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Homework 4

4-Homework-AlgebraB.pdf

Problem 1

Consider the linear operator \(f : \mathbb{R}^4 \to \mathbb{R}^4\) given by \(f(x, y, z, t) = (-4x - y - 5z + 5t, 9x + 2y + 9z - 9t, 4x + 2y + 5z - 4t, x + y + z).\)

  1. Find linear operators \(F_1, F_2, F_3\) such that:

    • \(f = \alpha_1 F_1 + \alpha_2 F_2 + \alpha_3 F_3\) for some \(\alpha_1, \alpha_2, \alpha_3 \in \mathbb{R}\)​
    • \(F_i^2 = F_i\) for \(1 \leq i \leq 3\),
    • \(\text{id}_{\mathbb{R}^4} = F_1 + F_2 + F_3\),
    • \(F_i F_j = 0\) for \(1 \leq i, j \leq 3\) and \(i \neq j\).

    Since \(f(x, y, z, t) = (-4x - y - 5z + 5t, 9x + 2y + 9z - 9t, 4x + 2y + 5z - 4t, x + y + z)\), then represent it in matrix

    \([f]_{E}=A= \begin{pmatrix} -4 & -1 & -5 & 5 \\ 9 & 2 & 9 & -9 \\ 4 & 2 & 5 & -4 \\ 1 & 1 & 1 & 0 \end{pmatrix}\)​

    Then \(\chi_{A}(x)=\det \begin{pmatrix} x+4 & 1 & 5 & -5 \\ -9 & x-2 & -9 & 9 \\ -4 & -2 & x-5 & 4 \\ -1 & -1 & -1 & x \end{pmatrix}=\left(-1\right)^{5}\cdot\left(-1\right)\cdot\det \begin{pmatrix} 1 & 5 & -5 \\ x-2 & -9 & 9 \\ -2 & x-5 & 4 \end{pmatrix}+\left(-1\right)^{6}\cdot\left(-1\right)\cdot\det \begin{pmatrix} x+4 & 5 & -5 \\ -9 & -9 & 9 \\ -4 & x-5 & 4 \end{pmatrix}+\left(-1\right)^{7}\cdot\left(-1\right)\cdot\det \begin{pmatrix} x+4 & 1 & -5 \\ -9 & x-2 & 9 \\ -4 & -2 & 4 \end{pmatrix}+(-1)^{8}\cdot x\cdot\det \begin{pmatrix} x+4 & 1 & 5 \\ -9 & x-2 & -9 \\ -4 & -2 & x-5 \end{pmatrix}\)

    \(=-5x^{2}+6x-1-\left(-9x^{2}+18x-9\right)+4x^{2}+6x-10+x(x^3-3x^2-7x+9)\)​

    \(=x^{4}-3x^{3}+x^{2}+3x-2=(x-1)(x+1)(x-2)(x-1)\)

    Then \(E_{1}=\ker \begin{pmatrix} 5 & 1 & 5 & -5 \\ -9 & -1 & -9 & 9 \\ -4 & -2 & -4 & 4 \\ -1 & -1 & -1 & 1 \end{pmatrix}=\ker \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & -1 \end{pmatrix}=\text{span}\{\left(1,0,0,1\right),\left(0,0,1,1\right)\}\)​

    Then \(E_{-1}=\ker \begin{pmatrix} 3 & 1 & 5 & -5 \\ -9 & -3 & -9 & 9 \\ -4 & -2 & -6 & 4 \\ -1 & -1 & -1 & -1 \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 1 & -1 \\ 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 3 \end{pmatrix}=\text{span}\{\left(1,-3,1,1\right)\}\)

    Then \(E_{2}=\ker \begin{pmatrix} 6 & 1 & 5 & -5 \\ -9 & 0 & -9 & 9 \\ -4 & -2 & -3 & 4 \\ -1 & -1 & -1 & 2 \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & -2 \\ 0 & 1 & 0 & -1 \end{pmatrix}=\text{span}\{\left(1,-1,-2,-1\right)\}\)

    Thus this is diagonalizable and \(B=\{\left(1,0,0,1\right),\left(0,0,1,1\right),\left(1,-3,1,1\right),\left(1,-1,- 2,-1\right)\}\)

