Operators

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6 Operators

All the space’s a stage,
and all functionals and operators merely players!

All our previous considerations were only a preparation of the stage and now the main actors come forward to perform a play. The vectors spaces are not so interesting while we consider them in statics, what really make them exciting is the their transformations. The natural first steps is to consider transformations which respect both linear structure and the norm.

6.1 Linear operators

Definition 1 A linear operator T between two normed spaces X and Y is a mapping T:X→ Y such that T(λ v + µ u)=λ T(v) + µ T(u). The kernel of linear operator kerT and image are defined by

kerT ={x∈ X: Tx=0} Im T={y∈ Y: y=Tx, for some x∈ X}.

Exercise 2 Show that kernel of T is a linear subspace of X and image of T is a linear subspace of Y.

As usual we are interested also in connections with the second (topological) structure:

Definition 3 A norm of linear operator is defined:

⎪⎪
⎪⎪

=sup{

⎪⎪
⎪⎪

_Y:

⎪⎪
⎪⎪

_X≤ 1}. (37)

T is a bounded linear operator if ||T||=sup{||Tx||: ||x||}<∞.

Exercise 4 Show that ||Tx||≤ ||T||·||x|| for all x∈ X.

Example 5 Consider the following examples and determine kernel and images of the mentioned operators.

On a normed space X define the zero operator to a space Y by Z: x→ 0 for all x∈ X. Its norm is 0.
On a normed space X define the identity operator by I_X: x→ x for all x∈ X. Its norm is 1.
On a normed space X any linear functional define a linear operator from X to ℂ, its norm as operator is the same as functional.
The set of operators from ℂⁿ to ℂ^m is given by n× m matrices which acts on vector by the matrix multiplication. All linear operators on finite-dimensional spaces are bounded.
On l₂, let S(x₁,x₂,…)=(0,x₁,x₂,…) be the right shift operator. Clearly ||Sx||=||x|| for all x, so ||S||=1.

On L₂[a,b], let w(t)∈ C[a,b] and define multiplication operator M_wf by (M_w f)(t)=w(t)f(t). Now:

⎪⎪
⎪⎪

M_w f

⎪⎪
⎪⎪

∫

⎪
⎪

w(t)

⎪
⎪

f(t)

⎪
⎪

² d t

≤

K²

∫

⎪
⎪

f(t)

⎪
⎪

² d t, where K=

⎪⎪
⎪⎪

_∞=

sup

[a,b]

⎪
⎪

w(t)

⎪
⎪

so ||M_w||≤ K.

Exercise 6 Show that for multiplication operator in fact there is the equality of norms ||M_w||₂= ||w(t)||_∞.

Theorem 7 Let T: X → Y be a linear operator. The following conditions are equivalent:

T is continuous on X;
T is continuous at the point 0.
T is a bounded linear operator.

Proof. Proof essentially follows the proof of similar Theorem 4. □

6.2 Orthoprojections

Here we will use orthogonal complement, see § 3.5, to introduce a class of linear operators—orthogonal projections. Despite of (or rather due to) their extreme simplicity these operators are among most frequently used tools in the theory of Hilbert spaces.

Corollary 8 (of Thm. 23, about Orthoprojection) Let M be a closed linear subspace of a hilbert space H. There is a linear map P_M from H onto M (the orthogonal projection or orthoprojection) such that

P_M²=P_M, kerP_M=M^⊥, P_M^⊥=I−P_M. (38)

Proof. Let us define P_M(x)=m where x=m+n is the decomposition from the previous theorem. The linearity of this operator follows from the fact that both M and M^⊥ are linear subspaces. Also P_M(m)=m for all m∈ M and the image of P_M is M. Thus P_M²=P_M. Also if P_M(x)=0 then x⊥ M, i.e. kerP_M=M^⊥. Similarly P_M^⊥(x)=n where x=m+n and P_M+P_M^⊥=I. □

Example 9 Let (e_n) be an orthonormal basis in a Hilber space and let S⊂ ℕ be fixed. Let M=CLin{e_n: n∈ S} and M^⊥=CLin{e_n:n∈ ℕ∖ S}. Then

∞

∑

k=1

a_k e_k =

∑

k∈ S

a_k e_k +

∑

k∉S

a_k e_k.

Remark 10 In fact there is a one-to-one correspondence between closed linear subspaces of a Hilber space H and orthogonal projections defined by identities (38).

