SPECTRAL THEOREM FOR COMPACT SELF-ADJOINT OPERATORS

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1 SPECTRAL THEOREM FOR COMPACT SELF-ADJOINT OPERATORS G. RAMESH Contents Introduction 1 1. Bounded Operators Examples 3 2. Compact Operators Properties 6 3. The Spectral Theorem Self-adjoint Operators Second form of the Spectral Theorem 14 Introduction Let T : V V be a normal matrix on a finite dimensional complex vector space V. Suppose that λ 1, λ 2,..., λ n are distinct eigenvalues of T and M i (i = 1, 2..., n) be the corresponding eigenspaces and P i : V V be the orthogonal projections onto M i. Then by the finite dimensional Spectral theorem, we have I = n P k and T = n λ kp k. In these lectures we see that this result can be extended to a particular class of operators on infinite dimensional Hilbert spaces, which resembles finite dimensional operators in some sense. 1. Bounded Operators In this section we define bounded linear operators between Hilbert spaces and discuss some properties and examples. Throughout we consider only Complex Hilbert spaces. Until other wise specified, all Hilbert spaces are assumed to be infinite dimensional. Definition 1.1. Let T : H 1 H 2 be linear. Then T is said to be bounded if and only if T (B) is bounded in H 2 for every bounded subset B of H 1. If H 1 and H 2 are Hilbert spaces and T : H 1 H 2 is a bounded operator, then we denote this by T B(H 1, H 2 ). If H 1 = H 2 = H, then B(H 1, H 2 ) is denoted by B(H). For T B(H), the null space and range space are denoted by N(T ) and R(T ) respectively. The unit sphere of a Hilbert space H is denoted by S H. The following conditions are equivalent for a linear operator to be bounded. Date: March 12,

2 2 G. RAMESH Theorem 1.2. Let T : H 1 H 2 be linear. Then the following are equivalent; (1) T is bounded (2) T is continuous at 0 (3) T is uniformly continuous (4) there exists an M > 0 such that T x M x for all x H 1 Definition 1.3. If T is bounded, then by Theorem 1.2, quantity is called the norm of T and is denoted by T. sup T x <. This x S H1 We have the following equivalent formulae for computing the norm of a bounded linear operator. Theorem 1.4. Let T B(H 1, H 2 ). Then the following are equivalent; (1) T = sup { T x : x S H1 } (2) T = sup { T x : x H 1, x 1} T x (3) T = sup { x : x H 1} (4) T = inf {k > 0 : T x k x for all x H 1 }. Remark 1.5. The statement (4) of the above Theorem gives a geometric interpretation of the norm a bounded operator as follows: T is the radius of the smallest ball contatining the image of the unit ball in H 1. For every bounded linear operator there is an another bounded linear operator associated with it in the following way. Definition 1.6. Let T B(H 1, H 2 ). Then there exists a unique operator from H 2 into H 1, denoted by T such that T x, y = x, T y for all x H 1, y H 2. This operator T is called the adjoint of T. We have the following properties of T. (1) (T ) = T (2) T = T (3) if S B(H 2, H 3 ), then (ST ) = T S (4) if R B(H 1, H 2 ), then (R + T ) = R + T (5) (αt ) = ᾱt Remark 1.7. let S, T B(H 1, H 2 ) and α C. Then (a) S + T S + T (b) α T = α T (c) T T = T 2 = T T. Exercise 1.1. Let T B(H). Then show that T = sup { T x, y : x, y S H }. Exercise 1.2. Let T B(H 1, H 2 ). Then (1) N(T ) = R(T ) (2) N(T ) = R(T ) (3) R(T ) = N(T ) (4) R(T ) = N(T )

