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Riesz projection for a matrix with separated spectrum

Example

Assume the Axiom of Choice (The Axiom of Choice). Let T:=diag⁡(1,2)∈M2(C), acting on the standard basis e1,e2 of C2. Its spectrum is σ(T)={1,2} (Spectrum in a finite-dimensional matrix algebra), the subset E:={1} is clopen in the spectrum, and the Riesz spectral projection (Riesz spectral projection) is

PE=diag⁡(1,0)∈M2(C),

the operator of orthogonal projection onto Ce1. Consequently ran⁡(PE)=Ce1 and ker⁡(PE)=Ce2 are the two invariant summands of Riesz spectral projection properties, and the restrictions of T to them have spectra {1} and {2} respectively.

Facts & Assumptions

Given: The Axiom of Choice, the diagonal matrix T=diag⁡(1,2), the spectral subset E={1}, and the circle γ(t):=1+12eit, 0≤t≤2π, which separates 1 from 2 and lies in the resolvent set of T.

[L1]

M2(C) with the operator norm is a unital Banach algebra and σ(A)={λ:det⁡(λI−A)=0} (Spectrum in a finite-dimensional matrix algebra).

[L2]

The Riesz projection is the calculus value of the locally constant function χE, equivalently the resolvent contour integral 12πi∫ΓχE(z)(zI−T)−1dz over a cycle with index 1 on E and 0 on σ(T)∖E (Riesz spectral projection).

[L3]

For a closed cycle, 12πi∫γ(z−a)mdz equals 1 for m=−1 and 0 otherwise when γ winds once around a (On a positively oriented circle about a, the integral of (z−a)m is zero for every integer m except −1, and is 2πi for m=−1).

[L4]

A uniformly convergent sequence of continuous functions on a contour may be integrated term by term (A uniformly convergent sequence of continuous integrands on a fixed contour permits passage of the limit through the complex line integral).

[L5]

The range and kernel of a Riesz projection are closed invariant summands, and the restriction spectra are the corresponding spectral parts (Riesz spectral projection properties).

Verification

technique · direct
1.1L1algebra

Resolvent: for z∉{1,2} one has (zI−T)−1=diag⁡((z−1)−1,(z−2)−1), and the circle γ of radius 1/2 about 1 avoids both spectral points; on it the resolvent is the diagonal pair of scalar functions 1/(z−1) and 1/(z−2).

2.1step 1.1L2L3L4algebra

Contour integral: by [L3], 12πi∫γdzz−1=1 because γ is the circle about 1. On this circle ∣z−1∣=1/2, and 1z−2=−11−(z−1)=−∑n≥0(z−1)n uniformly: the tail after degree N has modulus at most 2−N−1/(1−1/2). Each term has integral zero by [L3], so [L4] gives ∫γdz/(z−2)=0. Hence PE=12πi∫γ(zI−T)−1dz=diag⁡(1,0), the locally constant characteristic function of {1} evaluated on the diagonal.

3.1step 2.1L5L1algebra∎

The projection diag⁡(1,0) is idempotent, commutes with T and has ran⁡(PE)=Ce1, ker⁡(PE)=Ce2; both are T-invariant, T∣Ce1 is multiplication by 1 and T∣Ce2 is multiplication by 2, so the two restrictions have spectra {1} and {2}; this agrees with [L5] and the separation of the spectral parts.

Remarks

The circle below is the positively oriented contour γ(t)=1+12eit: it encloses 1 and excludes 2. The arrows on the right show the action of PE on the standard basis; T acts by 1 on Ce1 and by 2 on Ce2. These illustrate the contour computation and invariant summands in steps 2.1 and 3.1; the verification above is independent of the figure.

RezImz12°E=f1ge1e1e20PEPEran(PE)=Ce1ker(PE)=Ce2

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