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 {\displaystyle V} and a linear map T : V → V . {\displaystyle T:V\to V.} A subspace W ⊆ V {\displaystyle W\subseteq V} is called an invariant subspace for T {\displaystyle T} , or equivalently, T-invariant, if T transforms any vector v ∈ W {\displaystyle \mathbf {v} \in W} back into W. In formulas, this can be written v ∈ W ⟹ T ( v ) ∈ W {\displaystyle \mathbf {v} \in W\implies T(\mathbf {v} )\in W} or T W ⊆ W . {\displaystyle TW\subseteq W{\text{.}}}
In this case, T restricts to an endomorphism of W: T | W : W → W ; T | W ( w ) = T ( w ) . {\displaystyle 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 = [ T | W T 12 0 T 22 ] {\displaystyle T={\begin{bmatrix}T|_{W}&T_{12}\\0&T_{22}\end{bmatrix}}} for some T12 and T22, where T | W {\displaystyle T|_{W}} here denotes the matrix of T | W {\displaystyle T|_{W}} with respect to the basis C.
Examples Any linear map T : V → V {\displaystyle T:V\to V} admits the following invariant subspaces:
The vector space V {\displaystyle V} , because T {\displaystyle T} maps every vector in V {\displaystyle V} into V . {\displaystyle V.}
The set { 0 } {\displaystyle \{0\}} , because T ( 0 ) = 0 {\displaystyle 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 ∀ v ∈ U ∃ α ∈ R : T v = α v . {\displaystyle \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
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