Let ${\displaystyle T:V\rightarrow V}$  be a linear transformation of a vector space ${\displaystyle V}$  and let ${\displaystyle v}$  be a vector in ${\displaystyle V}$ . The ${\displaystyle T}$ -cyclic subspace of ${\displaystyle V}$  generated by ${\displaystyle v}$ , denoted ${\displaystyle Z(v;T)}$ , is the subspace of ${\displaystyle V}$  generated by the set of vectors ${\displaystyle \{v,T(v),T^{2}(v),\ldots ,T^{r}(v),\ldots \}}$ . In the case when ${\displaystyle V}$  is a topological vector space, ${\displaystyle v}$  is called a cyclic vector for ${\displaystyle T}$  if ${\displaystyle Z(v;T)}$  is dense in ${\displaystyle V}$ . For the particular case of finite-dimensional spaces, this is equivalent to saying that ${\displaystyle Z(v;T)}$  is the whole space ${\displaystyle V}$ . ^[1]

There is another equivalent definition of cyclic spaces. Let ${\displaystyle T:V\rightarrow V}$  be a linear transformation of a topological vector space over a field ${\displaystyle F}$  and ${\displaystyle v}$  be a vector in ${\displaystyle V}$ . The set of all vectors of the form ${\displaystyle g(T)v}$ , where ${\displaystyle g(x)}$  is a polynomial in the ring ${\displaystyle F[x]}$  of all polynomials in ${\displaystyle x}$  over ${\displaystyle F}$ , is the ${\displaystyle T}$ -cyclic subspace generated by ${\displaystyle v}$ .^[1]

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

Examples edit

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

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

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

form an ordered basis for ${\displaystyle V}$ . Let the characteristic polynomial for ${\displaystyle T}$  be

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

Then

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

Therefore, relative to the ordered basis ${\displaystyle B}$ , the operator ${\displaystyle T}$  is represented by the matrix

${\displaystyle {\begin{bmatrix}0&0&0&\cdots &0&-c_{0}\\1&0&0&\ldots &0&-c_{1}\\0&1&0&\ldots &0&-c_{2}\\\vdots &&&&&\\0&0&0&\ldots &1&-c_{n-1}\end{bmatrix}}}$ 

This matrix is called the companion matrix of the polynomial ${\displaystyle p(x)}$ .^[1]

• Companion matrix
• Krylov subspace

• PlanetMath: cyclic subspace