    Then \([f_{}]_{\mathcal{B}}= \begin{pmatrix} 1 & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & 1 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & -1 & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 2 \end{pmatrix}\Rightarrow A=\left\lbrack f\right\rbrack_{E}=C\left(B,E\right)\left \lbrack f_{A}\right\rbrack_{B}C\left(E,B\right)\)

    Thus \(A= \begin{pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & -3 & -1 \\ 0 & 1 & 1 & -2 \\ 1 & 1 & 1 & -1 \end{pmatrix} \begin{pmatrix} 2 & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & 2 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & -1 & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & -3 & -1 \\ 0 & 1 & 1 & -2 \\ 1 & 1 & 1 & -1 \end{pmatrix}^{-1}\)

    And we know \([(x,y,z,t)]_{\mathcal{B}}=C(E,B)(x,y,z,t)= \begin{pmatrix} 3 & 1 & 2 & -2 \\ -7 & -2 & -6 & 7 \\ 1 & 0 & 1 & -1 \\ -3 & -1 & -3 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}= \begin{pmatrix} 3x+y+2z-2t \\ -7x-2y-6z+7t \\ x+z-t \\ -3x-y-3z+3t \end{pmatrix}\)

    Then \((x,y,z,t)_{B}=\left(3x+y+2z-2t)(1,0,0,1\right)+(-7x-2y-6z+7t)(0,0,1,1)+(x+z-t)(1 ,-3,1,1)+(-3x-y-3z+3t)(1,-1,-2,-1)\)

    Then \(F(x,y,z,t)=\left(3x+y+2z-2t)(1,0,0,1\right)+(-7x-2y-6z+7t)(0,0,1,1)-(x+z-t)(1,-3 ,1,1)+2(-3x-y-3z+3t)(1,-1,-2,-1)\)

    Then \(F_{1}(x,y,z,t)=\left(3x+y+2z-2t)(1,0,0,1\right)+(-7x-2y-6z+7t)\left(0,0,1,1\right )=\left(3x+y+2z-2t,0,-7x-2y-6z+7t,-4x-y-4z+5t\right)\)

    \(F_{2}(x,y,z,t)=(x+z-t)(1,-3,1,1)=\left(x+z-t,-3x-3z+3t,x+z-t,x+z-t\right)\)

    \(F_{3}(x,y,z,t)=(-3x-y-3z+3t)(1,-1,-2,-1)=\left(-3x-y-3z+3t,3x+y+3z-3t,6x+2y+6z-6 t,3x+y+3z-3t\right)\)

    Thus \(F=F_{1}+(-1)F_{2}+(2)F_{3}\), \(Id=F_{1}+F_{2}+F_{3}\), \(F_{i}\circ F_{j}=0\), \(F_{i}^{2}=F_{i}\)

  2. Find a diagonalizable operator \(D\) and a nilpotent operator \(N\) such that

    • \(f = N + D\)
    • \(DN = ND\)

    Since \(f\) is diagonalizable with \(B=\{\left(1,0,0,1\right),\left(0,0,1,1\right),\left(1,-3,1,1\right),\left(1,-1,- 2,-1\right)\}\), then \(N=0\) and \(D=f\)​

    And we know \(D\circ 0=0=0\circ D\)​

    Thus \(N=0,D=f\) is valid

Problem 2

Consider the linear operator \(f : \mathbb{R}^5 \to \mathbb{R}^5\) given by \(f(x, y, z, u, v) = (z, -2x - 2y + 2u, x - u, z, -3x - 3y + 3u + v).\)

  1. Find \(f\)-invariant subspaces \(W_1, W_2, W_3\) such that

    • \(\mathbb{R}^5 = W_1 \oplus W_2 \oplus W_3\),
    • \(m_f = \prod_{1 \leq i \leq 3} m_f |_{W_i}\),
    • If \(i \neq j\), then \(m_{f} |_{W_i}\) and \(m_f |_{W_j}\) are coprime.

