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 $V$ . [1]

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 $v$ .[1]

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 $p(x)$ .[1]

External links

PlanetMath: cyclic subspace

References

Hoffman, Kenneth; Kunze, Ray (1971). Linear algebra (2nd ed.). Englewood Cliffs, N.J.: Prentice-Hall, Inc. p. 227. MR 0276251.

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[LA-1] Hoffman, Kenneth; Kunze, Ray (1971). Linear algebra (2nd ed.). Englewood Cliffs, N.J.: Prentice-Hall, Inc. p. 227. MR 0276251.