3 SPECTRAL THEOREM 3 (5) N(T T ) = N(T ) (6) R(T T ) = R(T ) Examples. (1) (identity operator) Let H be a complex Hilbert space. Let I be the identity map on H. Then I = 1. (2) (right shift operator) Let R : l 2 l 2 be given by R(x 1, x 2,... ) = (0, x 1, x 2,... ) for all (x 1, x 2,... ) l 2. Then we can show that Rx = x for all x l 2. Hence R is bounded and R = 1. One can check that R (y 1, y 2, y 3,... ) = (y 2, y 3,... ) for all (y n ) l 2. Note that R = 1. (3) (matrix) Let H 1 be a finite dimensional Hilbert space and H 2 be a Hilbert space. Let T : H 1 H 2 be linear. Then T is bounded. To see this, let {φ 1, φ 2, φ 3,... φ n } be an orthonormal basis for H 1. If x H 1, then x = x, φ k φ k. Hence T x = x, φ k T φ k. Using the Cauchy-Schwarz k=0 k=0 inequality, we can show that ( n ) 1 2 T x T φ j 2 x for all n That is T ( n T φ j 2) 1 2. Exercise 1.4. Solve the following. (a) (diagonal matrix) Let D : C n C n be a linear operator whose matrix with respect to the standard orthonormal basis of C n is the diagonal matrix with entries {λ 1, λ 2,, λ n }. Show that T = max { λ j }. 1 j n (b) (diagonal operator) Let H be a separable Hilbert space with an orthonormal basis {φ n } and (λ n ) be a bounded sequence of complex numbers. Define T : H H by T x = λ n x, φ n φ n, for all x H. n=1 Show that T is bounded and T = sup { λ n : n N}. (c) (multiplication operator)let g ( C[0, 1], ). Define Mg : L 2 [0, 1] L 2 [0, 1] by M g (f) = gf, for all f L 2 [0, 1]. Show that M g is bounded and M g = g. (d) find adjoint of each operator in (a), (b) and (c). Definition 1.8. Let T B(H). Then T is said to be (1) self-adjoint if T = T (2) normal if T T = T T (3) unitary if T T = T T = I (4) isometry if T x = x for all x H (equivalently T T = I ) (5) orthogonal projection if T 2 = T = T.

4 4 G. RAMESH Remark 1.9. If T B(H), then T T and T T are self-adjoint operators. Also T = A + ib, where A = T + T and B = T T. It can be easily checked that 2 2i A = A and B = B. This decomposition is called the cartesian decomposition of T. It can be easily verified that T is normal if and only if AB = BA. Definition Let H be a separable Hilbert space with an orthonormal basis φ 1, φ 2, φ 3,.... Then the matrix of T with respect to φ 1, φ 2, φ 3,... is given by (a ij ) where a ij = T φ j, φ i for i, j = 1, 2, 3..., n. Exercise 1.5. (1) Find the matrix of the right shift operator with respect to the standard orthonormal basis of l 2 (2) show that if (a ij ) is a matrix of T with respect to an orthonormal basis, then (a ji ) is the matrix of T with respect to the same basis. In order to get a spectral theorem analogous to the finite dimensional case, we have to look for bounded operators with the similar properties of the finite dimensional operators. One such property is that If H 1, H 2 are finite dimensional complex Hilbert spaces and T : H 1 H 2 is linear then: For every bounded set S H 1, T (S) is pre compact in H 2 under T.(*) For a set to be compact in a metric space, we have the following; Theorem Let X be a metric space and S X. Then the following conditions are equivalent. (1) S is compact (2) S is sequentially compact (3) S is totally bounded and complete. The following example shows that this property depends on the dimension of the range of the operator. Let H 1 and H 2 be infinite dimensional Hilbert spaces. Let w H 1 and z H 2 be a fixed vectors. Define T : H 1 H 2 by T x = x, w z, for all x H 1. As the range of the operator is one dimensional, T maps bounded sets into pre compact sets. This can be generalized to a bounded operator whose range is finite dimensional as follows: Let w 1, w 2,... w n H 1 and z 1, z 2,..., z n H 2 be fixed vectors. Define T : H 1 H 2 given by (1.1) T x = x, w j z j, for all x H 1. j=1 Then T is linear, bounded and has finite dimensional range. Such operators are called as finite rank operators. Definition Let T B(H). Then T is called a finite rank (rank n say) if R(T ) is finite dimensional. Note 1.6. We have seen that any operator given by Equation (1.1) is a finite rank operator and has the property ( ). Can we express every finite rank operator as in (1.1).

