Hilbert Spaces Fall 2001 Final Exam

1. (This is based on B, p. 171, 12(xii).)
   a) Completely describe the spectrum of the left shift operator on l². Which spectral values, if any, are eigenvalues?

   b) Show that the spectrum of the right shift operator on l² is the entire disc {|z| <= 1}. Which spectral values, if any, are eigenvalues?

   Solution

   a) Let L denote the left shift operator, so
   L[(s₁, s₂, ...)] = (s₂, s₃, ...).
   Since L is a contraction, the spectrum is a subset of {z : |z| <= 1} (see class notes). I claim that if |z| < 1, then z is an eigenvalue of L. Indeed, from the eigenvalue equation (L s = z s) and the formula given above,
   s_k+1 = z s_k.
   It follows that the eigenvectors corresponding to z are multiples of
   s = (1, z, z², ...), i.e., s_k = z².
   This is in l² iff |z| < 1.

   Every z with |z| < 1 is thus an eigenvalue. Because the spectrum is a closed set, sp(L) = {z : |z| <= 1}.

   b) It is shown in B that the right shift R = L* has no eigenvalues. However, its spectrum is the same as that of L for the following reason. If |z| < 1, then the closure of Ran(R - z) is the orthogonal complement of N(R* - z*) = N(L - z*) = the multiples of the eigenvector of L with eigenvalue z*.

   Since dim(N(L - z*)) = 1, Ran(R - z) is not all of l², so (R - z)^-1 is not in B(l²). This establishes that sp(R) contains {z : |z| < 1}. Since R is an isometry, sp(R) lies inside {z : |z| <= 1}, and since the spectrum is closed it is precisely {z : |z| <= 1}.

   Note: If you suspect that there is a general rule for bounded operators that sp(A*) = (sp(A))*, you are correct (and this is part (ii) of the same problem in B). This is not the situation for eigenvalues, as this problem teaches us..

2. Assume that A is positive and bounded. Show that sqrt(A) is compact if and only if all nonzero AP values of A are isolated eigenvalues of finite multiplicity.

   Solution

   The Riesz-Schauder theorem tells us that the nonzero part of the spectrum of a compact operator consists of isolated¹ eigenvalues of finite multiplicity, so for the "only if" part it suffices to show that A is compact. Since A = sqrt(A) sqrt(A)*, this follows form the first TFAE of the class notes on compact operators.

   The other direction is the tricky one. Suppose that the eigenvalues of A have the stated property. I shall show that A is compact, and it will follow that sqrt(A) is compact because of the first TFAE of the class notes on compact operators.

   Since A = A*, the spectrum consists entirely of AP values. By hypothesis these are eigenvalues z_n of finite multiplicity without accumulation (adherent) points other than possibly 0. If e_n are the orthonormalized eigenvectors, we would like to show that A = Sum(z_n P_n), where P_n f := <f | e_n> e_n; since there are either a finite number of z_n or else they have a limit of 0, this operator is compact.

   Let B := A - Sum(z_n P_n). B has no nonzero eigenvalues, because if it did, then A would have an eigenvalue we have not accounted for. Moreover, B e_n = 0. If z is a nonzero AP value for B, then we may assume by projecting away all the e_n that the vectors f_k, ||f_k|| = 1, such that

   (B - z) f_k = (A - Sum(z_n P_n) - z) f_k   ->   0
   are orthogonal to all e_n. Hence P_n f_k = 0, and it follows that
   (A - z) f_n  ->  0.
   Consequently z is also a nonzero AP value of A, contradicting our hypothesis that that we have accounted for all such values already with the z_n. Conclusion: The only AP value of B is 0. By B, p. 170, Thm. 2, ||B||_op = 0, so A = Sum(z_n P_n) as claimed, which is compact.
   1. Being "isolated" is the same as not being an accumulation (= adherent) point.
3. Consider the following sets of numbers and kinds of operators on a classical Hilbert space:
   1. {1/n}, n = 1, 2, ...
   2. The unit circle in the complex plane
   3. {1, 0}
   4. {x: 0 <= x <= 2}
   5. {n/(n² + i}, n = 0, 1, ...
   1. self-adjoint operators
   2. positive operators
   3. orthogonal projectors (what B calls "projections")
   4. unitary operators
   5. compact operators (what B calls "completely continuous" operators)
   6. contractions

   In the following grid write S if the set is possibly the spectrum of the kind of operator, and write E it the set is possibly the entire set of the eigenvalues of the operator:

   Solution

   	I	II	III	IV	V
   a)	E	X	E,S	S	X
   b)	E	X	E,S	S	X
   c)	X	X	E,S	X	X
   d)	X	S	X	X	X
   e)	E	X	E,S	X	E,S
   f)	E	S	E,S	X	E,S