Cyclic subspace

From Wikipedia, the free encyclopedia
  (Redirected from Cyclic decomposition theorem)
Jump to: navigation, search

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[edit]

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.[1]

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.[1]

Examples[edit]

  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[edit]

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


\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).[1]

See also[edit]

External links[edit]

References[edit]

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