5 SPECTRAL THEOREM 5 Theorem Let K : H 1 H 2 be a bounded operator of rank n. Then there exists vectors v 1, v 2,... v n H 1 and vectors φ 1, φ 2,..., φ n H 2 such that for every x H 1, we have Kx = x, v i φ i. j=1 The vectors φ 1, φ 2,..., φ n may be chosen to be any orthonormal basis for R(K). Proof. Let φ 1, φ 2,..., φ n be an orthonormal basis for R(K). Then Kx = Kx, φ i φ i, for every x H 1. i=1 For each i, the functionals f i (x) = Kx, φ i is a bounded linear functional on H 1. Now by the Riesz Representation theorem, there exists a unique v i H 1 such that f i (x) = x, v i and f i = v i for each i. Hence the result follows. Remark In the above theorem, the representation of K is not unique as it depends on the orthonormal basis and hence on the vectors {v j } n j=1. Can this happen for operators whose range and domain are infinite dimensional?. 2. Compact Operators In this section we discuss the properties of operators which are analogues of the finite dimensional operators. In other words we describe the infinite dimensional operators which have the property ( ). Definition 2.1. Let T : H 1 H 2 be a linear operator. Then T is said to be compact if for every bounded set S H 1, the set T (S) is compact in H 2. Example 2.2. Every m n matrix corresponds to a compact operator. Example 2.3. Every bounded finite rank operator is compact. Notation: The set of all compact operators from H 1 into H 2 is denoted by K(H 1, H 2 ) and if H 1 = H 2 = H, then K(H). Remark 2.4. (1) Every compact operator is bounded. The converse need not be true. For example consider the identity operator I : H H. Clearly I is bounded. Then I is compact if and only if dimension of H is finite. (2) An isometry is compact if and only if it is a finite rank operator. (3) Restriction of a compact operator to a closed subspace is again compact (4) An orthogonal projection onto a closed subspace of a Hilbert space is compact if and only if it is of finite rank. (5) Let T B(H) be a compact operator which is not a finite rank operator. Then R(T ) cannot be closed. In view of Theorem 1.11, the definition of a compact operator can be described as follows. Let T : H 1 H 2 be a bounded operator and B := {x H : x 1}. Then the following conditions are equivalent. (1) T (B) is compact

6 6 G. RAMESH (2) For every bounded sequence (x n ) H 1, (T x n ) has a convergent subsequence in H 2 (3) T maps bounded sets into totally bounded sets Properties. Theorem 2.5. Let T 1, T 2 : H 1 H 2 be compact operators and α C. Then (1) αt 1 is compact (2) T 1 + T 2 is compact. Proof. Proof of (1) is obvious. For the proof of (2), let (x n ) H be a bounded sequence. Since T 1 is compact, T 1 x n has a subsequence T 1 x nk which is convergent, say T 1 x nk y. Now x nk is bounded sequence. Since T 2 is compact, there exists a subsequence (x nkl ) of (x nk ) such that T 2 x nkl is convergent. Note that T 1 x nkl is convergent. Therefore (T 1 + T 2 )x nkl is convergent. Hence T 1 + T 2 is compact. From Theorem 2.5, we can conclude that K(H 1, H 2 ) is a vector subspace of B(H 1, H 2 ). Theorem 2.6. Let A : H 1 H 2 be compact and B : H 3 H 1, C : H 2 H 3 are bounded. Then CA and AC are compact. Proof. Let (x n ) H 1 be a bounded sequence. As A is compact, there exists a subsequence (x nk ) of (x n ) such that Ax nk is convergent. Since C is bounded, CAx nk is also convergent. Hence CA is compact. Now B is bounded (Bx n ) is bounded (ABx n ) has a convergent subsequence, since A is compact AB is compact. Conclusion: From the above two results one can conclude that the set of all compact operator on H is a two sided ideal in B(H), the space of all bounded operators on H. Remark 2.7. By Theorem 2.6, if T K(H), then T 2 K(H). But the converse need not be true. Exercise 2.2. Let T : l 2 l 2 l 2 l 2 given by is not compact (Note that T 2 = 0). T (x, y) = (0, x), (x, y) l 2 l 2 Theorem 2.8. Let T B(H 1, H 2 ). Then (1) T is compact T T or T T is compact (2) T is compact T is compact Proof. Proof of (1): If T is compact, then T T is compact by Theorem 2.6. To prove the converse, assume that T T : H 1 H 1 is compact. If (x n ) H 1 is a bounded sequence with bound M > 0, then T T x n has a convergent subsequence namely T T x nk,

