# Invariant subspace

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, T-invariant, if T transforms any vector $\mathbf {v} \in W$ back into W. In formulas, this can be written$\mathbf {v} \in W\implies T(\mathbf {v} )\in W$or $TW\subseteq W{\text{.}}$
In this case, T restricts to an endomorphism of W:$T|_{W}:W\to W{\text{;}}\quad T|_{W}(\mathbf {w} )=T(\mathbf {w} ){\text{.}}$
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 B, the operator T has form $T={\begin{bmatrix}T|_{W}&T_{12}\\0&T_{22}\end{bmatrix}}$ for some T$_{12}$ and T$_{22}$, 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 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 $\forall \mathbf {v} \in U\;\exists \alpha \in \mathbb {R} :T\mathbf {v} =\alpha \mathbf {v} {\text{.}}$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
    $P={\begin{bmatrix}1&0\\0&0\end{bmatrix}}:{\begin{matrix}W\\\oplus \\W'\end{matrix}}\rightarrow {\begin{matrix}W\\\oplus \\W'\end{matrix}}.$
A straightforward calculation shows that W is 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 $T={\begin{bmatrix}T_{11}&0\\0&T_{22}\end{bmatrix}}:{\begin{matrix}\operatorname {im} (P)\\\oplus \\\operatorname {im} (1-P)\end{matrix}}\rightarrow {\begin{matrix}\operatorname {im} (P)\\\oplus \\\operatorname {im} (1-P)\end{matrix}}\;.$
Colloquially, a projection that commutes with T "diagonalizes" T.

## 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 $\mathrm {Lat} ({\mathcal {T}})=\bigcap _{T\in {\mathcal {T}}}{\mathrm {Lat} (T)}{\text{.}}$

### 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

    Main article: 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 $Y$ of a Banach space $X$ is said to be almost-invariant under an operator $T\in {\mathcal {B}}(X)$ 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 $F\in {\mathcal {B}}(X)$ 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 $T\in {\mathcal {B}}(X)$ 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.