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 $T:V\to V$ 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 $\{v,T(v),T^2(v),\dots ,T^r(v),\dots \}$. 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 $T:V\to V$ 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 $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ relative to the standard ordered basis of $V$. Let $v=\left[\begin{array}{c}0\\ 1\end{array}\right]$. Then $Tv=\left[\begin{array}{c}1\\ 0\end{array}\right],T^{2}v=0,\dots ,T^{r}v=0,\dots$. Therefore $\{v,T(v),T^2(v),\dots ,T^r(v),\dots \}=\{\left[\begin{array}{c}0\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 0\end{array}\right]\}$ and so $Z(v;T)=V$. Thus $v$ is a cyclic vector for $T$.

Companion matrix

Let $T:V\to V$ 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

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

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

$\begin{array}{rl}T{v}_1& =v_2\\ T{v}_2& =v_3\\ T{v}_3& =v_4\\ ⋮& \\ T{v}_{n-1}& =v_n\\ T{v}_n& =-c_0{v}_1-c_1{v}_2-\cdots c_{n-1}{v}_n\end{array}$

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

$\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& -c_0\\ 1& 0& 0& \dots & 0& -c_1\\ 0& 1& 0& \dots & 0& -c_2\\ ⋮& & & & & \\ 0& 0& 0& \dots & 1& -c_{n-1}\end{array}\right]$

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

See also

• Companion matrix
• Krylov subspace

External links

• PlanetMath: cyclic subspace