7 SPECTRAL THEOREM 7 say T T x nk y. For notational convenience we denote the subsequence (x nk )by (x n ). For n > m, we have T x n T x m 2 = T (x n x m ), T (x n x m ) = T T (x n x m ), (x n x m ) T T (x n x m ) x n x m 2M T T (x n x m ). Hence (T x n ) is Cauchy and hence convergent. Similarly, T T K(H 2 ). Proof of (2): Let (z n ) H 2 be a bounded sequence. As T T is compact, (T T z n ) has a convergent subsequence, say T T z nk converging to z. Now for k > l, T z nk T z nl 2 = T T (z nk z nl ), z nk z nl T T (z nk z nl ) z nk z nl 0 as n. That is (T z nk ) is Cauchy, hence convergent. Thus T is compact. By the above argument, T is compact implies that T = T is compact. Theorem 2.9. K(H) is a closed in B(H). Proof. Let (K n ) be a sequence of compact operators converging to K. Let M > 0 be such that K n M for all n. Our aim is to show that K is compact. Let (x i ) be a bounded sequence in H. Let (x 1 i ) be a subsequence of (x i) be such that (K 1 x 1 i ) is convergent. Let (x 2 i ) (x1 i ) such that (K 2x 2 i ) is convergent. Let (x3 i ) x2 i be such that (K 3 x 3 i ) is convergent. Continuing this process, let (xn i ) be a subsequence of (x n 1 i ) such that (K n x n i ) is convergent. The sequence (z i ) = (x i i ) is a subsequence of (x i). Also for each n, except the first n terms, (z i ) is a subsequence of (x n i ) such that (K nz i ) is convergent. Now for all i, j and n we have, Kz i Kz i = (K K n )z i + K n z i K n z j (K K n )z j ( ) K K n z i + z j + K n (z i z j ). That is (Kz i ) is Cauchy, hence convergent as H is a Hilbert space. This concludes that K is compact. Lemma Let K be a compact operator on a separable Hilbert space H and suppose that (T n ) B(H) and T B(H) are such that for each x H, the sequence T n x T x. Then T n K T K in the norm of B(H). Proof. Suppose that T n K T K 0. Then there exists a δ > 0 and a subsequence {T nj K} such that T nj K T K > δ. Choose unit vectors (x ni ) of H such that (T nj K T K)(x ni ) > δ.

