Skip to content

4.3 Cyclic Decomposition Theorem

Definition

Let \(f:V\to V\) is a vector space over \(\mathbb{K}\) and let \(W\) be a subspace of \(V\), we say that \(f\)-admissible if:

  1. \(W\) is \(f\)-invariant

  2. If \(p\in \mathbb{K}[x]\) and \(p(f)v\in W\), then there exists \(w\in W\) such that \(p(f)v=p(f)w\)​

Theorem(Cyclic Decomposition Theorem)

Let \(f:V\to V\) be a linear operator over a finite dimensional vector space \(V\) over \(\mathbb{K}\) and let \(W_0\) be a proper \(f\)-admissible subspace of \(V\). There exists non-zero vector \(v_1,...,v_r\) in \(V\) with respective minimal polynomials \(m_{v_1},...,m_{v_r}\) such that

  1. \(V=W_{0}\oplus\langle v_{1}\rangle_{f}\oplus\langle v_{2}\rangle_{f}\oplus...\oplus \langle v_{r}\rangle_{f}\)​

  2. \(m_{v_k}\mid m_{v_{k-1}}\) with \(k=2,...,r\)​

Furthermore, the integer \(r\) and the minimal polynomials \(m_{v_1},...,m_{v_r}\) are uniquely determined by 1) and 2) and the fact that \(v_k\neq 0\)

Exercise

  1. \(f:\R^8\to \R^8\) given by \([f]_{\mathcal{E}}= \begin{pmatrix} 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & -1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & -2 & 0 & 0 & 1 \end{pmatrix}\)​

    \(\chi_{f}(x)=\det \begin{pmatrix} x-1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & x-1 & 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & x-1 & 0 & 0 & 0 & -1 & 0 \\ -1 & 1 & 0 & x-1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & x+1 & 0 & 0 & 0 \\ 0 & 0 & -2 & 0 & 0 & x+1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & x-1 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 & 0 & x-1 \end{pmatrix}\)

    \(=(x-1)\det \begin{pmatrix} x-1 & 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & x-1 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & x-1 & 0 & 0 & 0 & -1 \\ -1 & 1 & 0 & x-1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & x+1 & 0 & 0 \\ 0 & 0 & -2 & 0 & 0 & x+1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & x-1 \end{pmatrix}\)​

    \(=(x-1)\left(x+1\right)\det \begin{pmatrix} x-1 & 0 & 0 & -1 & 0 & 0 \\ 0 & x-1 & 0 & -1 & 0 & 0 \\ 0 & 0 & x-1 & 0 & 0 & -1 \\ -1 & 1 & 0 & x-1 & 0 & 0 \\ 0 & 0 & -2 & 0 & x+1 & -1 \\ 0 & 0 & 0 & 0 & 0 & x-1 \end{pmatrix}\)

    \(=(x-1)(x+1)^{2}\det \begin{pmatrix} x-1 & 0 & 0 & -1 & 0 \\ 0 & x-1 & 0 & -1 & 0 \\ 0 & 0 & x-1 & 0 & -1 \\ -1 & 1 & 0 & x-1 & 0 \\ 0 & 0 & 0 & 0 & x-1 \end{pmatrix}\)​

    \(=(x-1)^{2}(x+1)^{2}\det \begin{pmatrix} x-1 & 0 & -1 & 0 \\ 0 & x-1 & -1 & 0 \\ -1 & 1 & x-1 & 0 \\ 0 & 0 & 0 & x-1 \end{pmatrix}\)

    \(=(x-1)^{3}(x+1)^{2}\det \begin{pmatrix} x-1 & 0 & -1 \\ 0 & x-1 & -1 \\ -1 & 1 & x-1 \end{pmatrix}=(x-1)^{6}(x+1)^{2}\)

    • \(\ker(f-Id)=\text{span}\{e_{1}+e_{2},e_{3}+e_{6},e_{8}\}\) and \(\dim(\ker(f-Id))=3\)​
    • \(\ker(f-Id)^2=\text{span}\{e_{1}+e_{2},e_4, e_{3}+e_{6},e_7,e_{8}\}\) and \(\dim(\ker(f-Id)^2)=5\)
    • \(\ker(f-Id)^{3}=\text{span}\{e_{1},e_{2},e_{3}+e_{6},e_{4},e_{7},e_{8}\}\) and \(\dim(\ker(f-Id)^3)=6\)