6.3 B(H) as a Banach space (and even algebra)

Theorem 11 Let B(X,Y) be the space of bounded linear operators from X and Y with the norm defined above. If Y is complete, then B(X,Y) is a Banach space.

Proof. The proof repeat proof of the Theorem 8, which is a particular case of the present theorem for Y=ℂ, see Example 3. □

Theorem 12 Let T∈ B(X,Y) and S∈B(Y,Z), where X, Y, and Z are normed spaces. Then ST∈B(X,Z) and ||ST||≤||S||||T||.

Proof. Clearly (ST)x=S(Tx)∈ Z, and

⎪⎪
⎪⎪

STx

⎪⎪
⎪⎪

≤

⎪⎪
⎪⎪

≤

⎪⎪
⎪⎪

which implies norm estimation if ||x||≤1. □

Corollary 13 Let T∈ B(X,X)=B(X), where X is a normed space. Then for any n≥ 1, Tⁿ∈B(X) and ||Tⁿ||≤ ||T||ⁿ.

Proof. It is induction by n with the trivial base n=1 and the step following from the previous theorem. □

Remark 14 Some texts use notations L(X,Y) and L(X) instead of ours B(X,Y) and B(X).

Definition 15 Let T∈ B(X,Y). We say T is an invertible operator if there exists S∈ B(Y,X) such that

ST= I_X and TS=I_Y.

Such an S is called the inverse operator of T.

Exercise 16 Show that

for an invertible operator T:X→ Y we have ker T={0} and ℑ T=Y.
the inverse operator is unique (if exists at all). (Assume existence of S and S′, then consider operator STS′.)

Example 17 We consider inverses to operators from Exercise 5.

The zero operator is never invertible unless the pathological spaces X=Y={0}.
The identity operator I_X is the inverse of itself.
A linear functional is not invertible unless it is non-zero and X is one dimensional.
An operator ℂⁿ→ ℂ^m is invertible if and only if m=n and corresponding square matrix is non-singular, i.e. has non-zero determinant.
The right shift S is not invertible on l₂ (it is one-to-one but is not onto). But the left shift operator T(x₁,x₂,…)=(x₂,x₃,…) is its left inverse, i.e. TS=I but TS≠I since ST(1,0,0,…)=(0,0,…). T is not invertible either (it is onto but not one-to-one), however S is its right inverse.
Operator of multiplication M_w is invertible if and only if w⁻¹∈C[a,b] and inverse is M_w⁻¹. For example M_1+t is invertible L₂[0,1] and M_t is not.

6.4 Adjoints

Theorem 18 Let H and K be Hilbert Spaces and T∈B(H,K). Then there exists operator T^*∈ B(K,H) such that

⟨ Th,k ⟩_K=⟨ h,T^*k ⟩_H for all h∈ H, k∈ K.

Such T^* is called the adjoint operator of T. Also T^**=T and ||T^*||=||T||.

Proof. For any fixed k∈ K the expression h:→ ⟨ Th,k ⟩_K defines a bounded linear functional on H. By the Riesz–Fréchet lemma there is a unique y∈ H such that ⟨ Th,k ⟩_K=⟨ h,y ⟩_H for all h∈ H. Define T^* k =y then T^* is linear:

⟨ h,T^*(λ₁k₁+λ₂k₂) ⟩_H	=	⟨ Th,λ₁k₁+λ₂k₂ ⟩_K
	=	λ₁⟨ Th,k₁ ⟩_K+λ₂⟨ Th,k₂ ⟩_K
	=	λ₁⟨ h,T^k₁ ⟩_H+λ₂⟨ h,T^k₂ ⟩_K
	=	⟨ h,λ₁T^k₁+λ₂T^k₂ ⟩_H

So T^*(λ₁k₁+λ₂k₂)=λ₁T^*k₁+λ₂T^*k₂. T^** is defined by ⟨ k,T^**h ⟩=⟨ T^*k,h ⟩ and the identity ⟨ T^**h,k ⟩=⟨ h,T^*k ⟩=⟨ Th,k ⟩ for all h and k shows T^**=T. Also:

⎪⎪
⎪⎪

T^* k

⎪⎪
⎪⎪

⟨ T^*k,T^*k ⟩=⟨ k,TT^*k ⟩

≤

⎪⎪
⎪⎪

TT^*k

⎪⎪
⎪⎪

≤

⎪⎪
⎪⎪

T^*k

⎪⎪
⎪⎪

which implies ||T^*k||≤||T||·||k||, consequently ||T^*||≤||T||. The opposite inequality follows from the identity ||T||=||T^**||. □

Exercise 19

For operators T₁ and T₂ show that
(T₁T₂)^*=T₂^*T₁^*, (T₁+T₂)^*=T₁^*+T₂^* (λ T)^*=λT^*.
If A is an operator on a Hilbert space H then (kerA)^⊥= Im A^*.

