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; that is, T(W) ⊆ W. Consider a linear mapping An invariant subspace of has the property that all vectors are transformed by into vectors also contained in . This can be stated as Since maps every vector in into Since a linear map has to map A basis of a 1-dimensional space is simply a non-zero vector . Consequently, any vector can be represented as where is a scalar. If we represent by a matrix then, for to be an invariant subspace it must satisfy We know that with . Therefore, the condition for existence of a 1-dimensional invariant subspace is expressed as: where is a scalar (in the base field of the vector space. Note that this is the typical formulation of an eigenvalue problem, which means that any eigenvector of forms a 1-dimensional invariant subspace in . An invariant subspace of a linear mapping from some vector space V to itself is a subspace W of V such that T(W) is contained in W. An invariant subspace of T is also said to be T invariant. If W is T-invariant, we can restrict T to W to arrive at a new linear mapping This linear mapping is called the restriction of T on W and is defined by Next, we give a few immediate examples of invariant subspaces. Certainly V itself, and the subspace {0}, are trivially invariant subspaces for every linear operator T : V → V. For certain linear operators there is no non-trivial invariant subspace; consider for instance a rotation of a two-dimensional real vector space. Let v be an eigenvector of T, i.e. T v = λv. Then W = span{v} is T-invariant. 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 has a non-trivial invariant subspace. The fact that the complex numbers are an algebraically closed field is required here.