Lecture 21: The Spectrum of Self-Adjoint Operators and the Eigenspaces of Compact Self-Adjoint Operators

May 6, 2021 We’ll continue discussing properties of the spectrum of a bounded linear operator today: recall that the resolvent of an operator A is the set of complex numbers λ such that A − λ is an element of GL(H) (in other words, A − λ is bijective, meaning it has a bounded inverse), and the spectrum of A is the complement of the resolvent in C. While the spectrum is just the set of eigenvalues for matrices in a finite-dimensional vector space, there’s a more subtle distinction to be made now: we define λ ∈ Spec(A) to be an eigenvalue if there is some vector u with (A − λ)u = 0, so λ is in the spectrum because A − λ is not injective. But there are other reasons for why λ might be in the spectrum as well, for instance if the image is not closed. Last time, we proved that the spectrum is closed and is contained within the ball of radius ||A||, meaning that it is compact. We then focused our attention on self-adjoint operators, and that’s where we’ll be directing our study today. We proved last lecture that a self-adjoint bounded linear operator A always has hAu, ui real, and that it satisfies ||A|| = sup||u||=1 |hAu, ui|. Here’s our next result: Theorem 223 Let A = A∗ ∈ B(H) be a self-adjoint operator. Then the spectrum Spec(A) ⊂ [−||A||, ||A||] is contained within a line segment on the real line, and at least one of ±||A|| is in Spec(A). Proof. First, we’ll show the first property (that the spectrum is contained within this given line segment). We know from last time that Spec(A) ⊂ {|λ| ≤ ||A||}, so we just need to show that Spec(A) ⊂ R (in other words, any complex number with a nonzero imaginary part is in the resolvent). Write A = s + i t for s, t real and t 6= 0, so that A − λ = (A − s) − i t = à − i t, where à = A − s is another self-adjoint bounded linear operator because (A − sI)∗ = A∗ − (sI)∗ = A − sI. So it suffices to show that à − i t is bijective, and we’ll switch our notation back to using A instead of Ã. Note that because hAu, ui is real, Im(h(A − i t)u, ui) = Im(h−i tu, ui) = −t||u||2 , so (A − i t)u = 0 only if u = 0 (since that’s the only instance where the right-hand side is zero). Therefore, A − i t is injective, and we just need to show that it is surjective. Notice that (A − i t)∗ = A + i t is also injective by the same argument, so Range(A − i t)⊥ = Null((A − i t)∗ ) = {0}. And now we can use what we know about orthogonal complements: Range(A − it) = (Range(A − i t)⊥ )⊥ = {0}⊥ = H, so it suffices to show that the range of A − i t is closed. To show that, suppose we have a sequence of elements un such that (A − i t)un → v ; we want to show that v ∈ Range(A − i t). We know from the calculation above that |t| · ||un − um ||2 = |Im(h(A − i t)(un − um ), un − um i)| ≤ |h(A − i t)(un − um ), un − um i|, and by Cauchy-Schwarz this is bounded by ≤ ||(A − i t)un − (A − i t)um || · ||un − um ||. Simplifying the first and last expressions, we find that ||un − um || ≤ 1 ||(A − i t)un − (A − i t)um ||. |t| Since t is a fixed constant, and our sequence {(A − i t)un } converges, it is also Cauchy. In particular, for any ε > 0, we can find some N so that the right-hand side is smaller than ε as long as n, m ≥ N, and that same N shows that our sequence {un } is also Cauchy. Therefore, there exists some u ∈ H so that un → u by completeness of our Hilbert space, and now we’re done: since (A − i t) is a bounded and thus continuous linear operator, (A − i t)u = lim (A − i t)un = v . n→∞ So the range is closed, and combining this with our previous work, A − i t is surjective. This finishes our proof that A − i t is bijective and thus complex numbers with nonzero imaginary part are in the resolvent. Now for the second property, since we have shown that ||A|| = sup||u||=1 |hAu, ui|, there must be a sequence of unit vectors {un } such that |hAun , un i| → ||A||. Since each term in this sequence is real, there must be a subsequence of these {un } with hAun , un i converging to ||A|| or to −||A||, which means that we have h(A ∓ ||A||)un , un i → 0 as n → ∞ (this notation means one of − or +, depending on whether we had convergence to ||A|| or −||A||). We claim that this means A ∓ ||A|| is not invertible: assume for the sake of contradiction that it were invertible. Then 1 = ||un || = ||(A ± ||A||)−1 (A ∓ ||A||)un || ≤ ||(A ± ||A||)−1 || · ||(A ∓ ||A||)un ||, but the right-hand side converges to 0 as n → ∞, contradiction. So A ∓ ||A|| is not bijective, and thus one of ±||A|| must be in the spectrum of A, finishing the proof. We can in fact strengthen this bound even more: Theorem 224 If A = A∗ ∈ B(H) is a self-adjoint bounded linear operator, and we define a− = inf ||u||=1 hAu, ui and a+ = sup||u||=1 hAu, ui, then a± are both contained in Spec(A), which is contained within [a− , a+ ]. Proof. Applying a similar strategy as before, we know that because −||A|| ≤ hAu, ui ≤ ||A|| for all u, we must have −||A|| ≤ a− ≤ a+ ≤ ||A|| (by taking the infimum and supremum of the middle quantity). Now by the definition of a− , a+ , there exist two sequences {un± } of unit vectors so that hAun± , un± i → a± . And the argument we just gave works here very similarly: since we know that h(A − a± )un± , un± i → 0, this implies that a+ and a− are both in the spectrum because we have convergence to both points. It remains to show that the spectrum is contained within [a− , a+ ]. Let b = a− +a+ 2 be their midpoint, and let B = A − bI. Since b is a real number, B is also a bounded self-adjoint operator, so by Theorem 223, we know that Spec(B) ⊂ [−||B||, ||B||]. This means that (shifting by bI) Spec(A) ⊂ [−||B|| + b, ||B| + b], and we can finish by noticing that ||B|| = sup |hBu, ui| = sup hAu, ui − ||u||=1 ||u||=1 a+ + a− . 