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

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

The subspace $$Z(v;T)$$ is an invariant subspace for $$T$$, in the sense that $$ T Z(v;T) \subset Z(v;T)$$.

Examples

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

Companion matrix
Let $$T:V\rightarrow V $$ be a linear transformation of a $$n$$-dimensional vector space $$V$$ over a field $$F$$ and $$v$$ be a cyclic vector for $$T$$. Then the vectors


 * $$B=\{v_1=v, v_2=Tv, v_3=T^2v, \ldots v_n = T^{n-1}v\}$$

form an ordered basis for $$V$$. Let the characteristic polynomial for $$T$$ be


 * $$ p(x)=c_0+c_1x+c_2x^2+\cdots + c_{n-1}x^{n-1}+x^n$$.

Then



\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} $$

Therefore, relative to the ordered basis $$B$$, the operator $$T$$ is represented by the matrix



\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} $$

This matrix is called the companion matrix of the polynomial $$p(x)$$.