operator theory

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OPERATOR THEORY LECTURE NOTES UNIT III FOR M.SC., MATHEMATICS

INVITATION TO LINEAR OPERATORS FROM MATRICES TO BOUNDED LINEAR OPERATORS ON A HILBERT SAPCE BY TAKAYUKI FURUTA

Dr. D. SENTHILKUMAR DEPARTMENT OF MATHEMATICS GOVT. ARTS COLLEGE COIMBATORE – 18

Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

2.4. SPECTRUM OF AN OPERATOR 2.4.1 TWO KINDS OF CLASSIFICATION OF SPECTRUM DEFINITION 1: An operator T on a Hilbert space H is said to be an invertible operator if there exists an operator S such that ST = TS = I, where I is the identity operator, that is, Ix = x for all x ϵ H. We shall write S = T-1 and call T-1 the inverse of T. THEOREM 2: If T is an operator and c is a positive number such that ||Tx|| ≥ c ||x|| for every vector x Є H, then R(T), the range of T, is closed. Proof: fOr n=1,2,.. and yn→y0

Assume Since

– ≥ c ||

-

||

And, { } is a Cauchy sequence, { } is also Cauchy sequence and there exists → because H is a Hilbert space. –

Let, – – And

ϵ H such that

=T

– –

ϵ R(T), that is R(T) is closed.

THEOREM 3: An operator T on a Hilbert space H is invertible if and only if the following (i) and (ii) holds. (i) (ii)

There exists a positive number c such that R(T) is the range of T, is dense in H, that is

holds for any x ϵ H =H

Proof: If T is invertible and y ϵ H, there exists x ϵ H such that

and

.

It follows that R(T) is not only dense in H, but also coincides with H itself, that is (ii) holds. Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

We have,

 ,

where

Then we have (i). Suppose that (i) and (ii) hold, then R(T)=H holds If

= R(T) and

by (ii),

, then

– – – – So . This implies that every vector y has not only the form Tx for some x Є H, but also there is exactly one such x, and a single valued transformation S of H into itself is defined by . since S is easily verified to be linear, and since

It follows that S is an operator such that

and

,

i.e., S is the inverse of T Hence the proof. COROLLARY 4: If

for some

, then T is invertible

Proof: As

by Schwarz inequality, then


.

Let y be orthogonal to R(T), i.e.,

for all x, then

, so that

since

i.e., Hence T is invertible. DEFINITION 5: (i)

σ (T) of T is defined as σ(T) = {λ Є ℂ : T-λ is not invertible} and σ(T) is said to be the spectrum of T. ρ(T) of T is defined by ρ(T) = ℂ - σ(T) and ρ(T) is said to be the resolvent of T.

(ii)

Pσ(T) is said to be the point spectrum of T Pσ(T) = { λ Є ℂ : there exists x ≠ 0 such that Tx=λx}

(iii)

Cσ(T) is said to be the continuous spectrum of T Cσ(T) = { λ Є ℂ :

(iv)

is unbounded and

}

Rσ(T) is said to be the residual spectrum of T Rσ(T) = { λ Є ℂ :

exists and

}

PROPOSITION 6: σ(T) = Pσ(T) σ(T).

Cσ(T)

Rσ(T) holds, where Pσ(T) , Cσ(T) , Rσ(T) are mutually disjoint parts of


.

DEFINITION 7 : (i)

Aσ(T) is said to be the appropriate point spectrum of T, if

Aσ(T) = { λ Є ℂ : there exists a sequence of unit vector {xn} such that ||Txn – λxn||→0 as n→∞}

(ii)

Γ(T) is said to be the compression spectrum of T, if Γ (T) = { λ Є ℂ :

}

PROPOSITION 8 : σ(T) = Aσ(T)

Γ(T) holds, where Aσ(T) and Γ(T) are not necessarily disjoint parts of σ(T).

