Cyclic subspace

In mathematics, in linear algebra, a cyclic subspace is a certain special subspace of a finite-dimensional 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 $Z(v;T)$. If $V=Z(v;T)$, then $v$ is called a cyclic vector for $T$.

There is another equivalent definition of cyclic spaces. Let $T:V\rightarrow V$ be a linear transformation of a finite dimensional 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 $v$.

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

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \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 $p(x)$.

See also

• Companion matrix
• Krylov subspace

External links

• PlanetMath: cyclic subspace