invariant subspace

Consider a vector space $V$ and a linear map $T:\; V\; \backslash to\; V.$ A subspace $W\; \backslash subseteq\; V$ is called an invariant subspace for $T$, or equivalently, {{Mvar|T}}-invariant, if {{Mvar|T}} transforms any vector $\backslash mathbf\{v\}\; \backslash in\; W$ back into {{Mvar|W}}. In formulas, this can be written$$\backslash mathbf\{v\}\; \backslash in\; W\; \backslash implies\; T(\backslash mathbf\{v\})\; \backslash in\; W$$or{{harvnb|Roman|2008|loc=p. 73 §2}} $$TW\backslash subseteq\; W\backslash text\{.\}$$

In this case, {{Mvar|T}} restricts to an endomorphism of {{Mvar|W}}:{{harvnb|Roman|2008|loc=p. 73 §2}}$$T|\_W\; :\; W\; \backslash to\; W\backslash text\{;\}\backslash quad\; T|\_W(\backslash mathbf\{w\})\; =\; T(\backslash mathbf\{w\})\backslash text\{.\}$$

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 {{Mvar|B}}, the operator {{Mvar|T}} has form for some {{Math|T₁₂}} and {{Math|T₂₂}}, where $T|\_W$ here denotes the matrix of $T|\_W$ with respect to the basis C.

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

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

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

The equation above formulates an eigenvalue problem. Any eigenvector for {{Mvar|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.

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

:$$

P = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} : \begin{matrix}W \\ \oplus \\ W' \end{matrix} \rightarrow \begin{matrix}W \\ \oplus \\ W' \end{matrix}.

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

If 1 is the identity operator, then {{Math|1-P}} is projection onto {{Math|W′}}. The equation {{math|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 $$$$

T = \begin{bmatrix} T_{11} & 0 \\ 0 & T_{22} \end{bmatrix} : \begin{matrix} \operatorname{im}(P) \\ \oplus \\ \operatorname{im}(1-P) \end{matrix} \rightarrow \begin{matrix} \operatorname{im}(P) \\ \oplus \\ \operatorname{im}(1-P) \end{matrix} \;.

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

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 {{Mvar|T}}-invariant subspaces of {{Mvar|V}} is sometimes called the invariant-subspace lattice of {{Mvar|T}} and written {{Math|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 {{Math|Lat(T)}} in said to be a minimal invariant subspace.

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

Given a collection {{Math|{{mathcal|T}}}} of operators, a subspace is called {{Math|{{mathcal|T}}}}-invariant if it is invariant under each {{Math|T ∈ {{mathcal|T}}}}.

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

Let {{Math|End(V)}} be the set of all linear operators on {{Mvar|V}}. Then {{Math|1=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 {{Math|T ∈ End(V)}} and {{Mvar|Σ}} be the algebra generated by {1, T }, where 1 is the identity operator. Then Lat(T) = Lat(Σ).

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.

{{Math theorem

| math_statement = Assume {{mvar|V}} is a complex vector space of finite dimension. For every proper subalgebra {{mvar|Σ}} of {{math|End(V)}}, {{math|Lat(Σ)}} contains a non-trivial element.

| note = Burnside

}}

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 {{Math|End(V)}} is not commutative when {{Math|dim(V) ≥ 2}}.

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.

:{{main|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.

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 $T\; \backslash in\; \backslash mathcal\{B\}(X)$ if $TY\; \backslash subseteq\; Y+E$ for some finite-dimensional subspace $E$; equivalently, $Y$ is almost-invariant under $T$ if there is a finite-rank operator $F\; \backslash in\; \backslash mathcal\{B\}(X)$ such that $(T+F)Y\; \backslash subseteq\; Y$, 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 $T\; \backslash in\; \backslash mathcal\{B\}(X)$ 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.

• Invariant manifold
• Lomonosov's invariant subspace theorem

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Category:Linear algebra

Category:Operator theory

Category:Representation theory