Compact Operators in Hilbert Space Hart Smith Department of Mathematics University of Washington, Seattle Math 526/556, Spring 2015 Hilbert-Schmidt integral kernels on L2(A) 2 2 If K (x; y) 2 L (A × A), and H = L (A), define TK 2 B(H) by Z TK g(x) = K (x; y) g(y) dy : Then hf ; TK gi = hTK f ; gi iff Z Z Z Z f (x) K (x; y) g(y) dy dx = K (y; x)f (x) dx g(y) dy TK is self-adjoint iff K (x; y) = K (y; x) for a.a. x; y 2 A × A. TK is compact since it’s a Hilbert-Schmidt operator. Self-adjoint Hilbert-Schmidt integral kernels on L2(A) Corollary: assume K (x; y) = K (y; x) 2 9 orthonormal set fφj g ⊂ L (A)of eigenvectors for TK : Z K (x; y) φj (y) dy = λj φj (x) ; λj 6= 0 ; such that if f k g is an orthonormal basis for ker(TK ), then 2 fφj g [ f k g is an orthonormal basis for L (A). It follows that fφj (x)φk (y)g [ fφj (x) k (x)g [ f j (x)φk (y)g [ f j (x) k (y)g is an orthonormal basis for L2(A × A). Expand K (x; y) in basis: 1 X K (x; y) = λj φj (x) φj (y) j=1 Square root of T ∗T Suppose T is compact on H (not necessarily self-adjoint). T ∗T is compact, self-adjoint on H, and hT ∗Tx; xi = kTxk2 ≥ 0 Lemma There exists a unique compact, non-negative, self-adjoint operator S such that S2 = T ∗T . ∗ ? Proof. Let N0 = ker(T T ) = ker(T ) ; E = N0 ; so H = E ⊕ N0. T ∗T : E ! E is compact, 1-1, so in some orthonormal basis ∗ T T = diag(µ1; : : : ; µ1; µ2; : : : ; µ2; µ3;:::) where µ1 > µ2 > µ3 > ··· > 0 and µj ! 0 unless finitely many. Square root of T ∗T Define S : E ! E by the matrix p p p p p S = diag µ1 ;:::; µ1 ; µ2 ;:::; µ2 ; µ3 ;::: Extend S to H = E ⊕ N0 by Sy = 0 if y 2 N0 = ker(T ). 2 ∗ 2 ∗ S x = T Tx if x 2 E, and S y = T Ty = 0 if y 2 N0. 2 ∗ Any z 2 H of form x + y 2 E ⊕ N0, so S = T T on H. If S2 = T ∗T is as in the statement, then S commutes with 2 ∗ S = T T , so each Nµj , as well as N0, is invariant for S. p • S self-adjoint, non-negative, ) S = µj on Nµj . • Similarly, S = 0 on N0, so S unique. Note: hSx; Sxi = hS2x; xi = hT ∗Tx; xi = hTx; Txi therefore: kSxk = kTxk for all x 2 H. Polar Decomposition for compact T on H. Theorem Suppose T is compact on a Hilbert space H. Let E = (ker T )?. One can write T = US, where U : E !H is a norm preserving map, S is self-adjoint non-negative, and kSxk = kTxk 8 x 2 H. Proof. Let S2 = T ∗T , with S ≥ 0 self-adjoint, so kSxk = kTxk. S : E ! E is self-adjoint, 1-1, so dense range: S(E) = E. Define U : S(E) !H by U(Sx) = Tx for x 2 E. Since kUyk=kyk and S(E) = E, U extends uniquely to a map E !H with kUyk = kyk for all y 2 E. Indeed, U is an isometry of E onto T (E). Since U(Sx) = Tx for all x 2 E, and U(Sx) = Tx = 0 for x 2 E? = ker T , then U(Sx) = Tx for all x 2 H. Remark: U and S are unique. The condition kSxk = kTxk forces S : ker T ! f0g, and S = S∗ then implies S : E ! E. Then T ∗T = SU∗US = S2, so S is the unique non-negative self-adjoint square root of T ∗T . Example: 1 1 1 1 T (x1; x2; x3;:::) = (0; x1; 2 x2; 3 x3;:::) ; T = SR ◦ diag(1; 2 ; 3 ;:::) Corollary If T 2 B(H) is compact, there exists a sequence fTN g ⊂ B(H) of finite-rank operators such that limN!1 kT − TN kB(H) = 0. Proof. Let S be the truncation of S to Np ⊕ · · · ⊕ Np . N p µ1 µN Then kS − SN k ≤ µN+1 ! 0. Let TN = USN , then kT − TN k ≤ kUk kS − SN k ! 0 : Singular value decomposition Definition If µ > µ > µ ::: are the eigenvalues of T ∗T , the numbers 1 p 2 3 ρk = µk are called the singular values of T . Remarks. kT k = maxk ρk . 1 X 2 T is Hilbert-Schmidt iff ρk mult(ρk ) < 1 ; and k=1 1 2 X 2 kT kHS = ρk mult(ρk ) : k=1 ∗ Proof. Take fxng = eigenvector basis for T T . Then 1 1 1 X 2 X ∗ X 2 kTxnk = hT Txn; xni = ρk mult(ρk ) : n=1 n=1 k=1 Singular value decomposition Singular Value Decomposition Theorem If T 2 B(H) is compact with singular values fρk g, there exists ? an orthonormal basisfxngfor (ker T ) , and an orthonormal basis fyng for T (H) ; such that 1 X Tx = ρk(n) hxn; xi yn : j=0 ? Proof. Let fxng be O.N. basis of E = (ker T ) in which S = diag(ρ1; : : : ; ρ1; ρ2; : : : ; ρ2; ρ3;::: ) Since U : E ! T (H) is isometry, then fyng = fUxng is an P O.N. basis for T (H). Then Txn = ρk(n) yn, and x = nhxn; xixn with convergence in H, so the result holds by continuity..