Every eigenvalue of a bounded self-adjoint Hilbert-space operator is real.

A bounded operator T on a nonzero complex Hilbert space has ||T||<=3. Which scalar cannot be an eigenvalue of T?

• -2
• 0
• 4
• 3

For the zero operator on a nonzero Hilbert space, every nonzero vector is an eigenvector with eigenvalue zero.

A diagonal matrix has diagonal entries 2 and 5. What is the larger eigenvalue?

For a compact self-adjoint Hilbert-space operator, every nonzero spectral value is an eigenvalue of finite multiplicity.

Suppose T has orthonormal eigenvectors e_n with Te_n=2e_n for every n. Why can T not be compact?

• Its images remain distance 2sqrt(2) apart
• Its bounded input sequence has no convergent subsequence, which alone contradicts compactness
• Its eigenvalue is positive, which alone excludes compactness
• Its eigenvectors have norm 1, which alone excludes compactness

On complex l2, T(e_n)=(1/n)e_n for n>=1. How many eigenvalues, counted with multiplicity, are at least 1/4?

If distinct eigenvalues of a compact self-adjoint operator accumulate, where can their limit lie?

• At any positive real number, with no size restriction
• Only at zero
• At any negative real number, with no size restriction

Nina claims every bounded Hilbert-space operator has an orthonormal eigenbasis. Which response refutes her claim?

• The bounded right shift has no nonzero eigenvectors
• Every bounded operator has eigenvectors, although they need not span the space
• Every bounded operator has an eigenbasis, although it need not be orthonormal

For a bounded self-adjoint Hilbert-space operator, eigenvectors belonging to distinct eigenvalues are orthogonal, even if the operator is not compact.