The angle along a curve and range-kernel complementarity

The angle along a curve and range-kernelcomplementarity

Nikos Yannakakis (with D. Drivaliaris)

Department of MathematicsNational Technical University of Athens

Nikos Yannakakis (with D. Drivaliaris) The angle along a curve and range-kernel complementarity 1 / 28

Introduction The angle of an operator

Introduction

Let X be a Banach space and A : X → X be a bounded linear operator.We will denote the range of A by R(A) and the kernel of A by N(A).

The cosine of a A with respect to a semi-inner product [·, ·] compatiblewith the norm of X , is defined by

cosA = inf

{Re[Ax , x ]

‖Ax‖ ‖x‖: x /∈ N(A)

}.

Using the above one can define the angle φ(A) of the linear operator A by

φ(A) = arccos(cosA) .



Introduction



cosA = inf

{Re[Ax , x ]

‖Ax‖ ‖x‖: x /∈ N(A)

}.





Introduction



cosA = inf

{Re[Ax , x ]

‖Ax‖ ‖x‖: x /∈ N(A)

}.





The angle φ(A) of A has an obvious geometric interpretation; it measuresthe maximum (real) turning effect of A.

This concept was introduced by K. Gustafson and independently by M.Krein.

Example

If A is accretive i.e. Re[Ax , x ] ≥ 0, for all x ∈ X , then obviously cosA ≥ 0and hence φ(A) ≤ π

2 .

Note that K. Gustafson introduced the angle of A while trying to findsufficient conditions that guarantee that the product of two accretiveoperators is also accretive.





Example


2 .






Example


2 .






Example


2 .




The exact computation of cosA is difficult with the exception ofself-adjoint positive operators.

Example

If A is self-adjoint and positive, then

cosA = 2

√mM

m + M,

where m = minσ(A) and M = maxσ(A).



The exact computation of cosA is difficult with the exception ofself-adjoint positive operators.

Example

If A is self-adjoint and positive, then

cosA = 2

√mM

m + M,

where m = minσ(A) and M = maxσ(A).



Remark

If A has a negative eigenvalue then obviously φ(A) = π.

The converse is not true as the following example illustrates

Example

If A =

(0 10 0

)then φ(A) = π, whereas σ(A) = 0.



Remark

If A has a negative eigenvalue then obviously φ(A) = π.

The converse is not true as the following example illustrates

Example

If A =

(0 10 0

)then φ(A) = π, whereas σ(A) = 0.


Introduction Range-kernel complementarity

Range-kernel complementarity

Range-kernel complementarity i.e. the decomposition

X = R(A)⊕ N(A) ,

stands right next to the invertibility of A, since if it holds then A is of theform “invertible ⊕ 0”.

In finite dimensions R-K complementarity is equivalent toR(A) ∩ N(A) = {0}.

In infinite dimensions R-K complementarity is equivalent toR(A) ∩ N(A) = {0} and R(A) = R(A2).

Note that the latter is equivalent to both the ascent and the descent of Abeing ≤ 1.





X = R(A)⊕ N(A) ,









X = R(A)⊕ N(A) ,









X = R(A)⊕ N(A) ,






Introduction φ(A) < π

Theorem (D-Y JOT, 2016)

If φ(A) < π and both R(A) and R(A) + N(A) are closed then

X = R(A)⊕ N(A) .

QUESTION

Is the converse true? Is φ(A) < π necessary?

An immediate answer is no, since we may have X = R(A)⊕ N(A) and Ahas a negative eigenvalue.

Nevertheless if we can rotate the spectrum of A by taking e iθA, for someθ ∈ [0, 2π] so to avoid the negative axis and exploit the fact that bylinearity R(A) = R(e iθA) and N(A) = N(e iθA) we can have a converse.

ANSWER

If dimX <∞ and we allow “rotations”: YES.





X = R(A)⊕ N(A) .

QUESTION




ANSWER






X = R(A)⊕ N(A) .

QUESTION




ANSWER






X = R(A)⊕ N(A) .

QUESTION




ANSWER






X = R(A)⊕ N(A) .

QUESTION




ANSWER






X = R(A)⊕ N(A) .

