Invariant subspace

Template:Short description In mathematics, an invariant subspace of a linear mapping T : V → V i.e. from some vector space V to itself, is a subspace W of V that is preserved by T. More generally, an invariant subspace for a collection of linear mappings is a subspace preserved by each mapping individually.

For a single operator

Consider a vector space $V$ and a linear map $T:V\to V.$ A subspace $W\subseteq V$ is called an invariant subspace for $T$, or equivalently, Template:Mvar-invariant, if Template:Mvar transforms any vector $𝐯\in W$ back into Template:Mvar. In formulas, this can be written$$𝐯\in W⟹T(𝐯)\in W$$or $$TW\subseteq W\text{.}$$

In this case, Template:Mvar restricts to an endomorphism of Template:Mvar:$$T{|}_W:W\to W\text{;}T{|}_W(𝐰)=T(𝐰)\text{.}$$

The existence of an invariant subspace also has a matrix formulation. Pick a basis C for W and complete it to a basis B of V. With respect to Template:Mvar, the operator Template:Mvar has form $$T=\left[\begin{array}{cc}T{|}_W& T_{12}\\ 0& T_{22}\end{array}\right]$$ for some Template:Math and Template:Math, where $T{|}_W$ here denotes the matrix of $T{|}_W$ with respect to the basis C.

Examples

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

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

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

The equation above formulates an eigenvalue problem. Any eigenvector for Template:Mvar 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 Template:Mvar as the direct sum Template:Math; a suitable Template:Math can always be chosen by extending a basis of Template:Mvar. The associated projection operator P onto W has matrix representation

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

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

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

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 Template:Mvar-invariant subspaces of Template:Mvar is sometimes called the invariant-subspace lattice of Template:Mvar and written Template:Math. 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 Template:Math in said to be a minimal invariant subspace.

In the study of infinite-dimensional operators, Template:Math is sometimes restricted to only the closed invariant subspaces.

For multiple operators

Given a collection Template:Math of operators, a subspace is called Template:Math-invariant if it is invariant under each Template:Math.

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

Examples

Let Template:Math be the set of all linear operators on Template:Mvar. Then Template:Math.

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 Template:Math and Template:Mvar 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. Template:Math theorem 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 Template:Math is not commutative when Template:Math.

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

Template:Main

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 $Y$ of a Banach space $X$ is said to be almost-invariant under an operator $T\in ℬ(X)$ if $TY\subseteq Y+E$ for some finite-dimensional subspace $E$; equivalently, $Y$ is almost-invariant under $T$ if there is a finite-rank operator $F\in ℬ(X)$ such that $(T+F)Y\subseteq Y$, i.e. if $Y$ is invariant (in the usual sense) under $T+F$. In this case, the minimum possible dimension of $E$ (or rank of $F$) 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 $Y$ 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 $X$ is a complex infinite-dimensional Banach space and $T\in ℬ(X)$ then $T$ admits an AIHS of defect at most 1. It is not currently known whether the same holds if $X$ 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.

See also

• Invariant manifold
• Lomonosov's invariant subspace theorem

Sources

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