# 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 $TZ(v;T)\subset Z(v;T)$.

### Examples

* 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 $V$.
* If $I$ is the identity operator then every $I$-cyclic subspace is one-dimensional.
* $Z(v;T)$ is one-dimensional if and only if $v$ is a characteristic vector (eigenvector) of $T$.
* 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 $p(x)$.