# 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 $v$  is the subspace $W$  of $V$  generated by the set of vectors $\{v,T(v),T^{2}(v),\ldots ,T^{r}(v),\ldots \}$ . This subspace is denoted by $Z(v;T)$ . 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 $TZ(v;T)\subset Z(v;T)$ .

### 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 $Tv={\begin{bmatrix}1\\0\end{bmatrix}},\quad T^{2}v=0,\ldots ,T^{r}v=0,\ldots$ . 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

$B=\{v_{1}=v,v_{2}=Tv,v_{3}=T^{2}v,\ldots v_{n}=T^{n-1}v\}$

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

$p(x)=c_{0}+c_{1}x+c_{2}x^{2}+\cdots +c_{n-1}x^{n-1}+x^{n}$ .

Then

{\begin{aligned}Tv_{1}&=v_{2}\\Tv_{2}&=v_{3}\\Tv_{3}&=v_{4}\\\vdots &\\Tv_{n-1}&=v_{n}\\Tv_{n}&=-c_{0}v_{1}-c_{1}v_{2}-\cdots c_{n-1}v_{n}\end{aligned}}

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 $p(x)$ .