3.27 Cyclic vector
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Example of theorem: \(A= \begin{pmatrix} -1 & 3 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 2 & -1 & 1 & 0 \\ 1 & -2 & 1 & 2 \end{pmatrix}\) and \(f_A:\R^4\to \R^4\) given by \(f_A(x)=Ax\)
We compute: \(\chi_A(x)=(x-2)(x-1)(x+1)(x-2)\)
Then \(E_{2}=\ker(A-2I)=\ker \begin{pmatrix} -3 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 2 & -1 & -1 & 0 \\ 1 & -2 & 1 & 0 \end{pmatrix}=\ker \begin{pmatrix} 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}=\text{span}\{(1,1,1,0),(0,0,0,1)\}\)
\(E_{1}=\ldots E_{-1}=\ldots\)
Thus \(A\) is diagonalizable and \(B=\{(1,1,1,0),(0,0,0,1),(-1,0,1,0),(0,0,-1,1)\}\)
Then \([f_{A}]_{\mathcal{B}}= \begin{pmatrix} 2 & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & 2 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & -1 & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 \end{pmatrix}\Rightarrow A=\left\lbrack f\right\rbrack_{E}=C\left(B,E\right)\left \lbrack f_{A}\right\rbrack_{E}C\left(E,B\right)\)
Thus \(A= \begin{pmatrix} 1 & 0 & -1 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & 2 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & -1 & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & -1 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 \end{pmatrix}^{-1}\)
And we know \([(x,y,z,t)]_{\mathcal{B}}=C(E,B)(x,y,z,t)= \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & -2 & 1 & 1 \\ -1 & 1 & 0 & 0 \\ 1 & -2 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}= \begin{pmatrix} y \\ x-2y+z+t \\ -x+y \\ x-2y+z \end{pmatrix}\)
Then \((x,y,z,t)=y(1,1,1,0)+(x-2y+z+t)(0,0,0,1)+(-x+y)(-1,0,1,0)+(x-2y+z)(0,0,-1,1)\)
Then \(f(x,y,z,t)=2y(1,1,1,0)+2(x-2y+z+t)(0,0,0,1)+\left(-1)(-x+y\right)(-1,0,1,0)+\left (1)(x-2y+z\right)(0,0,-1,1)\)
Then \(f_{1}(x,y,z,t)=y(1,1,1,0)+(x-2y+z+t)(0,0,0,1)=\left(y,y,y,x-2y+z+t\right)\)
\(f_{2}(x,y,z,t)=(-x+y)(-1,0,1,0)=(x-y,0,-x+y,0)\)
\(f_{3}(x,y,z,t)=(x-2y+z)(0,0,-1,1)=\left(0,0,-x+2y-z,x-2y+z\right)\)
Thus \(f=2f_1+(-1)f_2+(1)f_3\), \(Id=f_1+f_2+f_3\), \(f_i\circ f_j=0\), \(f_i^2=f_i\), \(Imf_{i}=E_{\lambda_{i}}\)
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Theorem
\(f:\R^5\to \R^5\) such that \([f]_{\mathcal{E}}= \begin{pmatrix} -1 & 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 1 \end{pmatrix}\)
Find \(f\)-invariant subspace \(W_i\subseteq \R^5\) such that \(\R^{5}=\oplus W_{i}\) and \(m_{f}(x)=\prod m_{f|_{W_i}}(x)\)
Solution
After calculating \(\chi_{f}(x)=x^{2}\left(x-1\right)^{3}\), let's find \(m_f(x)\)
\(m_f(x)=x(x-1)\)
Then \([f]_{\mathcal{E}}\left([f]_{\mathcal{E}}-1\right)= \begin{pmatrix} -1 & 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} -2 & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \end{pmatrix}\neq0\)
.......