    Since \(\{0\}\subsetneq \ker(f-Id)\subsetneq \ker(f-Id)^{2}\subsetneq \ker(f-Id)^{3}\), thenimage​

    Then let callimage​

    Thus \(m_{e_1}=(x-1)^{3},m_{e_7}(x)=(x-1)^{2}=m_{e_4}(x),m_{e_8}(x)=(x-1)=m_{e_1+e_2}(x)=m_{e_3+e_4}(x)\)

    • \(\ker(f+Id)=\text{span}\{e_6,e_5+e_8\}\) and \(\dim\ker(f-Id)=mult(-1,\chi_{f})\)​

    Since \(\{0\}\subsetneq \ker(f+Id)\), then image​

    Let call image​

    Then \(m_{e_6}(x)=x+1=m_{e_5+e_8}\)

    Note that

    For \(e_1+e_6\), \(m_{e_1+e_6}(x)=(x-1)^{3}(x+1)=m_{f}(x)\neq \chi_{f}(x)\)

    Then \(\dim\langle e_{1}+e_{6}\rangle_{f}=4\) and basis for \(\lang e_1+e_6\rang_f\) is \(\{e_1+e_6,f(e_1+e_6),f^2(e_1+e_6),f^3(e_1+e_6)\}\)

    For \(e_{7}+e_{5}+e_{8}\), \(m_{e_7+e_5+e_8}(x)=(x-1)^{2}(x+1)\)

    Then \(\dim\lang e_{7}+e_{5}+e_{8}\rang_{f}=3\) and basis for \(\langle e_{7}+e_{5}+e_{8}\rangle_{f}\) is \(\{e_{7}+e_{5}+e_{8},f(e_{7}+e_{5}+e_{8}),f^{2}(e_{7}+e_{5}+e_{8})\}\)

    Then \(m_{e_7+e_5+e_8}(x)\mid m_{e_1+e_6}(x)\)

    \(\lang e_{8}\rang_{f}\), \(\dim\lang e_8\rang_f=1\) and \(m_{e_8}(x)=x-1\), then \(m_{e_8}(x)\mid m_{e_7+e_5+e_8}(x)\text{ and } m_{e_8}(x)\mid m_{e_1+e_6}(x)\)

    Thus \(\mathbb{R}^{8}=\langle e_{1}+e_{6}\rangle_{f}\oplus\langle e_{7}+e_{5}+e_{8}\rangle _{f}\oplus\langle e_{8}\rangle_{f}\)​

    image

    we choose this, then exists an preimage element

  2. \(f:\R^2\to \R^2\) given by \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\), this is a rotation

    \(f(x,y)=(-y,x)\), then \(\chi_f(x)=x^2+1\), then \(\ker(f^{2}+1)\supsetneq\{0\}\) where \(e_{1}\overset{f^2+1}{\longmapsto}0\)​

    Thus \(f\) has no \(f\)-admissible subspace of \(\dim 1\)​

  3. \(f:\R^3\to \R^3\) given by \(\begin{pmatrix} 0 & \placeholder{} & \placeholder{} \\ 1 & 0 & \placeholder{} \\ \placeholder{} & 1 & 0 \end{pmatrix}\), then \(\chi_f(X)=x^3=m_f(x)\)​

    Then \(\ker f^3\supsetneq \ker f^2\supsetneq \ker f\supsetneq \{0\}\) where \(e_1\mapsto e_2\mapsto e_3\)​

    Then \(W=\text{span}\{e_3,e_2\}\) is not \(f\)-admissible.

    \(W\) is \(f\)-invariant, let \(p(X)=x,v=e_1:p(f)(e_1)=e_2\in W\)​

    If \(v\in W\) such that \(p(f)(v)=e_{2}\Rightarrow\) \(v=\alpha e_3+\beta e_2\Rightarrow p(f)(v)=f(\alpha e_3)+f(\beta e_2)=\beta e_3\) is not equal \(e_2\)

‍

‍