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Cyclic subspace

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(Redirected from Cyclic decomposition theorem)

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 V {\displaystyle T:V\rightarrow V} be a linear transformation of a vector space V {\displaystyle V} and let v {\displaystyle v} be a vector in V {\displaystyle V} . The T {\displaystyle T} -cyclic subspace of V {\displaystyle V} generated by v {\displaystyle v} , denoted Z ( v ; T ) {\displaystyle Z(v;T)} , is the subspace of V {\displaystyle V} generated by the set of vectors { v , T ( v ) , T 2 ( v ) , , T r ( v ) , } {\displaystyle \{v,T(v),T^{2}(v),\ldots ,T^{r}(v),\ldots \}} . In the case when V {\displaystyle V} is a topological vector space, v {\displaystyle v} is called a cyclic vector for T {\displaystyle T} if Z ( v ; T ) {\displaystyle Z(v;T)} is dense in V {\displaystyle V} . For the particular case of finite-dimensional spaces, this is equivalent to saying that Z ( v ; T ) {\displaystyle Z(v;T)} is the whole space V {\displaystyle V} .

There is another equivalent definition of cyclic spaces. Let T : V V {\displaystyle T:V\rightarrow V} be a linear transformation of a topological vector space over a field F {\displaystyle F} and v {\displaystyle v} be a vector in V {\displaystyle V} . The set of all vectors of the form g ( T ) v {\displaystyle g(T)v} , where g ( x ) {\displaystyle g(x)} is a polynomial in the ring F [ x ] {\displaystyle F} of all polynomials in x {\displaystyle x} over F {\displaystyle F} , is the T {\displaystyle T} -cyclic subspace generated by v {\displaystyle v} .

The subspace Z ( v ; T ) {\displaystyle Z(v;T)} is an invariant subspace for T {\displaystyle T} , in the sense that T Z ( v ; T ) Z ( v ; T ) {\displaystyle TZ(v;T)\subset Z(v;T)} .

Examples

  1. For any vector space V {\displaystyle V} and any linear operator T {\displaystyle T} on V {\displaystyle V} , the T {\displaystyle T} -cyclic subspace generated by the zero vector is the zero-subspace of V {\displaystyle V} .
  2. If I {\displaystyle I} is the identity operator then every I {\displaystyle I} -cyclic subspace is one-dimensional.
  3. Z ( v ; T ) {\displaystyle Z(v;T)} is one-dimensional if and only if v {\displaystyle v} is a characteristic vector (eigenvector) of T {\displaystyle T} .
  4. Let V {\displaystyle V} be the two-dimensional vector space and let T {\displaystyle T} be the linear operator on V {\displaystyle V} represented by the matrix [ 0 1 0 0 ] {\displaystyle {\begin{bmatrix}0&1\\0&0\end{bmatrix}}} relative to the standard ordered basis of V {\displaystyle V} . Let v = [ 0 1 ] {\displaystyle v={\begin{bmatrix}0\\1\end{bmatrix}}} . Then T v = [ 1 0 ] , T 2 v = 0 , , T r v = 0 , {\displaystyle 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 ) , , T r ( v ) , } = { [ 0 1 ] , [ 1 0 ] } {\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 Z ( v ; T ) = V {\displaystyle Z(v;T)=V} . Thus v {\displaystyle v} is a cyclic vector for T {\displaystyle T} .

Companion matrix

Let T : V V {\displaystyle T:V\rightarrow V} be a linear transformation of a n {\displaystyle n} -dimensional vector space V {\displaystyle V} over a field F {\displaystyle F} and v {\displaystyle v} be a cyclic vector for T {\displaystyle T} . Then the vectors

B = { v 1 = v , v 2 = T v , v 3 = T 2 v , v n = T n 1 v } {\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 V {\displaystyle V} . Let the characteristic polynomial for T {\displaystyle T} be

p ( x ) = c 0 + c 1 x + c 2 x 2 + + c n 1 x n 1 + x n {\displaystyle p(x)=c_{0}+c_{1}x+c_{2}x^{2}+\cdots +c_{n-1}x^{n-1}+x^{n}} .

Then

T v 1 = v 2 T v 2 = v 3 T v 3 = v 4 T v n 1 = v n T v n = c 0 v 1 c 1 v 2 c n 1 v n {\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 B {\displaystyle B} , the operator T {\displaystyle T} is represented by the matrix

[ 0 0 0 0 c 0 1 0 0 0 c 1 0 1 0 0 c 2 0 0 0 1 c n 1 ] {\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 p ( x ) {\displaystyle p(x)} .

See also

External links

References

  1. ^ Hoffman, Kenneth; Kunze, Ray (1971). Linear algebra (2nd ed.). Englewood Cliffs, N.J.: Prentice-Hall, Inc. p. 227. ISBN 9780135367971. MR 0276251.
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