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 $\$. 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 /math> 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 $T Z(v;T) \subset Z(v;T)$.

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 $\begin 0&1\\ 0&0\end$ relative to the standard ordered basis of $V$. Let $v=\begin 0 \\ 1 \end$. Then $Tv = \begin 1 \\ 0 \end, \quad T^2v=0, \ldots, T^rv=0, \ldots$. Therefore $\ = \left\$ 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=\$

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

::$p(x)=c_0+c_1x+c_2x^2+\cdots + c_x^+x^n$.

Then

::$\begin Tv_1 & = v_2\\ Tv_2 & = v_3\\ Tv_3 & = v_4\\ \vdots & \\ Tv_ & = v_n\\ Tv_n &= -c_0v_1 -c_1v_2 - \cdots c_v_n \end$

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

::$\begin 0 & 0 & 0 & \cdots & 0 & -c_0 \\ 1 & 0 & 0 & \ldots & 0 & -c_1 \\ 0 & 1 & 0 & \ldots & 0 & -c_2 \\ \vdots & & & & & \\ 0 & 0 & 0 & \ldots & 1 & -c_ \end$

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

See also

* Companion matrix
* Krylov subspace

External links

* PlanetMath
cyclic subspace

References

{{reflist

Linear algebra