Cyclic subspace

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 ${\displaystyle TZ(v;T)\subset Z(v;T)}$.

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