8 8 G. RAMESH is conver- Since K is compact, we get a subsequence (x nj ) of (x ni ) such that Kx nj gent. Assume that Kx nj y. Then (2.1) δ < (T nj K T K)x nj (T ni T )(Kx nj y) + (T nj T )y Since Kx nj y, there exists n such that for n j > n, Kx nj y < δ 8C. Also as T nj y T y for ach y H, there exists m such that n j > m implies (T T nj )y < δ 4. Since (T n ) B(H) is bounded, then T n C and T x = lim n T n x C. Hence T T nj 2C. Now from Equation 2.1, δ < (T nj K T K)x nj < δ 4 + δ 4 = δ 2, a contradiction. Theorem Every compact operator on a separable Hilbert space H is a norm limit of a sequence of finite rank operators. In other words, the set of finite rank operators is dense in the space of compact operators. Proof. Let {φ n : n N} be an orthonormal basis for H and H n := span{φ k } n. Then the orthogonal projections P n : H H defined by P n x = x, φ j φ j j=1 has the property that P n x x for each x H. Now, if K is compact, then by Lemma 2.10, it follows that P n K K in the operator norm of B(H). Here R(P n K) R(P n ) = H n is finite dimensional. Example Let H = l 2 and {e n } be the standard orthonormal basis of H. Define D : H H by D(x 1, x 2, x 3,... ) = (x 1, x 2 2, x 3 3,... ), for all (x n) H. Then D is bounded. Next, we show that D K(H). Define D n : H H by D n x = x, e j e j. Then D n is finite rank bounded operator and D n D as j=1 n. Hence by theorem 2.11, D is compact. Example Let T = RD and S = DR where R is the right shift operator and D is as in example Then both T and S are compact. Example Let k(, ) L 2 [a, b]. Define K : L 2 [a, b] L 2 [a, b] by (Kf)(s) = b a k(s, t)f(t)dt, for all f L 2 [a, b].

9 SPECTRAL THEOREM 9 It can be verified that K B(L 2 [a, b]). Let {φ n : n N} be an orthonormal basis for L 2 [a, b]. Then ψ m,n (s, t) = φ n (s)φ m (t) for all s, t [a, b] and for all m, n N forms an orthonormal basis for L 2 ([a, b] [a, b]). Hence k(s, t) = k(s, t), ψ m,n (s, t) ψ m,n (s, t). Let k N (s, t) = m,n=1 N m,n=1 Now define K N : L 2 [a, b] L 2 [a, b] by (K N f)(s) = b a k(s, t), ψ m,n (s, t) ψ m,n (s, t). k N (s, t)f(t)dt, for all f L 2 [a, b]. Note that K N is a finite rank operator and K N K as N. Hence by Theorem 2.11, K K(L 2 [a, b]). 3. The Spectral Theorem Definition 3.1. A complex number λ C is called an eigenvalue of T B(H) if there exists a vector 0 x H such that T x = λx. The vector x is called an eigenvector for T corresponding to the eigenvalue λ. Equivalently λ is an eigenvalue of T iff T λi is not one-to-one. Example 3.2. Let H = l 2. Define T : H H by T (x 1, x 2, x 3,... ) = (x 1, x 2 2, x 3 3,... ), (x 1, x 2, x 3,... ) H. Let {e n : n N} denote the standard orthonormal basis for H. Then T e n = 1 n e n. Hence { 1 n } is a set of eigenvalues of T with corresponding eigenvectors e n. Exercise 3.1. Let R : l 2 l 2 be given by R(x 1, x 2,... ) = (0, x 1, x 2,... ), (x 1, x 2,... ) l 2. Find the eigenvalues and eigenvectors of R. Exercise 3.2. Show that the operator T : l 2 l 2 defined by T (x 1, x 2,... ) = (0, x 1 2, x 2 3,... ), (x 1, x 2,... ) is compact but has no eigenvalues Self-adjoint Operators. Definition 3.3. Let T B(H). If T = T, then T is called self-adjoint. The operators in Example 3.2 is self-adjoint, where as the operator in Exercise 3.1 is not. Exercise 3.4. Let M : L 2 [0, 1] L 2 [0, 1] be given by (Mf)(t) = tf(t), f L 2 [0, 1], t [0, 1]. Show that M is self-adjoint and has no eigenvalue. Proposition 3.4. Let T B(H) be self-adjoint. Then (1) eigenvalues of T are real

