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 operatorEdit
Consider a vector space <math>V</math> and a linear map <math>T: V \to V.</math> A subspace <math>W \subseteq V</math> is called an invariant subspace for <math>T</math>, or equivalently, Template:Mvar-invariant, if Template:Mvar transforms any vector <math>\mathbf{v} \in W</math> back into Template:Mvar. In formulas, this can be written<math display="block">\mathbf{v} \in W \implies T(\mathbf{v}) \in W</math>or<ref>Template:Harvnb</ref> <math display="block">TW\subseteq W\text{.}</math>
In this case, Template:Mvar restricts to an endomorphism of Template:Mvar:<ref>Template:Harvnb</ref><math display="block">T|_W : W \to W\text{;}\quad T|_W(\mathbf{w}) = T(\mathbf{w})\text{.}</math>
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 <math display=block> T = \begin{bmatrix} T|_W & T_{12} \\ 0 & T_{22} \end{bmatrix} </math> for some Template:Math and Template:Math, where <math>T|_W</math> here denotes the matrix of <math>T|_W</math> with respect to the basis C.
ExamplesEdit
Any linear map <math>T : V \to V</math> admits the following invariant subspaces:
- The vector space <math>V</math>, because <math>T</math> maps every vector in <math>V</math> into <math>V.</math>
- The set <math>\{0\}</math>, because <math>T(0) = 0</math>.
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 subspacesEdit
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 <math display="block"> \forall\mathbf{v}\in U\;\exists\alpha\in\mathbb{R}: T\mathbf{v}=\alpha\mathbf{v}\text{.}</math>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 projectionsEdit
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
- <math>
P = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} : \begin{matrix}W \\ \oplus \\ W' \end{matrix} \rightarrow \begin{matrix}W \\ \oplus \\ W' \end{matrix}. </math>
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 <math display=block> 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} \;. </math>
Colloquially, a projection that commutes with T "diagonalizes" T.
Lattice of subspacesEdit
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 operatorsEdit
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 <math display="block">\mathrm{Lat}(\mathcal{T})=\bigcap_{T\in\mathcal{T}}{\mathrm{Lat}(T)}\text{.}</math>
ExamplesEdit
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 algebraEdit
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 idealsEdit
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 problemEdit
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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 halfspacesEdit
Related to invariant subspaces are so-called almost-invariant-halfspaces (AIHS's). A closed subspace <math>Y</math> of a Banach space <math>X</math> is said to be almost-invariant under an operator <math>T \in \mathcal{B}(X)</math> if <math>TY \subseteq Y+E</math> for some finite-dimensional subspace <math>E</math>; equivalently, <math>Y</math> is almost-invariant under <math>T</math> if there is a finite-rank operator <math>F \in \mathcal{B}(X)</math> such that <math>(T+F)Y \subseteq Y</math>, i.e. if <math>Y</math> is invariant (in the usual sense) under <math>T+F</math>. In this case, the minimum possible dimension of <math>E</math> (or rank of <math>F</math>) 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 <math>Y</math> 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 <math>X</math> is a complex infinite-dimensional Banach space and <math>T \in \mathcal{B}(X)</math> then <math>T</math> admits an AIHS of defect at most 1. It is not currently known whether the same holds if <math>X</math> 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.