Another way is to find
\(E_0=\ker f=\ker \begin{pmatrix} -1 & 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 1 \end{pmatrix}=\text{span}\{(1,1,1,1,1)\}\) and \(\dim E_0\neq mult(0,\chi_f)\) then \(r_1=2\)
\(\ker f^{2}=\ker \begin{pmatrix} -1 & 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 1 \end{pmatrix}^{2}=\ker \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 1 \end{pmatrix}=\text{span}\{(1,0,0,0,0),(0,1,1,1,1)\}\)
\(\mathcal{B}_{E_0}=\{(1,1,1,1,1)\},\mathcal{B}_{E_0^2}=\{(1,0,0,0,0),(0,1,1,1,1)\}\)
Since \(\mathcal{B}_{E_0^2}^{\prime}=\{(1,0,0,0,0),(-1,-1,-1,-1,-1)\}\supseteq B_{E_0}\), then \(\{0\}\subsetneq\ker f\subsetneq\ker f^{2}\)
\(E_{1}=\ker(f-Id)=\ker \begin{pmatrix} -2 & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \end{pmatrix}=\text{span}\{e_{3},e_{4},e_{5}\}\) and \(\dim E_1=3\)
Thus \(m_f(x)=x^2(x-1)\), \(W_{1}=\ker f^{2}=E_{0}^{2}=\text{span}\{(1,0,0,0,0),(-1,-1,-1,-1,-1)\}\) and \(W_{2}=\ker(f-Id)=E_{1}=\text{span}\{e_{3},e_{4},e_{5}\}\)
We can check that \(\R^5=W_1\oplus W_2\) and \(W_i\) is \(f\)-invariant and \(m_{f|_{W_1}}(x)=x^2\) and \(m_{f|_{W_1}}(x)=\left(x-1\right)\)
Then \([f]_\mathcal{B}=\)
\(= \begin{pmatrix} 0 & \placeholder{} & \placeholder{} & \placeholder{} & \placeholder{} \\ 1 & 0 & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & 0 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 0 & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & \placeholder{} & 0 \end{pmatrix}\text{nilpotent call it }N^{\prime}+ \begin{pmatrix} 0 & \placeholder{} & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & 0 & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & 1 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & \placeholder{} & 1 \end{pmatrix}\text{diagonalizable call it }D'\) -
Theorem
\(D,N\) is uniquely determined by 1 and 2
\([f]_E=C(B,E)[f]_BC(E,B)\) and \([f]_B=D'+N'\Rightarrow [f]_E=C(B,E)D'C(B,E)^{-1}+C(B,E)N'C(B,E)^{-1}\)
\(= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 1 \end{pmatrix}\text{call it }D+ \begin{pmatrix} -1 & 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \end{pmatrix}\text{call it }N\)
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Definition
Def. If \(v \in V\), the \(f\)-cyclic subspace generated by \(v\) is $\langle v \rangle_{f} = { w \in V \mid w = p(f)v \text{ for any }p \in K[x] } $
If \(\langle v \rangle_f = V\), then we say that \(V\) is cyclic.
Theorem and Theorem: \(f: V \to V\) a linear transformation. \(f\) has a cyclic vector \(v\) if and only if there exists an ordered basis \(B\) such that the companion matrix of \(M_{f,v}\) is \([f]_B\)
Corollary: If \(A\) is the companion matrix of a monic polynomial \(p\), then \(p=m_A=\chi_A\)
Prove that if \(f\) is a linear operator on a finite dimensional vector space over \(\mathbb{K}\), then \(m_f=\chi_f\) iff \(f\) has a cyclic vector
First we do following exercise
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\(A= \begin{pmatrix} 1 & \placeholder{} & \placeholder{} & \placeholder{} & \placeholder{} \\ 1 & 1 & 0 & \placeholder{} & \placeholder{} \\ 0 & 1 & 1 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 2 & 0 \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 & 2 \end{pmatrix}\), find a cyclic vector of \(f_A\)