6.5 Hermitian, unitary and normal operators

Definition 20 An operator T: H→ H is a Hermitian operator or self-adjoint operator if T=T^*, i.e. ⟨ Tx,y ⟩=⟨ x,Ty ⟩ for all x, y∈ H.

Example 21

On l₂ the adjoint S^* to the right shift operator S is given by the left shift S^*=T, indeed:
⟨ Sx,y ⟩ = ⟨ (0,x₁,x₂,…),(y₁,y₂,…) ⟩

= x₁ȳ₂+x₂y_3+⋯=⟨ (x₁,x₂,…),(y₂,y₃,…) ⟩

= ⟨ x,Ty ⟩.

Thus S is not Hermitian.
Let D be diagonal operator on l₂ given by
D(x₁,x₂,…)=(λ₁ x₁, λ₂ x₂, …).

where (λ_k) is any bounded complex sequence. It is easy to check that ||D||=||(λ_n)||_∞=sup_k| λ_k | and
D^* (x₁,x₂,…)=(λ₁ x₁, λ₂ x₂, …),

thus D is Hermitian if and only if λ_k∈ℝ for all k.
If T: ℂⁿ→ ℂⁿ is represented by multiplication of a column vector by a matrix A, then T^* is multiplication by the matrix A^*—transpose and conjugate to A.

Exercise 22 Show that for any bounded operator T operators T_r=1/2(T+ T^*), T_i=1/2i(T− T^*), T^*T and TT^* are Hermitians. Note, that any operator is the linear combination of two hermitian operators: T=T_r+i T_i (cf. z= ℜ z + i ℑ z for z∈ℂ).

To appreciate the next Theorem the following exercise is useful:

Exercise 23 Let H be a Hilbert space. Show that

For x∈ H we have ||x||= sup { | ⟨ x,y ⟩ | for all y∈ H such that ||y||=1}.
For T∈B(H) we have
⎪⎪
⎪⎪ T ⎪⎪
⎪⎪ = sup {  ⎪
⎪ ⟨ Tx,y ⟩ ⎪
⎪ for all x,y∈ H such that ⎪⎪
⎪⎪ x ⎪⎪
⎪⎪ = ⎪⎪
⎪⎪ y ⎪⎪
⎪⎪ =1}. (39)

The next theorem says, that for a Hermitian operator T the supremum in (39) may be taken over the “diagonal” x=y only.

Theorem 24 Let T be a Hermitian operator on a Hilbert space. Then

⎪⎪
⎪⎪

sup

⎪⎪
⎪⎪

= 1

⎪
⎪

⟨ Tx,x ⟩

⎪
⎪

Proof. If Tx=0 for all x∈ H, both sides of the identity are 0. So we suppose that ∃ x∈ H for which Tx≠ 0.

We see that | ⟨ Tx,x ⟩ |≤ ||Tx||||x|| ≤ ||T||||x²||, so sup_{||x|| =1} | ⟨ Tx,x ⟩ |≤ ||T||. To get the inequality the other way around, we first write s:=sup_{||x|| =1} | ⟨ Tx,x ⟩ |. Then for any x∈ H, we have | ⟨ Tx,x ⟩ |≤ s||x²||.

We now consider

⟨ T(x+y),x+y ⟩ =⟨ Tx,x ⟩ +⟨ Tx,y ⟩+⟨ Ty,x ⟩ +⟨ Ty,y ⟩ = ⟨ Tx,x ⟩ +2ℜ ⟨ Tx,y ⟩ +⟨ Ty,y ⟩

(because T being Hermitian gives ⟨ Ty,x ⟩=⟨ y,Tx ⟩ =⟨ Tx,y ⟩) and, similarly,

⟨ T(x−y),x−y ⟩ = ⟨ Tx,x ⟩ −2ℜ ⟨ Tx,y ⟩ +⟨ Ty,y ⟩.