10 10 G. RAMESH (2) eigenvectors corresponding to distinct eigenvalues are orthogonal. Proof. Proof of 1: Let λ be an eigenvalue of T and x be the corresponding eigen vector. Then T x = λx. λ x 2 = λ x, x = λx, x = T x, x = x, T x = x, λx = λ x 2. Since x 0, we have λ = λ. Proof of 2: Let λ and µ be distinct eigenvalues of T and x and y be the corresponding eigenvectors. Then T x = λx and T y = µy. Now µ x, y = x, µy = x, T y = T x, y = λ x, y. Since λ, µ R and distict, it follows that x, y = 0. Theorem 3.5. Let T B(H) be self-adjoint. Then T = sup T x, x. x =1 Proof. Let m = sup x =1 T x, x. Note that T x, x m for each x H with x = 1. If x = 1, then by the Cauchy-Schwartz-Bunyakovsky inequality, T x, x T x x = T x. Hence m T. To prove the other way, let x, y H. Then T (x ± y), x ± y = T x, x + 2Re T x, y + T y, y. Therefore, 4Re T x, y = T (x + y), x + y T (x y), x y T (x + y), x + y + T (x y), x y m ( x + y 2 + x y 2) = 2m ( x 2 + y 2 ) (by the parallelogram law). Now T x, y = T x, y exp(iθ) for some real θ. Substituting x exp(iθ) in the above equation, we get Substituting y = Hence T = m. 4Re T x exp( iθ), y 2m( x 2 + y 2 ) 4 T x, y 2m( x 2 + y 2 ). x T x in place of y in the above equation, we get T x m x. T x Corollary 3.6. Let T B(H) be self-adjoint. If T x, x = 0, for each x H. Then T = 0. Theorem 3.7. Let T B(H) be self-adjoint. (1) Let λ = inf x =1 T x, x. If there exists an x 0 H such that x 0 = 1 and λ = T x 0, x 0, then λ is an eigenvalue of T with corresponding eigenvector x 0. (2) Let µ = sup x =1 T x, x. If there exists a x 1 H, x 1 = 1 such that µ = T x 1, x 1, then µ is eigenvalue value of T with corresponding eigenvector x 1.

11 SPECTRAL THEOREM 11 Proof. Let α C and v H, by defintion of λ, we have Hence T (x 0 + αv), x 0 + αv λ x 0 + αv, x 0 + αv. T x 0, x 0 + ᾱ T x 0, v + α T v, x 0 + α 2 T v, v That is λ x 0, x 0 + λᾱ x 0, v + λα x 0, v + λ α 2 v, v. T x 0, x 0 +2 αre T x 0, v + α 2 T v, v λ x 0, x 0 +λ α 2 v, v +2 Re αλ x 0, v. Substituting λ = T x 0, x 0 and taking α = r v, (T λi)x 0, r R, we can conclude that v, (T λi)x 0 = 0 for each v H. Hence T x 0 = λx 0. The proof of the other statement follows by taking T in place of T. Theorem 3.8. If T B(H) is compact and self-adjoint, then at least one of the numbers T or T is an eigenvalue. Proof. By Theorem 3.7, there exists a sequence (x n ) H with x n = 1 for each n such that and λ R such that T x n, x n λ, where λ = + T or λ = T. Now T x n λx n 2 = T x n 2 + λ 2 2λ T x n, x n 2λ 2 2λ T x n, x n 0 as n. Since T is compact, there exists a sub sequence (T x nj ) of (T x n ) such that T x nj y. So T x nj λx nj 0 as n. That is x n 1 λ y. Hence y = lim T x n j = 1 λ T y. Hence λ is an eigen value. Corollary 3.9. If T K(H) be self-adjoint, then max T x, x = T. x =1 Exercise 3.5. Let K : L 2 [0, 1] L 2 [0, 1] be defined by (Kf)(t) = t 0 f(s)ds. Check that K is non self-adjoint, compact and does not possesses eigenvalues. Definition A closed subspace M of H is said to be invariant under T B(H) if and only if T (M) M. If both M and M are invariant under T, then we say that M is a reducing subspace for T. Exercise 3.6. M is invariant under T iff M is invariant under T. Exercise 3.7. Let T B(H). Let M be a closed subspace of a Hilbert space and P : H H be an orthogonal projection such that R(P ) = M. Then (1) M is invariant under T iff T P = P T P (2) M reduces T iff T P = P T. Observations (1) If T is a self-adjoint operator, then every invariant subspace is reducing. (2) every eigen space corresponding to a particular eigenvalue is invariant under the operator. In particular, if the operator is self-adjoint every eigen space reduce the operator.

