Editing Invariant subspace (section)

===1-dimensional subspaces===
If {{Mvar|U}} is a 1-dimensional invariant subspace for operator {{Mvar|T}} with vector {{Math|'''v''' &isin; ''U''}}, then the vectors {{Math|'''v'''}} and {{Math|''T'''''v'''}} 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 {{Mvar|&alpha;}} does not depend on {{Math|'''v'''}}.  

The equation above formulates an [[eigenvalue]] problem.  Any [[eigenvector]] for {{Mvar|T}} spans a 1-dimensional invariant subspace, and vice-versa.  In particular, a nonzero '''invariant vector''' (i.e. a [[Fixed point (mathematics)|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 [[dimension (vector space)|finite-dimensional]] [[complex number|complex]] vector space has an eigenvector. Therefore, every such linear operator in at least two dimensions has a proper non-trivial invariant subspace.