Subtracting gives

4ℜ ⟨ Tx,y ⟩

= ⟨ T(x+y),x+y ⟩−⟨ T(x−y),x−y ⟩

≤ s(

⎪⎪
⎪⎪

x+y

⎪⎪
⎪⎪

² +

⎪⎪
⎪⎪

x−y

⎪⎪
⎪⎪

²)

= 2s(

⎪⎪
⎪⎪

² +

⎪⎪
⎪⎪

²),

by the parallelogram identity.

Now, for x∈ H such that Tx≠ 0, we put y=||Tx||⁻¹||x|| Tx. Then ||y|| =||x|| and when we substitute into the previous inequality, we get

⎪⎪
⎪⎪

=4ℜ⟨ Tx,y ⟩ ≤ 4s

⎪⎪
⎪⎪

x²

⎪⎪
⎪⎪

So ||Tx||≤ s||x|| and it follows that ||T||≤ s, as required. □

Definition 25 We say that U:H→ H is a unitary operator on a Hilbert space H if U^*=U⁻¹, i.e. U^*U=UU^*=I.

Example 26

If D:l₂→l₂ is a diagonal operator such that D e_k=λ_k e_k, then D^* e_k=λ_k e_k and D is unitary if and only if | λ_k |=1 for all k.
The shift operator S satisfies S^*S=I but SS^*≠ I thus S is not unitary.

Theorem 27 For an operator U on a complex Hilbert space H the following are equivalent:

U is unitary;
U is surjection and an isometry, i.e. ||Ux||=||x|| for all x∈ H;
U is a surjection and preserves the inner product, i.e. ⟨ Ux,Uy ⟩=⟨ x,y ⟩ for all x, y∈ H.

Proof. 1⇒2. Clearly unitarity of operator implies its invertibility and hence surjectivity. Also

⎪⎪
⎪⎪

²=⟨ Ux,Ux ⟩=⟨ x,U^*Ux ⟩=⟨ x,x ⟩=

⎪⎪
⎪⎪

².

2⇒3. Using the polarisation identity (cf. polarisation in equation (16)):

4⟨ Tx,y ⟩

⟨ T(x+y),x+y ⟩+i⟨ T(x+iy),x+iy ⟩

−⟨ T(x−y),x−y ⟩−i⟨ T(x−iy),x−iy ⟩.

∑

k=0

i^k⟨ T(x+i^ky),x+i^ky ⟩

Take T=U^*U and T=I, then

4⟨ U^*Ux,y ⟩

∑

k=0

i^k⟨ U^*U(x+i^ky),x+i^ky ⟩

∑

k=0

i^k⟨ U(x+i^ky),U(x+i^ky) ⟩

∑

k=0

i^k⟨ (x+i^ky),(x+i^ky) ⟩

4⟨ x,y ⟩.

3⇒1. Indeed ⟨ U^*U x,y ⟩=⟨ x,y ⟩ implies ⟨ (U^*U−I)x,y ⟩=0 for all x,y∈ H, then U^*U=I. Since U is surjective, for any y∈ H there is x∈ H such that y=Ux. Then, using the already established fact U^*U=I we get

UU^* y = UU^*(Ux) = U(U^*U)x = Ux= y.

Thus we have UU^*=I as well and U is unitary. □

Definition 28 A normal operator T is one for which T^*T=TT^*.

Example 29

Any self-adjoint operator T is normal, since T^*=T.
Any unitary operator U is normal, since U^*U=I=UU^*.
Any diagonal operator D is normal , since D e_k=λ_k e_k, D^* e_k=λ_k e_k, and DD^*e_k=D^*D e_k=| λ_k |² e_k.
The shift operator S is not normal.
A finite matrix is normal (as an operator on l₂ⁿ) if and only if it has an orthonormal basis in which it is diagonal.

Remark 30 Theorems 24 and 2 draw similarity between those types of operators and multiplications by complex numbers. Indeed Theorem 24 said that an operator which significantly change direction of vectors (“rotates”) cannot be Hermitian, just like a multiplication by a real number scales but do not rotate. On the other hand Theorem 2 says that unitary operator just rotate vectors but do not scale, as a multiplication by an unimodular complex number. We will see further such connections in Theorem 17.


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Last modified: November 6, 2024.