12 12 G. RAMESH Note 3.8. If T is compact operator and M is a closed subspace of H, then the restriction operator T M is also compact. Theorem 3.11 (The Spectral theorem). Suppose T K(H) be self-adjoint. Then there exists a system of orthonormal vectors φ 1, φ 2, φ 3,... of eigenvectors of T and corresponding eigenvalues λ 1, λ 2, λ 3,... such that λ 1 λ 2 λ 3..., T x = λ k x, φ k φ k for all x H. If (λ n ) is infinite, then λ n 0 as n. converges in the operator norm of B(H). Proof. We prove the theorem step by step. Step 1: Construction of eigen vectors The series on the right hand side We use Theorem 3.8 repeatedly for constructing eigenvalues and eigenvectors. Let H 1 = H and T 1 = T. Then by Theorem 3.8, there exists an eigenvalue λ 1 of T 1 and an eigenvector φ 1 such that φ 1 and λ 1 = T 1. Now span{φ 1 } is a closed subspace of H 1, hence by the projection theorem, H 1 = span{φ 1 } span{φ 1 }. Let H 2 = span{φ 1 }. Clearly H 2 is a closed subspace of H 1 and T (H 2 ) H 2. Now let T 2 = T 1 H2. Then T 2 is a compact and self-adjoint operator in B(H 2 ). If T 2 = 0, then there is nothing to prove. Assume that T 2 0. Then by Theorem 3.8, there exists an eigenvalue λ 2 of T 2 with λ 2 = T 2 and a corresponding eigenvector φ 2 with φ 2 = 1. Since T 2 is a restriction of T 1, λ 2 = T 2 T 1 = λ 1. By the construction φ 1 and φ 2 are orthonormal. Now let H 3 = span{φ 1, φ 2 }. Clearly H 3 H 2. It is easy to show that T (H 3 ) H 3. The operator T 3 = T H3 is compact and self-adjoint. Hence by Theorem 3.8, there exists an eigenvalue λ 3 of T 3 and a corresponding eigenvector φ 3 with φ 3 = 1. Here λ 3 = T 3. Hence λ 3 λ 2 λ 1. Proceeding in this way, either after some stage n, T n = 0 or there exists a sequence λ n of eigenvalues of T and corresponding vector φ n with φ n = 1 and λ n = T n. Also λ n+1 λ n for each n. Step 2: If (λ n ) is infinite, then λ n 0 If λ n 0, there exists an ɛ > 0 such that λ n ɛ for infinitely many n s. If n m, T φ n T φ m 2 = λ n φ n λ m φ m 2 = λ 2 m + λ 2 n > ɛ 2. This shows that (T φ n ) has no convergent subsequence, a contradiction to the compactness of T. Hence λ n 0 as n. Step 3: Representation of T Here we consider two cases Case 1: T n = 0 for some n