QUESTION




ANSWER



Introduction am(A) < π

The “amplitude” of M. Krein

Related to the above discussion is the so called “amplitude” of A

am(A) = min{φ(e iθA) : θ ∈ [0, 2π]

}introduced also by M. Krein.

The“amplitude” of A compares the angles of A along all possible“directions” and provides the smallest one.


If am(A) < π and both R(A) and R(A) + N(A) are closed then

X = R(A)⊕ N(A) .





am(A) = min{φ(e iθA) : θ ∈ [0, 2π]





X = R(A)⊕ N(A) .





am(A) = min{φ(e iθA) : θ ∈ [0, 2π]





X = R(A)⊕ N(A) .



QUESTION (improved)

Is the converse true? Is am(A) < π necessary?

M. Krein noted that φ(A) < π implies that the spectrum of A σ(A) iscontained in the sector

S = 0 ∪ {z ∈ C : |argz | ≤ φ(A)} .

Hence the angle am(A) cannot characterize range-kernel complementarityin more general cases and thus

ANSWER

If dimX =∞ NO.

Example

The bilateral shift.



QUESTION (improved)



S = 0 ∪ {z ∈ C : |argz | ≤ φ(A)} .


ANSWER

If dimX =∞ NO.

Example




QUESTION (improved)



S = 0 ∪ {z ∈ C : |argz | ≤ φ(A)} .


ANSWER

If dimX =∞ NO.

Example




QUESTION (improved)



S = 0 ∪ {z ∈ C : |argz | ≤ φ(A)} .


ANSWER

If dimX =∞ NO.

Example




Note that the inclusion of σ(A) in some sector of the complex planeimplies that there exists a ray emanating from the origin that DOES NOTintersect σ(A).

As we will later see in cases like the bilateral shift (i.e. when 0 lies in ahole in the spectrum) even a generalized notion of angle seems unsuitable.

So we have tried to deal with the cases between these two extremes. Inparticular we deal with operators for which there exists a curve emanatingfrom the origin that DOES NOT intersect the spectrum. To do so

we define the angle of a linear operator A, along an unbounded curveemanating from the origin

we show that X = R(A)⊕ N(A) if and only if R(A) is closed andsome such angle (the corresponding “amplitude” ) of A is less than π.






























The angle along a curve Some motivation

Some motivation

Let H be a complex Hilbert space. If x , y ∈ H are non-zero vectors thenthe number

θ(x , y) = arccosRe〈x , y〉‖x‖‖y‖

is usually called the angle between x and y .

Recall that

infλ≥0

‖x + λy‖‖x‖

= 1

is equivalent to θ(x , y) being acute.

Moreover it can be easily seen that if this is not the case then

( infλ≥0

‖x + λy‖‖x‖

)2 + (Re〈x , y〉‖x‖‖y‖

)2 = 1 .



Some motivation




Recall that

infλ≥0

‖x + λy‖‖x‖

= 1



( infλ≥0

‖x + λy‖‖x‖

)2 + (Re〈x , y〉‖x‖‖y‖

)2 = 1 .



Some motivation




Recall that

infλ≥0

‖x + λy‖‖x‖

= 1



( infλ≥0

‖x + λy‖‖x‖

)2 + (Re〈x , y〉‖x‖‖y‖

)2 = 1 .



Hence taking

infλ≥0

‖x + λy‖‖x‖

as our starting point we can say that θ(x , y) is acute if the infimum is 1and is equal to

π − arcsin infλ≥0

‖x + λy‖‖x‖

otherwise.

Note that taking into account that H is a complex Hilbert space maybe amore appropriate name for this θ(x , y) is “the angle between x and yalong the positive semi-axis”.

In a complex normed space, even in a Hilbert space, there is no uniqueway to assign an angle between two vectors. We may have angles withrespect to every direction (recall Krein’s “amplitude”)



Hence taking

infλ≥0

‖x + λy‖‖x‖



‖x + λy‖‖x‖

otherwise.





Hence taking

infλ≥0

‖x + λy‖‖x‖



‖x + λy‖‖x‖

otherwise.




