Invariant Subspaces and Where to Find Themwrobel/GIGEM2018/Talks/... · All operators on a...

The ProblemPositive resultsNegative results

Open questions and generalizations

Invariant Subspaces and Where to Find ThemA history of the invariant subspace problem

Amudhan Krishnaswamy-Usha

Texas A&M

TAMU AMU GIGEM, March 4,2018

Amudhan Krishnaswamy-Usha Invariant Subspaces and Where to Find Them



Let X be a (complex) vector space, and T : X → X a linear map.Is there a subspace V ⊂ X , such that T (V ) ⊂ V ?

[V is said to be an invariant subspace for T if this happens]

Trivial answer: yes. {0} and X . What about non-trivial invariantsubspaces?




Let X be a (complex) vector space, and T : X → X a linear map.Is there a subspace V ⊂ X , such that T (V ) ⊂ V ?

[V is said to be an invariant subspace for T if this happens]

Trivial answer: yes. {0} and X . What about non-trivial invariantsubspaces?




Theorem

Let X be finite dimensional, with dim(X ) > 1. Then T has a nontrivial invariant subspace.

Proof: Eigenvalues exist, since the characteristic polynomial has aroot.




What about infinite dimensions?

... It’s dangerous to go alone, take this:

|| . ||

From now on, X is an infinite dimensional Banach space (acomplete normed linear space), and T is a continuous (bounded)linear map on X .





...

It’s dangerous to go alone, take this:

|| . ||







|| . ||







|| . ||





Let x ∈ X , and Vx = span{x ,Tx ,T 2x ...}. Clearly Vx is invariant.

Theorem

If x 6= 0, then Vx is a non-trivial invariant subspace.

Proof: Vx is the countable union of finite dimensional spaces, eachof which has empty interior. The Baire category theorem thenimplies Vx 6= X .




Make the problem non-trivial: demand non-trivial closed invariantsubspaces.

The invariant subspace problem

Let X be an infinite dimensional Banach space, and T : X → X abounded linear map. Is is true that T has a non-trivial, closedinvariant subspace?




All operators on a non-separable Banach space.

Normal (TT ∗ = T ∗T ) operators on a Hilbert space. (fromthe spectral theorem)

Compact operators on a Banach space [von Neumann forHilbert spaces (1920s), Aronszajn and Smith (1054) forBanach spaces]

’polynomially compact operators’ : p(T ) compact for somepolynomial p. [Bernstein, Robinson 1960s, using non-standardanalysis ]

























Theorem (Lomonosov,1973)

If T is not a scalar multiple of the identity, and T commutes witha non-zero compact operator, then there is a non trivial closedsubspace of X which is invariant for every operator commutingwith T .

Such subspaces are said to be hyperinvariant




Theorem (Lomonosov,1973)

If T is not a scalar multiple of the identity, and T commutes witha non-zero compact operator, then there is a non trivial closedsubspace of X which is invariant for every operator commutingwith T .

Such subspaces are said to be hyperinvariant




Theorem (Argyros and Hayden, 2009)

There exists a Banach space such that every bounded operator isof the form scalar + compact. In particular, every operator has aninvariant subspace.




Per Enflo, (discovered in ’76, published in ’87): An operatoron a (non-reflexive) Banach space.

C. Read, ’84,’85: An operator on `1.




The Invariant Subspace Problem*

Does every bounded operator on a reflexive Banach space have anon trivial closed invariant subspace?

The Hyperinvariant Subspace problem

Does every bounded operator on a Hilbert space have ahyperinvariant subspace?




The Invariant Subspace Problem*

Does every bounded operator on a reflexive Banach space have anon trivial closed invariant subspace?

The Hyperinvariant Subspace problem

Does every bounded operator on a Hilbert space have ahyperinvariant subspace?




An algebra A of operators is transitive if there is no common(closed,non trivial) invariant subspace.

Theorem (Burnside)

If A is a subalgebra of n × n matrices, then A is transitive iffA = Mn.

Is this true in the infinite dimensional case? No, consider thealgebra of finite rank operators.

Is this true if the algebra is closed? No, consider the compactoperators.





Theorem (Burnside)


Is this true in the infinite dimensional case?

No, consider thealgebra of finite rank operators.






Theorem (Burnside)








Theorem (Burnside)







Replace the operator norm topology by the strong operatortopology (pointwise convergence)

Transitive algebra problem

Let X be an infinite dimensional Banach space, and A is an algebraof bounded operators closed in the strong operator topology. IfA 6= B(X ), does A have a non trivial invariant subspace?


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