of this theorem. In fact, advanced analytical machinery is only required to establish that lim sup ||xⁿ||^1/n ≤ r(x). However, the theorem has been proved in full generality on account of the importance of the infinite dimensional case.

Application to the Solution of Polynomial Equations. If the space of n×n complex matrices is provided with a suitable norm, the spectral radius is the modulus of the largest eigenvalue of a matrix. Taking the companion matrix of a monic polynomial, we see that the spectral radius is the modulus of the largest root of the polynomial. Finding the largest root therefore reduces to finding the spectral radius of the companion matrix and plotting the values of the polynomial along the circle of that radius.

Choice of Norm. We consider the space of n×n matrices over C. For a matrix A = [a_ij]_{1 ≤ i, j ≤ n} we take ||A|| = (∑_{i, j} |a_ij|²)^1/2. This is a special case of the norm on the subalgebra of Hilbert-Schmidt operators on a Hilbert space, which will be explained in the Appendix below. There may be other norms which are easier to compute, but we will not go into this matter.

Note.This theorem was proved by Israel M. Gelfand in his paper Normierte Ringe, Matematiceskii Sbornik, N.S. 9, !941, pp.3 - 23, which developed the theory of Banach algebras.

APPENDIX: HILBERT-SCHMIDT OPERATORS

Preliminaries.

Let (H, (.,.)) be a Hilbert space. H is said to be separable if it contains a countable dense subset. An orthonormal basis of H is an indexed collection {x_α}_{α ∈ A} of elements of H such that

(x_α, x_β) = δ_{α, β}
The linear span of the {x_α} is dense in H.

Lemma 1. Suppose that {x_n} is an orthonormal sequence in H so that f ∈ H and (f, x_n) = 0 for all n then f = 0. Then the series

∑_{n = 1, ... , ∞} (f, x_n) x_n

converges to f.

Proof. Set c_n = (f, x_n). Set S_N = ∑_{n = 1, ... , N} c_nx_n. Set h_N = f - S_N. Then (h_N, S_N) = 0 and f = S_N + h_N. Thus ||f||² = ||S_N||² + ||h_N||². Hence ||S_N||² ≤ ||f||². Since ||S_N||² = ∑_{i = 1, ... , N} |c_i|² we see that the series ∑_{i = 1, ... , ∞} |c_i|² converges and ∑_{i = 1, ... , ∞} |c_i|² ≤ ||f||². Now if M > N then ||S_M - S_N||² = ∑_{j = N + 1, ... , M} |c_j|². This implies that the sequence {S_N} is Cauchy. Hence lim_{N → ∞} S_N = f₀ exists in H since H is complete. Now (f₀, x_n) = lim_{N → ∞} (S_N, x_n) = c_n = (f, x_n) for all n. Thus (f - f₀, x_n) = 0 for all n, proving the lemma.

Lemma 2. H has a countable orthonormal basis if and only if it is separable.

Proof. Let H have an orthonormal basis {x_n}_{n = 1, ... , ∞}. Let P ⊂ H be the subset consisting of all linear combinations ∑_{n = 1, ... , N} (a_n + ib_n)x_n, a_n, b_n rational. Then P is dense in H. Hence H is separable. Suppose H is separable. Let P = {z_n} be a countable dense subset. Define Q₁₁ if z₁ ≠ 0, otherwise Q₁ = ∅. suppose that Q_N has been defined. Let Q_{N + 1} = Q_N ∪ {z_{N + 1}} if z_{N + 1} is independent of Q_N; else Q_{N + 1} = Q_N. Let Q = ⋃_{N = 1, ... , ∞} Q_N. Then label the elements of Q as Q = {y_n}_{n = 1, ... , ∞}. The space of linear combinations of the {y_n} is dense in H. Furthermore {y₁, ... , y_N} is linearly independent for each N. We may therefore apply the Gram-Schmidt orthogonalisation process to {y₁, ... , y_N} for each N (that is, z₁ = y₁, z₂ = (y₂ - (y₂, z₁ ) z₁)/||y₂ - (y₂, z₁) z₁||, ...). We then get an orthonormal sequence {z_n} whose linear span is dense in H. This proves the lemma.

Basic Properties of Operators

Adjoints. Given an operator T on H, we can define another operator T^* so that

(Tx, y) = (x, T^*y)

for all x, y ∈ H. it is left to the reader to verify that this actually defines an operator on H. T^* is called the adjoint of T. The following proposition is easy to prove.

Proposition 1. The adjoint operation T |→ T^* has the following properties:

(T₁ + T₂)^* = T₁^* + T₂^*
(αT)^* = α^- T^*
(T₁ T₂)^* = T₂^* T₁^*
T^** = T
||T^*|| = ||T||
||T^* T|| = ||T||²

Definition. T∈ B(H) is said to be self-adjoint if T^* = T.

Proposition 2. The self-adjoint operators in B(H) form a closed linear subspace which contains the identity transformation.

Proposition 3. If A₁ and A₂ are self-adjoint operators on H, then A₁A₂ is self-adjoint if and only if A₁A₂ = A₂A₁.

Proof. This is an obvious consequence of

(A₁A₂)^* = A₂^* A₁^* = A₂A₁

Definition. An operator U on H is said to be unitary if it satisfies the equation U U^* = U^* U = I. That is to say U^-1 = U^*.

Proposition 3. If T is an operator on H, the following are equivalent:

T^* T = I
(Tx, Ty) = (x, y) for all x and y
||Tx|| = ||x|| for all x.

Lemma. If T is an operator on H for which (Tx, x) = 0 for all x, then T = 0.