The angle along a curve The angle along a curve

For the rest of this presentation C will denote an unbounded curve in thecomplex plane emanating from the origin.

Definition

If X is a complex Banach space and x , y ∈ X , with x 6= 0 then we definesC (x , y) by

sC (x , y) = infλ∈C

‖x + λy‖‖x‖

.

Considering the above discussion we say that the angle θC (x , y) of x , yalong C is acute if sC (x , y) = 1 and is equal to π − arcsin sC (x , y)otherwise.




Definition



‖x + λy‖‖x‖

.





Definition



‖x + λy‖‖x‖

.





Definition



‖x + λy‖‖x‖

.



The angle along a curve The sine of A

The sine of the operator A along C is then

sinC A = infx /∈N(A)

sC (Ax , x) .

If C is the positive axis instead of sC (Ax , x) and sinC A we will just writes(Ax , x) and sinA.


The angle along a curve The sine of A

The sine of the operator A along C is then

sinC A = infx /∈N(A)

sC (Ax , x) .

If C is the positive axis instead of sC (Ax , x) and sinC A we will just writes(Ax , x) and sinA.


The angle along a curve A connection with cos A

Recall that A is accretive if and only if

infλ≥0

‖Ax + λx‖‖Ax‖

= 1 , for all x /∈ N(A)

Proposition

If A ∈ B(H) is not accretive, then

sin2 A + cos2 A = 1 .

Remark

K. Gustafson has shown, in his well-known min-max theorem, that if A isstrongly accretive then another sine satisfying the basic trigonometricidentity may be defined as minε>0 ‖εA− I‖.




infλ≥0


= 1 , for all x /∈ N(A)

Proposition


sin2 A + cos2 A = 1 .

Remark





infλ≥0


= 1 , for all x /∈ N(A)

Proposition


sin2 A + cos2 A = 1 .

Remark



The angle along a curve The angle of A revisited

The angle of A along C is defined to be

φC (A) = supx /∈N(A)

θC (Ax , x) .

As long as A is not accretive along C , the angle φC (A) may be given ageometric interpretation; it measures the maximum turning effect of Aalong C .


The angle along a curve The angle of A revisited

The angle of A along C is defined to be

φC (A) = supx /∈N(A)

θC (Ax , x) .

As long as A is not accretive along C , the angle φC (A) may be given ageometric interpretation; it measures the maximum turning effect of Aalong C .


The angle along a curve Main aim

Our aim in this talk is to show that if 0 faces the unbounded componentof the resolvent set ρ(A), then

X = R(A)⊕ N(A)

if and only if R(A) is closed and some angle φC (A) of A is less than π.


Results A first lemma

Our starting point is the following Lemma.

Lemma

Let A : X → X be a bounded linear operator. If φC (A) < π for some curveC , then

R(A) + N(A)

is closed and R(A) ∩ N(A) = {0}.


Results A first lemma

Our starting point is the following Lemma.

Lemma

Let A : X → X be a bounded linear operator. If φC (A) < π for some curveC , then

R(A) + N(A)

is closed and R(A) ∩ N(A) = {0}.


Results “Filling the hole”

In what follows by D∞ we denote the unbounded component of theresolvent set of A.

In the proof of our main result we need the following corollary of the socalled “filling the hole” theorem.

Proposition

If M is a closed invariant subspace of A, then

σ(A|M) ∩ D∞ = ∅ .


Results “Filling the hole”

In what follows by D∞ we denote the unbounded component of theresolvent set of A.

In the proof of our main result we need the following corollary of the socalled “filling the hole” theorem.

Proposition

If M is a closed invariant subspace of A, then

σ(A|M) ∩ D∞ = ∅ .


Results Main result

Our main result is the following.

Theorem

Let A : X → X be a bounded linear operator and assume that 0 ∈ ∂D∞.Then

X = R(A)⊕ N(A)

if and only if R(A) is closed and φC (A) < π, for some C .


Results Proof of the main result

Proof.

If X = R(A)⊕ N(A) then there exists δ > 0 such that

‖Ax + y‖ ≥ δ‖Ax‖ , for all x ∈ X , y ∈ N(A) . (1)

By the above Proposition we have that

σ(A|R(A)) ∩ D∞ = ∅ .

