Invariant subspace

In mathematics, an invariant subspace of a linear mapping T : V &rarr; 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 $T_{22}$.

Examples
Any linear map $$T : V \to V$$ admits the following invariant subspaces: 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.
 * The vector space $$V$$, because $$T$$ maps every vector in $$V$$ into $$V.$$
 * The set $$\{0\}$$, because $$T(0) = 0$$.

1-dimensional subspaces
If $U$ is a 1-dimensional invariant subspace for operator $T$ with vector $v &isin; U$, 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 $&alpha;$ 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 $W &oplus; W&prime;$; a suitable $W&prime;$ 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 =&thinsp;TP.

If 1 is the identity operator, then $1-P$ is projection onto $W&prime;$. The equation $TP = PT$ holds if and only if both im(P) and im(1&thinsp;− 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 $\mathcal{T}$ of operators, a subspace is called $\mathcal{T}$-invariant if it is invariant under each $T &isin; \mathcal{T}$.

As in the single-operator case, the invariant-subspace lattice of $\mathcal{T}$, written $Lat(\mathcal{T})$, is the set of all $\mathcal{T}$-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 $T &isin; End(V)$ and $&Sigma;$ be the algebra generated by {1,&thinsp;T&thinsp;}, 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.

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 $Lat(Σ)$.

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 >&thinsp;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.