Cyclic subspace

In mathematics, in linear algebra and functional analysis, a cyclic subspace is a certain special subspace of a vector space associated with a vector in the vector space and a linear transformation of the vector space. The cyclic subspace associated with a vector v in a vector space V and a linear transformation T of V is called the T-cyclic subspace generated by v. The concept of a cyclic subspace is a basic component in the formulation of the cyclic decomposition theorem in linear algebra.

Definition

Let [math]\displaystyle{ T:V\rightarrow V }[/math] be a linear transformation of a vector space [math]\displaystyle{ V }[/math] and let [math]\displaystyle{ v }[/math] be a vector in [math]\displaystyle{ V }[/math]. The [math]\displaystyle{ T }[/math]-cyclic subspace of [math]\displaystyle{ V }[/math] generated by [math]\displaystyle{ v }[/math], denoted [math]\displaystyle{ Z(v;T) }[/math], is the subspace of [math]\displaystyle{ V }[/math] generated by the set of vectors [math]\displaystyle{ \{ v, T(v), T^2(v), \ldots, T^r(v), \ldots\} }[/math]. In the case when [math]\displaystyle{ V }[/math] is a topological vector space, [math]\displaystyle{ v }[/math] is called a cyclic vector for [math]\displaystyle{ T }[/math] if [math]\displaystyle{ Z(v;T) }[/math] is dense in [math]\displaystyle{ V }[/math]. For the particular case of finite-dimensional spaces, this is equivalent to saying that [math]\displaystyle{ Z(v;T) }[/math] is the whole space [math]\displaystyle{ V }[/math].

There is another equivalent definition of cyclic spaces. Let [math]\displaystyle{ T:V\rightarrow V }[/math] be a linear transformation of a topological vector space over a field [math]\displaystyle{ F }[/math] and [math]\displaystyle{ v }[/math] be a vector in [math]\displaystyle{ V }[/math]. The set of all vectors of the form [math]\displaystyle{ g(T)v }[/math], where [math]\displaystyle{ g(x) }[/math] is a polynomial in the ring [math]\displaystyle{ F[x] }[/math] of all polynomials in [math]\displaystyle{ x }[/math] over [math]\displaystyle{ F }[/math], is the [math]\displaystyle{ T }[/math]-cyclic subspace generated by [math]\displaystyle{ v }[/math].

The subspace [math]\displaystyle{ Z(v;T) }[/math] is an invariant subspace for [math]\displaystyle{ T }[/math], in the sense that [math]\displaystyle{ T Z(v;T) \subset Z(v;T) }[/math].

Examples

1. For any vector space [math]\displaystyle{ V }[/math] and any linear operator [math]\displaystyle{ T }[/math] on [math]\displaystyle{ V }[/math], the [math]\displaystyle{ T }[/math]-cyclic subspace generated by the zero vector is the zero-subspace of [math]\displaystyle{ V }[/math].
2. If [math]\displaystyle{ I }[/math] is the identity operator then every [math]\displaystyle{ I }[/math]-cyclic subspace is one-dimensional.
3. [math]\displaystyle{ Z(v;T) }[/math] is one-dimensional if and only if [math]\displaystyle{ v }[/math] is a characteristic vector (eigenvector) of [math]\displaystyle{ T }[/math].
4. Let [math]\displaystyle{ V }[/math] be the two-dimensional vector space and let [math]\displaystyle{ T }[/math] be the linear operator on [math]\displaystyle{ V }[/math] represented by the matrix [math]\displaystyle{ \begin{bmatrix} 0&1\\ 0&0\end{bmatrix} }[/math] relative to the standard ordered basis of [math]\displaystyle{ V }[/math]. Let [math]\displaystyle{ v=\begin{bmatrix} 0 \\ 1 \end{bmatrix} }[/math]. Then [math]\displaystyle{ Tv = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad T^2v=0, \ldots, T^rv=0, \ldots }[/math]. Therefore [math]\displaystyle{ \{ v, T(v), T^2(v), \ldots, T^r(v), \ldots\} = \left\{ \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\} }[/math] and so [math]\displaystyle{ Z(v;T)=V }[/math]. Thus [math]\displaystyle{ v }[/math] is a cyclic vector for [math]\displaystyle{ T }[/math].

Companion matrix

Let [math]\displaystyle{ T:V\rightarrow V }[/math] be a linear transformation of a [math]\displaystyle{ n }[/math]-dimensional vector space [math]\displaystyle{ V }[/math] over a field [math]\displaystyle{ F }[/math] and [math]\displaystyle{ v }[/math] be a cyclic vector for [math]\displaystyle{ T }[/math]. Then the vectors

[math]\displaystyle{ B=\{v_1=v, v_2=Tv, v_3=T^2v, \ldots v_n = T^{n-1}v\} }[/math]

form an ordered basis for [math]\displaystyle{ V }[/math]. Let the characteristic polynomial for [math]\displaystyle{ T }[/math] be

[math]\displaystyle{ p(x)=c_0+c_1x+c_2x^2+\cdots + c_{n-1}x^{n-1}+x^n }[/math].

Then

[math]\displaystyle{ \begin{align} Tv_1 & = v_2\\ Tv_2 & = v_3\\ Tv_3 & = v_4\\ \vdots & \\ Tv_{n-1} & = v_n\\ Tv_n &= -c_0v_1 -c_1v_2 - \cdots c_{n-1}v_n \end{align} }[/math]

Therefore, relative to the ordered basis [math]\displaystyle{ B }[/math], the operator [math]\displaystyle{ T }[/math] is represented by the matrix

[math]\displaystyle{ \begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & -c_0 \\ 1 & 0 & 0 & \ldots & 0 & -c_1 \\ 0 & 1 & 0 & \ldots & 0 & -c_2 \\ \vdots & & & & & \\ 0 & 0 & 0 & \ldots & 1 & -c_{n-1} \end{bmatrix} }[/math]

This matrix is called the companion matrix of the polynomial [math]\displaystyle{ p(x) }[/math].

See also

• Companion matrix
• Krylov subspace

External links

• PlanetMath: cyclic subspace