Hence D∞ lies in the resolvent of A|R(A) and thus there exists anunbounded path C emanating from the origin with

C ⊆ ρ(A|R(A)) .

Since A|R(A) is invertible we may show that there exists k > 0 such that

‖Ax + λx‖ ≥ k‖x‖ , for all x ∈ R(A), λ ∈ C . (2)



Proof.




σ(A|R(A)) ∩ D∞ = ∅ .


C ⊆ ρ(A|R(A)) .





Proof.




σ(A|R(A)) ∩ D∞ = ∅ .


C ⊆ ρ(A|R(A)) .





We then may show that there c > 0 such that sC (Ax , x) ≥ c for allx /∈ N(A) and so φC (A) < π.


Results The converse

Conversly, assume that φC (A) < π, for some C .

Then by the Lemma we have that R(A) ∩ N(A) = {0} and hence theascent α(A) of A is lesser or equal to 1.

To conclude the proof we have to show that the descent δ(A) is finite.

In particular we will show that R(A2) = R(A) and hence δ(A) ≤ 1.

Also by the Lemma, using the fact that R(A) is closed, we have thatR(A) + N(A) is a closed subspace of X which implies that R(A2) is alsoclosed.































Hence since R(A) ∩ N(A) = {0} we have that

‖Ax‖ ≥ ‖x‖ , for all x ∈ R(A) . (3)

If 0 ∈ σ(A|R(A)) and since by the“filling the hole theorem” we have that

σ(A|R(A)) ∩ D∞ = ∅

we get that 0 ∈ ∂σ(A|R(A)) ⊆ σapp(A|R(A)) which contardicts (3).

So 0 /∈ σ(A|R(A)) and the proof is complete.




‖Ax‖ ≥ ‖x‖ , for all x ∈ R(A) . (3)


σ(A|R(A)) ∩ D∞ = ∅






‖Ax‖ ≥ ‖x‖ , for all x ∈ R(A) . (3)


σ(A|R(A)) ∩ D∞ = ∅




Results Holes in the spectrum

Recall that a hole in a compact subset of the complex plane is a boundedconnected component of its complement.

Using the above we can show that if we have range-kernelcomplementarity then 0 faces the unbounded component of the resolventset if and only if some angle of A is less than π.

Proposition

Let A ∈ B(X ) and assume that X = N(A)⊕ R(A). Then 0 ∈ ∂D∞ if andonly if φC (A) < π, for some C .

Remark

The above verifies what we have already pointed out in the introduction:if 0 lies in a hole of the spectrum our generalized angle is unsuitable tocharacterize R-K complentarity.





Proposition


Remark






Proposition


Remark






Proposition


Remark



Possible extension to Banach algebras

Let A be a Banach algebra

If and a ∈ A we may define the angle of a along the curve C using the leftregular representation Ta defined as usual by

Tab = ab , for all b ∈ A .

The angle of a along C is defined to be the angle of Ta along C .















This definition directly on the algebra A is

If there exists 0 ≤ c < 1 such that

‖ab + λb‖ ≥ c‖ab‖, for all λ ∈ C and b ∈ A

we say that the angle of a along C is equal to

π − arcsin infλ∈C

infab 6=0

‖ab + λb‖‖ab‖

If the above double infimum is 1 we just say that the angle of a is acute.








infab 6=0










infab 6=0




Possible extension to Banach algebras An application

Using the above we may have the following

Proposition

If a ∈ A is an idempotent then there exists c > 0 such that

‖ab + λb‖ ≥ c‖ab‖, for all λ > 0 and b ∈ A

Proof.

Since Ta is a projection we have that A = R(Ta)⊕ N(Ta) and the resultfollows from our main theorem.


Possible extension to Banach algebras An application

Using the above we may have the following

Proposition

If a ∈ A is an idempotent then there exists c > 0 such that

‖ab + λb‖ ≥ c‖ab‖, for all λ > 0 and b ∈ A

Proof.

Since Ta is a projection we have that A = R(Ta)⊕ N(Ta) and the resultfollows from our main theorem.


The angle along a curve and range-kernel complementarity

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