Proof. It is easily verified that

(T(αx + βy), αx + βy) = |α|² (Tx, x) + |β|² (Ty, y) = αβ^- (Tx, y) + α^-β (Ty, x)

By hypothesis, the left side equals 0 for all α and β. Putting α = 1, β = 1; we get

(Tx, y) + (Ty, x) = 0

Putting α = i, β = 1

i (Tx, y) - i (Ty, x) = 0

Thus (Tx, y) = 0 for all x and y, proving the lemma.

Proof of Proposition. (1) implies (2): If T^* T = I then (Tx, Tx) = (T^* Tx, x) = (x, x), proving (2).

(2) implies (3): Taking y = x in (2), (Tx, Tx) = (x, x) or ||Tx|| = ||x||, establishing (3).

(3) implies (1) is a consequence of the lemma and the chain of implications

||Tx|| = ||x|| => ||Tx||² = ||x||² => (Tx, Tx) = (x, x) => (T^* Tx, x) = (x, x) => ([T^* T - I] x, x) = 0

Proposition 4. An operator T on H is an unitary if and only if it is an isometric isomorphism.

Proof. If T is unitary, then we know from the definition that it is onto; and since by Proposition 3 it preserves norms, it is an isometric isomorphism of H onto itself. Conversely, if T is an isometric isomorphism, then T^-1 exists, and by Proposition 3 we have T^* T = I. Thus T is unitary.

Trace Class Operators

Definition. Let T be a continuous linear operator on H. T is said to be of trace class if for each orthonormal basis {e_n} of H the sum ∑_n (Te_n, e_n) converges and is independent of the choice of basis. If T is of trace class define

tr T = ∑_n (Te_n, e_n)

Proposition. If T is a continuous linear operator on H and if for a fixed orthonormal basis {e_n} of H, ∑_i,j |(Te_i, e_j)| < ∞ then for any A, B continuous operators on H, ATB, TBA, and BAT are of trace class and tr ATB = tr TBA = tr BAT.

Proof. Recall that ||A|| = sup {||Av||: ||v|| = 1}. Set a_ij = (Ae_i, e_j), b_ij = (Be_i, e_j). Then

∑_{i, j, n ≤ N} |a_in b_nj (Te_i, e_j)| ≤ ∑_{i, j = 1, ... , N} ((∑_{n = 1, ... , N} |a_in|²)^1/2 (∑_{n = 1, ... , N} |b_nj|²)^1/2) |(Te_i, e_j)|

by Schwarz's inequality. But ∑_{n = 1, ... , N} |a_in|² ≤ ||A^* e_n||² = ||Ae_n||² ≤ ||A||². Thus we have

∑ |a_in b_nj (Te_i, e_j)| ≤ ||A|| ||B|| ∑_{i, j ≤ N} |(Te_i, e_j)|

Thus the sum

∑_{i, j, n = 1, ... , ∞} a_in b_nj (Te_i, e_j) is absolutely convergent.

But ∑_i (ATBe_i, e_i), ∑_i (BATe_i, e_i), and ∑_i (TBAe_i, e_i) are just rearrangements of the above. They therefore have the same sum. Let now U: H → H be a unitary operator. Then

∑_n (ATBUe_n, Ue_n) = ∑_n (U^-1 ATBUe_n, e_n) = ∑_n (UU^-1 ATBe_n, e_n) = ∑_n (ATBe_n, e,sub.n)

Thus ATB, BAT, TAB are all trace class and have the same trace.

Hilbert -Schmidt Operators

More complete information can be found in Chapter XI of Baggett's textbook.

Definition. An element T ∈ B(H) is said to be a Hilbert-Schmidt operator if ∑_i ||Te_i|| exists for all orthonormal bases {e_i} of H and is independent of the choice of basis. In that case we define the Hilbert-Schmidt norm ||T||_HS by

||T||_HS = [∑_i ||Te_i||²]^1/2

The set of Hilbert-Schmidt operators is denoted by B_HS(H).

Proposition 1. T is Hilbert-Schmidt if ∑_i ||Te_i||² exists for some orthonormal basis. Further, the set of all Hilbert-Schmidt operators is a two-sided self-adjoint ideal in B(H).

Proof. Suppose T ∈ B(H) and that there exists an orthonormal basis {e_i} such that

∑_i ||Te_i||² < ∞

Let {f_i} be another orthonormal basis. Then

∑_i ||Tf||² = ∑_i ∑_j |(Tf_i, e_j)|²

= ∑_i ∑_j |(f_i, T^* e_j)|²

= ∑_j ||T^* f_j||²

= ∑_j ∑_i |(T^* f_j, f_i)|²

= ∑_j ∑_i |(f_j, Tf_i)|²

= ∑_i ||Tf_i||²

Now, if A is any bounded operator on H

∑_i ||ATe_i||² ≤ ∑_i ||A||² ||Te_i||²

= ||A||² ∑_i ||Te_i||²

= ||A||² ||T||_HS²

and

∑_i ||TAe_i||² = ∑_i ||A^* T^* e_i||²

since (AT)^* = T^* A^*

≤ ||A^*|| ∑_i ||T^* e_i||²

= ||A^*|| ∑_i ||Te_i||²

showing that AT, TA ∈ B_HS(H).

Exercise. On B_HS(H) × B_HS(H) define

(T, S) = ∑_i (S^* T e_i, e_i)

where {e_i} is an orthonormal basis. Verify that (T, S) is a well defined inner product on B_HS(H). Show further that if T ∈ B_HS(H) then ||T|| ≤ ||T||_HS and if S ∈ B(H) is arbitrary, then

||ST||_HS ||S|| ||T||_HS

Proposition 2. An operator T is a trace class operator if and only if there exist two Hilbert-Schmidt operators S₁ and S₂ such that T = S₁ S₂

Proof. See Baggett's textbook.

[DRAFT INCOMPLETE]

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