We know \(\chi_f(x)=(x-1)^3(x-2)^2\), then find minimal polynomial, use method 2, find kernel
\(\ker(A-I)=\ker \begin{pmatrix} 0 & 0 & 0 & \placeholder{} & \placeholder{} \\ 1 & 0 & 0 & \placeholder{} & \placeholder{} \\ 0 & 1 & 0 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 & 0 \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 & 1 \end{pmatrix}=\text{span}\{e_{3}\}\)
And \(\ker(A-I)^{2}=\ker \begin{pmatrix} 0 & 0 & 0 & \placeholder{} & \placeholder{} \\ 0 & 0 & 0 & \placeholder{} & \placeholder{} \\ 1 & 0 & 0 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 & 0 \\ \placeholder{} & \placeholder{} & \placeholder{} & 2 & 1 \end{pmatrix}=\text{span}\{e_{2},e_{3}\}\)
\(\ker(A-I)^{3}=\ker \begin{pmatrix} 0 & 0 & 0 & \placeholder{} & \placeholder{} \\ 0 & 0 & 0 & \placeholder{} & \placeholder{} \\ 0 & 0 & 0 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 & 0 \\ \placeholder{} & \placeholder{} & \placeholder{} & 3 & 1 \end{pmatrix}=\text{span}\{e_{1},e_{2},e_{3}\}\)
\(\ker(A-2I)=\ker \begin{pmatrix} -1 & 0 & 0 & \placeholder{} & \placeholder{} \\ 1 & -1 & 0 & \placeholder{} & \placeholder{} \\ 0 & 1 & -1 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 0 & 0 \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 & 0 \end{pmatrix}=\text{span}\{e_{5}\}\)
\(\ker(A-2I)^{2}=\ker \begin{pmatrix} 1 & 0 & 0 & \placeholder{} & \placeholder{} \\ -2 & 1 & 0 & \placeholder{} & \placeholder{} \\ 1 & -2 & 1 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 0 & 0 \\ \placeholder{} & \placeholder{} & \placeholder{} & 0 & 0 \end{pmatrix}=\text{span}\{e_{4},e_{5}\}\)
Thus \(m_f(x)=(x-1)^3(x-2)^2=\chi_f(x)\)
And we know \(\{0\}\subsetneq \ker(f-Id)\subsetneq \ker(f-Id)^2\subsetneq \ker(f-Id)^3\)
Then \(\exists v_1\in \ker(f-Id)^3\setminus \ker(f-Id)^2\)
Since \(\{0\}\subsetneq\ker(f-2Id)\subsetneq\ker(f-2Id)^{2}\), then \(\exists v_{2}\in\ker(f-2Id)^{2}\setminus\ker(f-2Id)^{1}\)
Use primary decomposition theorem, \(V=\left\lbrack\ker(f-Id)^{3}=W_{1}\right\rbrack\oplus\left\lbrack\ker(f-2Id)^{2} =W_{2}\right\rbrack\) where \(v=u_{1}+u_{2},u_{i}\in W_{i}\)
Define \(v=v_1+v_2=e_1+e_4\)
If \(g\) is linear operator, then \(g(e_1+e_4)=g(e_1)+g(e_4)\)
If \(g(x)=(f(x)-Id)^{3}(f(x)-2Id)^{2}=(f(x)-2Id)^{2}(f(x)-Id)^{3}\)
Thus \(g(e_{1}+e_{4})=g(e_{1})+g(e_{2})=(f-2Id)^{2}(f-Id)^{3}\left(e_{1}\right)+(f-Id)^{3} (f-2Id)^{2}\left(e_{4}\right)=0\)
Suppose power less than 2 or 3 since the relation with kernel , then it different from zero. Similarly
Then \(m_{f,v}(x)=m_f(x)\), then \(v\) is a cyclic vector and \(B = \{ e_1 + e_4, (1,1,0,2,1), (1,2,1,4,4), (1,3,3,8,12), (1,4,6,16,32) \}\)
Where \(v \rightarrow f(v) \rightarrow f^{2}(v) \rightarrow f^{3}(v) \rightarrow f^{4}(v )\)
If \(m_f = \chi_f\), then there exists a vector \(v\) such that \(m_v = m_f = \chi_f\).
\(\Rightarrow \{ v, f(v), f^2(v), \dots, f^{n-1}(v) \}\)is a basis for \(V\).
Primary Decomposition Theorem (P.D.Th.):
\(\chi_f(X) = \prod_{i=1}^{l} p_i^{r_i}(X)\) where \(p_i\) are irreducible polynomials.
Define subspaces: \(W_i = \ker p_i(f)^{r_i}\)and elements: \(W_i \in \ker p_i(f)^{r_i} \setminus \ker p_i(f)^{r_i-1}\)
The space \(V\) is decomposed as: \(V = \sum_{i=1}^{l}W_{i}\)Thus, \(V\) is cyclic.