Invariant subspace

Examples

Any linear map ${\displaystyle T:V\to V}$ admits the following invariant subspaces:

• The vector space ${\displaystyle V}$, because ${\displaystyle T}$ maps every vector in ${\displaystyle V}$ into ${\displaystyle V.}$
• The set ${\displaystyle \{0\}}$, because ${\displaystyle T(0)=0}$.

These are the improper and trivial invariant subspaces, respectively. Certain linear operators have no proper non-trivial invariant subspace: for instance, rotation of a two-dimensional real vector space. However, the axis of a rotation in three dimensions is always an invariant subspace.

1-dimensional subspaces

If U is a 1-dimensional invariant subspace for operator T with vector v ∈ U, then the vectors v and Tv must be linearly dependent. Thus $${\displaystyle \mathrm{\forall }\mathbf{v}\in U\mathrm{\exists }\alpha \in \mathbb{R}:T\mathbf{v}=\alpha \mathbf{v}\text{.}}$$In fact, the scalar α does not depend on v.

The equation above formulates an eigenvalue problem. Any eigenvector for T spans a 1-dimensional invariant subspace, and vice-versa. In particular, a nonzero invariant vector (i.e. a fixed point of T) spans an invariant subspace of dimension 1.

As a consequence of the fundamental theorem of algebra, every linear operator on a nonzero finite-dimensional complex vector space has an eigenvector. Therefore, every such linear operator in at least two dimensions has a proper non-trivial invariant subspace.

Diagonalization via projections

Determining whether a given subspace W is invariant under T is ostensibly a problem of geometric nature. Matrix representation allows one to phrase this problem algebraically.

Write V as the direct sum W ⊕ W′; a suitable W′ can always be chosen by extending a basis of W. The associated projection operator P onto W has matrix representation

${\displaystyle P=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]:\begin{array}{c}W\\ \oplus \\ W^{\prime }\end{array}\to \begin{array}{c}W\\ \oplus \\ W^{\prime }\end{array}.}$

A straightforward calculation shows that W is T-invariant if and only if PTP = TP.

If 1 is the identity operator, then 1-P is projection onto W′. The equation TP = PT holds if and only if both im(P) and im(1 − P) are invariant under T. In that case, T has matrix representation $${\displaystyle T=\left[\begin{array}{cc}{T}_{11}& 0\\ 0& T_{22}\end{array}\right]:\begin{array}{c}\mathrm{im}(P)\\ \oplus \\ \mathrm{im}(1-P)\end{array}\to \begin{array}{c}\mathrm{im}(P)\\ \oplus \\ \mathrm{im}(1-P)\end{array}.}$$

Colloquially, a projection that commutes with T "diagonalizes" T.

Lattice of subspaces

As the above examples indicate, the invariant subspaces of a given linear transformation T shed light on the structure of T. When V is a finite-dimensional vector space over an algebraically closed field, linear transformations acting on V are characterized (up to similarity) by the Jordan canonical form, which decomposes V into invariant subspaces of T. Many fundamental questions regarding T can be translated to questions about invariant subspaces of T.

The set of T-invariant subspaces of V is sometimes called the invariant-subspace lattice of T and written Lat(T). As the name suggests, it is a (modular) lattice, with meets and joins given by (respectively) set intersection and linear span. A minimal element in Lat(T) in said to be a minimal invariant subspace.

In the study of infinite-dimensional operators, Lat(T) is sometimes restricted to only the closed invariant subspaces.

For multiple operators

Given a collection T of operators, a subspace is called T-invariant if it is invariant under each T ∈ T.

As in the single-operator case, the invariant-subspace lattice of T, written Lat(T), is the set of all T-invariant subspaces, and bears the same meet and join operations. Set-theoretically, it is the intersection $${\displaystyle \mathrm{L}\mathrm{a}\mathrm{t}(\mathcal{T})=\bigcap _{T\in \mathcal{T}}\mathrm{L}\mathrm{a}\mathrm{t}(T)\text{.}}$$

Examples

Let End(V) be the set of all linear operators on V. Then Lat(End(V))={0,V}.