2 Since hAu, ui always lies in the line segment [a− , a+ ] (getting arbitrarily close to the endpoints), and a+ +a− 2 is their midpoint, this supremum will be half the length of that line segment, meaning that ||B|| = a+ − a− =⇒ Spec(A) ⊂ [−||B|| + b, ||B| + b] = [a− , a+ ], 2 as desired, completing the proof. Corollary 225 Let A∗ = A ∈ B(H) be a self-adjoint linear operator. Then hAu, ui ≥ 0 for all u if and only if Spec(A) ⊂ [0, ∞). (This can be shown by basically walking through the logic for what a− needs to be under either of these conditions.) We’ll now move on to the spectral theory for self-adjoint compact operators: the short answer is that we essentially see just the eigenvalues, with the exception of zero being a possible accumulation point. And in particular, the spectrum will be countable, and this should make sense because compact operators are the limit of finite rank operators – we don’t expect to end up with wildly different behavior in the limit. Definition 226 Let A ∈ B(H) be a bounded linear operator. We denote Eλ to be the nullspace of A − λ, or equivalently the set of eigenvectors {u ∈ H : (A − λ)u = 0}. Theorem 227 Suppose A∗ = A ∈ B(H) is a compact self-adjoint operator. Then we have the following: 1. If λ 6= 0 is an eigenvalue of A, then λ ∈ R and dim Eλ is finite. 2. If λ1 6= λ2 are eigenvalues of A, then Eλ1 and Eλ2 are orthogonal to each other (every element in Eλ1 is orthogonal to every element in Eλ2 ). 3. The set of nonzero eigenvalues of A is either finite or countably infinite, and if it is countably infinite and given by a sequence {λn }n , then |λn | → 0. Proof. For (1), let λ be a nonzero eigenvalue. Suppose for the sake of contradiction that Eλ is infinite-dimensional. Then by the Gram-Schmidt process, there exists a countable collection {un }n of orthonormal elements of Eλ . Since A is a compact operator, this means that {Aun }n must have a convergent subsequence, and in particular that means we have a Cauchy sequence {Aunj }j . But we can calculate ||Aunj − Aunk ||2 = ||λunj − λunk ||2 = |λ|2 ||unj − unk ||2 = 2|λ|2 , so the distance between elements of the sequence does not go to 0 for large j, k, a contradiction. Thus Eλ is finitedimensional. To show that λ must be real, notice that we can pick a unit-length eigenvector u satisfying Au = λu, and then we have λ = λhu, ui = hλu, ui = hAu, ui, and we’ve already shown that this last inner product must be real, so λ is real. For (2), suppose λ1 6= λ2 , and supppose u1 ∈ Eλ1 , u2 ∈ Eλ2 . Then λ1 hu1 , u2 i = hλ1 u1 , u2 i = hAu1 , u2 i, and now because A is self-adjoint, this is = hu1 , Au2 i = hu1 , λ2 u2 i = λ2 hu1 , u2 i (no complex conjugate because eigenvalues are real). Therefore, we must have (λ1 − λ2 )hu1 , u2 i = 0, so (because λ1 − λ2 6= 0) hu1 , u2 i = 0 and we’ve shown the desired orthogonality. Finally, for (3), let Λ = {λ 6= 0 : λ eigenvalue of A}. We need to show that Λ is either finite or countably infinite, and we claim that we can actually prove both parts of (3) simultaneously by showing that if {λn }n is a sequence of distinct eigenvalues of A, then λn → 0. This is because the set ΛN = {λ ∈ Λ : |λ| ≥ 1 } N is a finite set for each N (otherwise we could take any sequence of distinct elements in ΛN , and that can’t converge S to 0), and thus Λ = N∈N ΛN is a countable union of finite sets and thus countable. In order to prove this claim, let {un }n be the associated unit-length eigenvectors of our eigenvalues λn . Then |λn | = ||λn un || = ||Aun ||, so we further reduce the problem to showing that ||Aun || → 0. But showing this is a consequence of us having an orthonormal sequence of vectors and A being compact: suppose that ||Aun || does not converge to 0. Then there exists some ε0 > 0 and a subsequence {Aunj } so that for all j, ||Aunj || ≥ ε0 . Then because A is a compact operator, there exists a further convergent subsequence ek = unjk , meaning that {Aek }k converges in H. Since ek and e` are eigenvectors that correspond to distinct eigenvalues, they are orthogonal, and therefore Aek and Ae` are also orthogonal. But now if f = limk→∞ Aek , then ||f || = lim ||Aek || ≥ ε0 , k→∞ meaning that by continuity of the inner product, ε20 ≤ ||f ||2 = hf , f i = lim hAek , f i = lim hek , Af i. k→∞ k→∞ And because the sequence hek , Af i gives us the Fourier coefficients of Af , the sum of their squares should be finite (by Bessel’s inequality, it’s at most ||Af ||2 < ∞). This contradicts the fact that the limit of the Fourier coefficients is at least ε20 . So our original assumption is wrong, and ||Aun || must converge to 0, proving the claim.