THEOREM 9: If T is an operator such that ||I-T|| < 1 , then T is invertible. Proof: Let ||I-T|| = 1-α, where 0 < α ≤ 1 Then, –

We have only to show

inorder to prove the invertible of T.

Let S=inf{||y-x||: x Є R(T)} for an arbitrary vector y Є H. We have only to show S=0 for the invertibility of T. If δ >0, then there exists a vector x Є R(T) such that

. Since x,


.

and also

, i.e.,

–

This contradiction proves δ=0 Remark 10 : If a is a non-zero real number such that |1-a| < 1, then a-1 can be expressed as follows: = =

THEOREM 11: If T is an operator, then σ(T) is a compact subset of the complex number plane if

, then

Proof: If λ0 Then ,

σ(T) so that (T-λ0)is invertible, –

And therefore – henever is sufficiently small. If T is an operator such that , then T is invertible. That is is invertible and is also invertible whenever is sufficiently small.this yields that the complement of σ(T) is an open subset of the complex plane. That is σ(T) is closed subset of complex plane. Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

Next we show the second assertion of the theorem. It follows that λ  σ(T) and hence the contraposition asserts that if λ Є σ(T), then |λ|≤║T║. Hence proved. THEOREM 12: If T is an operator, then Aσ (T) is a compact of the complex number PROOF: If λ0  Aσ(T), then there exists a positive number Є such that

for all unit vectors x.

therefore if x is aunit vector and if

,then

So that , this means that the complement of Aσ(T) is an open set,that is Aσ(T) is  closed set, and Aσ(T) is a compact subset of the complex plane. Since Aσ(T) is a subset of σ(T). THEOREM 13 If T is a self adjoint operator on a Hilbert space H, then all the eigen values of T are real numbers. PROOF: If holds for all x H , then

=λ

( self adjoint T*=T)

(

)

THEOREM 14: If T is a self adjoint operator on a Hilbert space H, then

has a bounded inverse operator.

PROOF: Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

– – – –

And we have only to prove

= H,

Let yϵH such that y perpendicular Then

( T*=T) – for all , so that we have – ,(i.e) and this equation does not hold because the eigenvalues of the self adjoint operator T must be real. Therefore , so that =H. Hence proved. (1) An operator T is said to be range of T(R(T)) if R(T) = {Tx : x Є H} (2) An operator T is said to be kernel of T(N(T)) if N(T) = {x Є H : Tx = 0} THEOREM 15: If T is any operator on a Hilbert space H. Then the following (i) and (ii) holds. (i) (ii)

H= H=

 N(T*) and  N(T)

PROOF: Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

Let H = H=

 N(T*) and  N(T) the two relations and H = M  M˔

Then Where

=M

Hence proved. THEOREM 16: If

then

PROOF: Since

by the definition of holds. There exists a units vector x such that

and

because That is,

.

COROLLARY 17: If λ Є Rσ(T),then

Є Pσ(T*)

PROOF: Let that is

by the definition of and holds.there exists a unit vector x such that T*x= x because (i,e) .

THEOREM 18: If an operator T is normal, then

hold

PROOF: Let σ(T) Є Aσ(T) holds, we have only to prove σ(T) such that

Aσ(T).if λ  Aσ (T).then there exists > 0

for all holds for any z by the normality of T and,

and since

To prove  to show

for all

, we have only to show that (T-λ) has a bounded inverse, that is we have only = H. If y perpendicular then we have


.

for all x H and therefore . If

and

, so that we have

=H

THEOREM 19: If an operator T is normal, then PROOF: If λ= Rσ(T),then Pσ(T). Since ║T*y- y║ = ║Ty-λy║ holds for any y by normality of T,λ Pσ(T) hold but is a contradiction since Rσ(T)  Pσ(T)=φ (i,e)Rσ(T)=φ THEOREM 20: If an operator T is self adjoint, then σ(T) is a subset of the real line. PROOF: If λ is not a real number ,then for all non zero vector x,