13 SPECTRAL THEOREM 13 Let x n = x x, φ k φ k. Then x n φ i for 1 i n. Hence That is T x = λ k x, φ k φ k. Case 2: T n 0 for infinitely many n 0 = T n x n = T x λ k x, φ k φ k. For x H, by Case 1, we have T x λ k x, φ k φ k = T n x n T n x n = λ n x n λ n x 0. Hence T x = λ k x, φ k φ k. Observations: (a) N(T ) = span{φ 1, φ 2,...} (b) By the projection theorem H = N(T ) N(T ) = N(T ) R(T ). The spectral theorem guarantees the existstence of the orthonormal basis for R(T ). Hence R(T ) is separable. In addition if N(T ) is separable, then H is separable. (c) H = N(T ) G i, where G n = Hn = n 1 i=1 N(T λ ii). (d) T N(T ) = diag(λ 1, λ 2,..., λ n,... ). (e) each λ j is repeated in the sequence {λ n } exactly p j = dimn(t λ j I) times. Since the sequence λ n 0, each λ j appears finitely many times. Let λ j = λ ni, i = 1, 2,... p. Then N(T λ j I) = span{φ n1,..., φ np }. If this is not possible, then there exists a 0 v N(T λ j I) such that v φ ni where 1 i p. If k n i 1 i p, v φ k as λ j λ k. Hence λ j v = T v = k λ k v, φ k φ k = 0 which is impossible because λ j 0 and v 0. Example Define D : l 2 l 2 by D(x 1, x 2,... ) = (x 1, x 2 2,... ) (x 1, x 2,... ) l 2. Then T K(H) and self-adjoint. Note that De n = 1 n e n. Hence 1 is an eigenvalue n with an eigen vector e n. Also 1 n 0 as n. We can also represent D as 1 D(x) = n x, e n e n for all x = (x n ) l 2. n=0

14 14 G. RAMESH Exercise 3.9. (1) Let K K(H). Show that R(K λi) is closed for each λ C \ {0}. The converse of the spectral theorem is also true. Theorem Suppose that φ 1, φ 2,... is a sequence of orthogonal vectors in H and (λ k ) be a sequence of real numbers such that (λ k ) is finite or converges to 0. Then the operator defined by T x = λ k x, φ k φ k is compact and self-adjoint. Proof. We consider the following cases. Case 1: (λ k ) is finite Consider T x = λ k x, φ k φ k. Then T x 2 = T x, T x = λ k 2 x, φ k 2 max λ k x 2. k This shows that T B(H) and T is a finite rank operator. Hence T is compact. Case 2: (λ k ) is infinite and λ k 0 as n. By the Spectral Theorem, T x 2 = n λ n 2 x, φ n 2 max k x 2. k n Hence T <. Now define T n x = λ k x, φ k φ k. Then T T n 2 = sup λ k x, φ k φ k 2 x =1 k=n+1 sup λ k 2 0 as n. k>n As each T n is finite rank and hence compact, T is compact. It easy to verify that T = T Second form of the Spectral Theorem. Definition An orthonormal system φ 1, φ 2,... of eigenvectors of T B(H with corresponding non zero eigen values λ 1, λ 2,... is called a basic system of eigenvalues and eigenvectors of T if T x = k λ k x, φ k φ k. The Spectral Theorem guarantees the existence of a basic system of eigenvalues and eigenvectors for a compact self-adjoint operator. Theorem Let T be a compact self-adjoint operator and µ j be the set of all non zero eigenvalues of T and let P j be the orthogonal projection onto N(T µ j I). Then

15 SPECTRAL THEOREM 15 (1) P j P k = 0, j k (2) T = j µ jp j, where the convergence of the series is with respect to the norm of B(H). (3) For each x H, x = P 0 x + P j x, j where P 0 is the orthogonal projection onto N(T ). Proof. Let {φ n }, {λ n } be a basic system of eigen vectors and eigen values of T. For each k, a subset of {φ n } is an orthonormal basis for N(T µ k I) say, φ ni 1 i p. Then P k x = p i=1 x, φ n i φ ni since each λ n is µ k, it follows that x = P 0 x + P k x and T x = λ k P k x. k Furthermore P j P k = 0, j k, since N(T µ j I) N(T µ k I). If (λ n ) is an infinite sequence, then T µ k P k 2 = sup T x µ k P k x 2 x =1 sup x =1 j n λ 2 j x, φ j 2 sup λ j 2 0 as n. j>n