Compact Self-Adjoint Operators and the Spectral Theorem
The spectrum contains all eigenvalues and can contain other values. A compact self-adjoint Hilbert-space operator has an orthonormal eigenbasis with real eigenvalues. Nonzero eigenvalues have finite multiplicity; if there are infinitely many, they tend to zero.
What a learner can do afterwards
- Separate the spectrum from the point spectrum with a concrete operator
- State the spectral theorem for compact self-adjoint operators
- Explain why the eigenvalues can accumulate only at zero
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For a bounded operator T on a complex Banach space, lambda is in its spectrum when T-lambda I has no bounded inverse. Its point spectrum consists of eigenvalues: Tx=lambda x for some nonzero x. Point spectrum is contained in spectrum; inclusion can be strict or equal. The operator norm is sup_{||x||=1}||Tx||. For a unit eigenvector, ||Tx||=|lambda|<=||T||. Bounded operators on a Banach space form a complete space in operator norm. Summable term norms give norm-convergent operator series.
On complex l2, the square-summable sequences, the right shift is S(x1,x2,...)=(0,x1,x2,...). It is a bounded isometry with spectrum {|lambda|<=1}. It has no eigenvalues. For lambda=0, injectivity excludes nonzero eigenvectors. For lambda nonzero, Sx=lambda x first forces x1=0, then x2=0 and every later coordinate zero.
On a Hilbert space, a bounded T is compact if every bounded sequence has a subsequence whose images converge. Self-adjoint means <Tx,y>=<x,Ty>, with the first argument linear. Then <Tx,x> is real, so eigenvalues are real. The compact self-adjoint spectral theorem gives an orthonormal eigenbasis including a basis of ker T. Every nonzero spectral value is an eigenvalue with finite multiplicity, meaning the dimension of its eigenspace.
Self-adjointness gives (a-b)<u,v>=0 for eigenvectors with distinct real eigenvalues, so they are orthogonal. If orthonormal eigenvectors e_n have |lambda_n|>=epsilon>0, their images satisfy ||Te_n-Te_m||^2=|lambda_n|^2+|lambda_m|^2>=2epsilon^2. No image subsequence converges, contradicting compactness. Thus finitely many eigenvalues, counted with multiplicity, exceed any positive size. An infinite nonzero list tends to zero.
Compactness and self-adjointness are sufficient hypotheses for this eigenbasis theorem, not necessary for each operator with an eigenbasis. On infinite-dimensional l2, the identity has an orthonormal eigenbasis but is not compact: its images of unit basis vectors have no convergent subsequence. The bounded right shift has no eigenvectors, refuting a claim about every bounded operator. For the zero operator, every nonzero vector is an eigenvector with eigenvalue zero.
For bounded diagonal T(e_n)=d_n e_n on complex l2, ||T||=sup|d_n|. Its truncation T_N to N coordinates has finite rank and ||T-T_N||=sup_{n>N}|d_n|. Finite-rank operators and their operator-norm limits are compact. Thus d_n tending to zero proves compactness; an eigenvalue list alone does not. Diagonal entries d_n are eigenvalues; zero entries count the kernel dimension. A finite matrix's trace is its diagonal sum and eigenvalue sum. Example: diag(4,7) has eigenvalues 4 and 7.
On C([0,1];C) with sup norm, (Vf)(t)=integral from 0 to t of f(s). Vf=0 gives f=0 by differentiation. If Vf=lambda f, lambda!=0, then lambda f'=f and f(0)=0. Thus (e^(-t/lambda)f)'=0 forces f=0: no eigenvalues. Put R_N=sum_{n=0}^N V^n/lambda^(n+1). Then (lambda I-V)R_N=R_N(lambda I-V)=I-(V/lambda)^(N+1). Since ||V^n||<=1/n!, R_N converges in norm and the remainder vanishes for lambda!=0, giving an inverse. Vf(0)=0 prevents surjectivity. Its spectrum is {0}.
Spectrum can exceed point spectrum. Compact self-adjoint operators have an orthonormal eigenbasis; any infinite list of nonzero eigenvalues tends to zero.
Original lesson, questions and flashcards on operator spectra and compact self-adjoint operators, written for this learning dataset.
Original readings, questions and flashcards written for this learning dataset. Khan Academy videos provide prerequisite review of finite-dimensional eigenvalues.
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