=|((T-λ)x,x)-((T-λ)x,x)| (S-I) Therefore λ  Aσ(T) and the proof is complete. Since σ(T)= Aσ(T) holds for a self adjoint operator T. THEOREM 21: Let T be a normal operator, Tx=λx and Ty=μy where λ≠μ .then (x,y)=0 PROOF: Let Ty=μy  T*y= y Since ║Ty-μy║ = ║ T*y= y ║ by normality of T. Then,


.



where

Theorem.22: The following two conditions on an operator T are equivalent. (i) T has an approximate point spectrum µ

such that

(ii) sup Proof: Case 1: (i) => (ii) Let a sequence {xn} of unit vectors such that,

→µ,

So that

(i.e)

as

n→∞

Since We have

.

Case 2: (ii) =>(i) Let

→µ such that

Since,


.

Therefore µ is an approximate an point spectrum.

1. An operator T is said to be Numerical Radius If 2. operator T is said to be Numerical Range If .

Theorem.23: For any operators A and B, of σ(AB) and σ(BA) are the same.

holds, that is the non-zero elements

Proof: We have only to prove that if

,

Then is invertible if and only if BA-λ is invertible. To prove that is invertible, Then

is invertible. −−−−−−→ 1

If there exists C the inverse of

, −−−−−−−→2

(i.e)

−−−−−−−→3

Then, From 1 and 3 We get,

Remark.24: The inverse of

say C, then


.

The inverse of ( I -BA) say D,then

2.4.2. Spectral Mapping Theorem:[SMT] Theorem[SMT].25: Let σ(T) be the spectrum of an operator T, and P(t) be any polynomial of a complex number t. Then Proof: If an arbitrary so that

,then there exists g(λ), such that

λ0

,and

If i.e,

.

Conversely, Let

,then there exists λ1,λ2,λ3,……,λn.

Such that and

So that there exists some natural number k. such that λk€σ(T) satisfying λ0=P(λk)ϵP(σ(T)), i.e,

σ(P(T))≤P(σ(T))


.

consequently we have σ(P(T))=P(σ(T)). Example.26: find σ(T)=? And σ(T2)=?

If T= Soln: |T-λ|=0

=0 (2-λ)(1-λ)-6=0 2-2λ-λ+λ2-6=0 λ2-3λ-4=0 (λ-4)(λ+1)=0 (λ-4)=0, λ=4 (i) (ii)

(λ+1)=0 λ=-1 σ(T)={4,-1} σ(T2)=[ σ(T)]2 ={42,(-1)2} ={16,1}

Theorem.27: Let σ(T) be the spectrum of an invertible operator. then Then σ(T-1)=[σ(T)]-1 Proof: The invertible of T ensures 0₡σ(T),so that {σ(T)}-1. i.e, T-1-λ-1=(λ-T)λ-1T-1 asserts the following result λ₡σ(T) if and only if λ-1₡σ(T)-1. Then σ(T)={σ(T)}-1. Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

Hence the proof.

Theorem.28: Let σ(T) be the spectrum of an operator T, Then σ(T*)={ σ(T)}*={λ*: λ€ σ(T)}. Proof: If λ₡ σ(T), so (T-λ) is invertible, Then T*-λ* is also invertible, that is λ*₡ σ(T*), Equivalently, σ(T*)≤{ σ(T)}*

−−−−−→ 1

If λ₡σ(T), so (T-λ) is invertible, Then

T*-λ* is also invertible, i.e, λ*₡ σ(T*)

which is implies, σ(T*)≥{ σ(T)}*

−−−−−→ 2

From 1 and 2 we get, σ(T*)={ σ(T)}*

2.5.Numerical Range Of An Operator 2.5.1. Numerical range is a convex set: Definition.29:An operator Ton a Hilbert H is said to be numerical range W(T) if

W(T)={

:||x||=1}.