Given a representation of a group G on a vector space V, we have a linear transformation T(g) : V → V for every element g of G. If a subspace W of V is invariant with respect to all these transformations, then it is a subrepresentation and the group G acts on W in a natural way. The same construction applies to representations of an algebra.

As another example, let T ∈ End(V) and Σ be the algebra generated by {1, T }, where 1 is the identity operator. Then Lat(T) = Lat(Σ).

Fundamental theorem of noncommutative algebra

Just as the fundamental theorem of algebra ensures that every linear transformation acting on a finite-dimensional complex vector space has a non-trivial invariant subspace, the fundamental theorem of noncommutative algebra asserts that Lat(Σ) contains non-trivial elements for certain Σ.

Theorem (Burnside) — Assume V is a complex vector space of finite dimension. For every proper subalgebra Σ of End(V), Lat(Σ) contains a non-trivial element.

One consequence is that every commuting family in L(V) can be simultaneously upper-triangularized. To see this, note that an upper-triangular matrix representation corresponds to a flag of invariant subspaces, that a commuting family generates a commuting algebra, and that End(V) is not commutative when dim(V) ≥ 2.

Left ideals

If A is an algebra, one can define a left regular representation Φ on A: Φ(a)b = ab is a homomorphism from A to L(A), the algebra of linear transformations on A

The invariant subspaces of Φ are precisely the left ideals of A. A left ideal M of A gives a subrepresentation of A on M.

If M is a left ideal of A then the left regular representation Φ on M now descends to a representation Φ' on the quotient vector space A/M. If [b] denotes an equivalence class in A/M, Φ'(a)[b] = [ab]. The kernel of the representation Φ' is the set {a ∈ A | ab ∈ M for all b}.

The representation Φ' is irreducible if and only if M is a maximal left ideal, since a subspace V ⊂ A/M is an invariant under {Φ'(a) | a ∈ A} if and only if its preimage under the quotient map, V + M, is a left ideal in A.

Invariant subspace problem

The invariant subspace problem concerns the case where V is a separable Hilbert space over the complex numbers, of dimension > 1, and T is a bounded operator. The problem is to decide whether every such T has a non-trivial, closed, invariant subspace. It is unsolved.

In the more general case where V is assumed to be a Banach space, Per Enflo (1976) found an example of an operator without an invariant subspace. A concrete example of an operator without an invariant subspace was produced in 1985 by Charles Read.

Almost-invariant halfspaces

Related to invariant subspaces are so-called almost-invariant-halfspaces (AIHS's). A closed subspace ${\displaystyle Y}$ of a Banach space ${\displaystyle X}$ is said to be almost-invariant under an operator ${\displaystyle T\in \mathcal{B}(X)}$ if ${\displaystyle TY\subseteq Y+E}$ for some finite-dimensional subspace ${\displaystyle E}$; equivalently, ${\displaystyle Y}$ is almost-invariant under ${\displaystyle T}$ if there is a finite-rank operator ${\displaystyle F\in \mathcal{B}(X)}$ such that ${\displaystyle (T+F)Y\subseteq Y}$, i.e. if ${\displaystyle Y}$ is invariant (in the usual sense) under ${\displaystyle T+F}$. In this case, the minimum possible dimension of ${\displaystyle E}$ (or rank of ${\displaystyle F}$) is called the defect.

Clearly, every finite-dimensional and finite-codimensional subspace is almost-invariant under every operator. Thus, to make things non-trivial, we say that ${\displaystyle Y}$ is a halfspace whenever it is a closed subspace with infinite dimension and infinite codimension.

The AIHS problem asks whether every operator admits an AIHS. In the complex setting it has already been solved; that is, if ${\displaystyle X}$ is a complex infinite-dimensional Banach space and ${\displaystyle T\in \mathcal{B}(X)}$ then ${\displaystyle T}$ admits an AIHS of defect at most 1. It is not currently known whether the same holds if ${\displaystyle X}$ is a real Banach space. However, some partial results have been established: for instance, any self-adjoint operator on an infinite-dimensional real Hilbert space admits an AIHS, as does any strictly singular (or compact) operator acting on a real infinite-dimensional reflexive space.