Theorem.30: [Toeplitz-hausdorff theorem] The numerical range W(T) of an operator T is a convex set in the complex plane. Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

Proof: Suppose that T is an operator on a Hilbert space H. ᶓ=

and η=

,

where x and y are unit vector xϵH It suffices to prove that every point of segment joining ᶓ and η is in W(T). If ᶓ=η, the problem is obvious, If ᶓ≠η, then there exists complex numbers α and β.such that, αᶓ+β=1 and αη+β=0. It is sufficient to prove that the unit interval [0,1]is included in

If

,

Then,

Consequently there is no loss of generality is assuming that ξ = 1 and η = 0. Write T = A+iB with self -adjoint operator A and B . Since , and = 0 are real ,then (Bx,x) = (By,y) = 0.if x is replaced by λx,where = 1, then remains the same and (Bx,y) becomes λ(Bx,y). Consequently there is no loss of generality is assuming that (Bx,y) is purely imaginary. Let , where t is in [0,1]. If x and y are linearly dependent ,then y=μx where which contradicts to and independent ,so that h(t) ≠ 0.

,it would then follow that , hence the vectors x and y are linearly

Now, The relations (Bx,x) = (By,y) = 0 and Re (Bx,y) = 0 Iimply that hence is real for all t, whence the function

for all t,and


.

is real –valued and continuous on the closed interval [0,1].since f(0)=0 and f(1) = 1, we obtain .

EXAMBLE: 31 If T =

, Then W(T) = [0,1] (the closed unit interval).

EXAMBLE :32 If T =

, Then W(T) = { z : |z| < ½ }

EXAMBLE:33 If T = axis

, then W(T) is closed elliptical disc with foci at 0 and 1, minor axis 1 and major .

THEOREM:34 1.If T is a two –by –two matrix with distinct eigenvalues α and β ,and corresponding normalized eigen values x and y ,then W(T) is a closed elliptical disc with foci at αand β if and , then the minor axis and major axis can be expressed respectively as follows, The minor axis =

and major axis =

2. If T has only one eigenvalues α, then W(T) is the disc with center α and radius EXAMBLE.35 If T = 1, that is 1,

, Then W(T) is the equilateral triangle whose vertices are three cubic roots of and

.

EXAMBLE:36. If T =

Then W(T) is union of all the closed segments that join the point 1 to all

points of the closed disc with center 0 and radius ½. Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

REMARK: 37 Assume that = 1 Is important , not in the definition of W(T) = { (Tx,x): }. The reason is that W’(T) = { (Tx,x): } is easily expressed by W(T), but not vice versa. In fact, W’(T) is the union of all the closed segments that join the origin to all points of W(T).

2.5.2: Numerical Radius Is Equivalent To Operator Norm: Definition 38: 1. An Operator T on Hilberts space H Is said to be numerical radius w(t) if

An operator T on Hilberts space H is said to be Spectral radius r(T) If r

THEOREM 39: for any operator T. Proof: By Generalized polarization identity – –

Hence we have Since inequality holds. The second inequality is obvious since

– –

and the first .

THEOREM: 40 (Power inequality of w(T)). For any operator T, The following power inequality holds: number n.

for any natural


.

Proof: By the homogeneity of , it suffices to prove that ensures natural number n. First observe that for any complex number z such that ensures

On the other hand , when We have ,taking and letting whenever I – Z T is invertible, we have

for any ,

for all complex number z such that , Re( ) and

–

for any

. Thus, it suffices to prove that for all

, . Thus,

by using

,

for any complex number z such that

.

We recall the following identity:

Where w is a primitive nth root of 1. Since for each of the operators

, we have

for all We deduce that for all Hence we have Consequently

for all

and

.

.

2.5.3 THE CLOUSER OF NUMERICAL RANGE INCLUDES THE SPECTRUM:

THEOREM: 41 For any operator T, then the following holds. 1. The closure of the numerical range of an operator includes spectrum, i.e., co σ (T). Where co σ (T) means the convex hull of σ (T). Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

2. r(T) ≤ w(T) ≤ ≤

3.

≤

For all μ ϵ σ (T) to the first inequality and for all μ 

to the second inequality.

PROOF: 1. We have only to prove Let μ 

σ (T). Since W (T) is a convex set in a convex plane.

.then for any unit vector x, 0 ˂ d = d (μ,

) ≤ │ ( Tx, x ) - μ | =| ((T – μ) x, x ) | ≤ ||TX – μx || by Schwarz inequality.

So that we have ||Ty –μy|| ≥ d ||y|| for any vector y. If follows that μ ϵ ρ (T) or μ ϵ If μ ϵ , then ϵ .there exists an unit vector x such that = x, so that ϵ ( => μ ϵ W (T).which contradicts to μ , so that μ But μ ϵ ρ (T) = that is μ σ(T). Hence we have 1 by the contraposition. 2. It is follows by the theorem ||T|| ≤ w (T) ≤ ||T|| and by 1. 3.

Let ||Tx - μy|| ≥ d||y|| for any vector y, holds for μ = sup

, for any μ

σ (T)

= sup │

=sup| = r( r(

≤ ||

||

Hence proved.

2.5.4. NORMALOID OPERATOR AND SPECTRALIOD OPERATOR: Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

DEFINITION: An operator T is said to be a normaloid operator if ||T|| = r (T). An operator T is said to be spectraloid operator if w(T) = r (T) THEOREM: 1 r (T) =

holds for any operator T.

r (T) ≤

holds since r

PROOF: = r(

)≤

||.

THEOREM: 2 CHARACTERIZATIONS OF NORMALOID OPERATOR. The following assertions are mutually equivalent: 1. T is normaloid operator, that is ||T|| = r (T). 2. || = for all natural number n. 3. = w(T) PROOF: 1 2 We have only prove

|| ≥

Since the inequality

|| ≤

always holds.

= = ≤ ≤

2

3 Let prove

|| =


.

= ||T|| =||T|| r (T) =||T|| W (T) =||T|| 1

3 Let

||T|| = r (T) ensures ||T||=w (T) since r (T )≤ w (T )≤ ||T||,

3

1 Let ||T||=w (T) =1 ensures ||T||= r (T) =1

By the homogeneity of w (T) and r (T). Assume a sequence { } such that | without loss of generality by multiplying a suitable constant of modules 1.

|

1.

Since, | ≤ ||

|

≤ 1 and

1.

= = =

- 2Re

-

+

+1

0

So that 1 is an approximate point spectrum of T. Therefore we obtain r(T)=1.

THEOREM: 3 If T is a self- ad joint operator, then T is normaloid, that is ||T||= r (T). PROOF: If T is self-ad joint operator (T*=T). Then

|| =


.

 ||T*T||=| By using induction method, we have, || = So that = ||T|| = ||T|| r(T)

=||T||.

COROLLARY: If T is normal operator, then T is normaloid, that is ||T|| = r (T). PROOF: Let (T*

=

Since T commutes with T* and T*T is self-ad joint. so that, ||T|| = || =

= = r(T).

THEOREM: 4 CHACTERIZATION OF SPECTRALOID OPERATORS Dr. D. SENTHILKUMAR, DEPARTMENT OF MATHEMATICS, GOVT. ARTS COLLEGE,CBE -18.

.

The following assertions are mutually equivalent. 1. T is spectraloid operator w (T) = r (T). 2. w ( = w for all natural number n. PROOF: 1

2

We have only to proof that w ( w(

2

Assume

w(

≥w

since the reverse inequality always holds.

=r =r ≤w( =w . Since r(S) ≤ w(S) ≤ ||S|| for any operator S.

r (T) = r ≤w = w (T) ≤ So that w (T) = r (T). Since

r (T) as n