    First calculate \(\chi_{f}(x)=\det \begin{pmatrix} x & 0 & -1 & 0 & 0 \\ 2 & x+2 & 0 & -2 & 0 \\ -1 & 0 & x & 1 & 0 \\ 0 & 0 & -1 & x & 0 \\ 3 & 3 & 0 & -3 & x-1 \end{pmatrix}=\left(x-1\right)\det \begin{pmatrix} x & 0 & -1 & 0 \\ 2 & x+2 & 0 & -2 \\ -1 & 0 & x & 1 \\ 0 & 0 & -1 & x \end{pmatrix}=\left(x-1\right)\left(x+2\right)\det \begin{pmatrix} x & -1 & 0 \\ -1 & x & 1 \\ 0 & -1 & x \end{pmatrix}\)

    \(=(x-1)(x+2)\left\lbrack x^{3}+x-x\right\rbrack=x^{3}\left(x-1\right)\left(x+2\right )\)​

    Then let's find \(m_f(x)\)​

    \(E_{0}=\ker \begin{pmatrix} 0 & 0 & -1 & 0 & 0 \\ 2 & 2 & 0 & -2 & 0 \\ -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 3 & 3 & 0 & -3 & -1 \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}=\text{span}\{(1,0,0,1,0)\}\)

    And \(\dim E_{0}<mult(0,\chi_{f}(x))\), thus this is impossible

    Then \(E_{0}^{2}=\ker \begin{pmatrix} 0 & 0 & -1 & 0 & 0 \\ 2 & 2 & 0 & -2 & 0 \\ -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 3 & 3 & 0 & -3 & -1 \end{pmatrix}^{2}=\ker \begin{pmatrix} 1 & 0 & 0 & -1 & 0 \\ 4 & 4 & 0 & -4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 & 0 \\ 3 & 3 & 0 & -3 & 1 \end{pmatrix}=\ker \begin{pmatrix} 1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}=\text{span}\{(1,0,0,1,0),\left(0,0,1,0,0\right)\}\)

    And \(\dim E_{0}^{2}<mult(0,\chi_{f}(x))\), thus this is impossible

    Then \(E_{0}^{3}=\ker \begin{pmatrix} 0 & 0 & -1 & 0 & 0 \\ 2 & 2 & 0 & -2 & 0 \\ -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 3 & 3 & 0 & -3 & -1 \end{pmatrix}^{3}=\ker \begin{pmatrix} 1 & 0 & 0 & -1 & 0 \\ 4 & 4 & 0 & -4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 & 0 \\ 3 & 3 & 0 & -3 & 1 \end{pmatrix}=\ker \begin{pmatrix} 1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}=\text{span}\{(-1,1,0,0,0),(1,0,0,1,0),\left(0,0,1,0,0\right)\}\)

    Then \(E_{1}=\ker \begin{pmatrix} 1 & 0 & -1 & 0 & 0 \\ 2 & 3 & 0 & -2 & 0 \\ -1 & 0 & 1 & 1 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 3 & 3 & 0 & -3 & 0 \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}=\text{span}\{(0,0,0,0,1)\}\)

    Then \(E_{-2}=\ker \begin{pmatrix} -2 & 0 & -1 & 0 & 0 \\ 2 & 0 & 0 & -2 & 0 \\ -1 & 0 & -2 & 1 & 0 \\ 0 & 0 & -1 & -2 & 0 \\ 3 & 3 & 0 & -3 & -3 \end{pmatrix}=\ker \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1 \end{pmatrix}=\text{span}\{(0,1,0,0,1)\}\)

    Thus \(m_{f}(x)=\chi_{f}(x)=x^{3}\left(x-1\right)\left(x+2\right)\)​

    Let's \(W_{1}=\ker f^{3},W_{2}=\ker\left(f-1\right),W_{3}=\ker\left(f+2\right)\)​

    Then by theorem, \(\mathbb{R}^5 = W_1 \oplus W_2 \oplus W_3\) and \(W_1, W_2, W_3\) are \(f\)-invariant subspaces and \(m_{f}|_{W_1}=x^{3},m_{f}|_{W_2}=x-1,m_{f}|_{W_3}=x+2\)​

    Then \(m_f = \prod_{1 \leq i \leq 3} m_f |_{W_i}\), then clearly, If \(i \neq j\), then \(m_{f} |_{W_i}\) and \(m_f |_{W_j}\) are coprime.

  2. Find a cyclic vector for \(f\).

    \(\exists v_{1}\in\ker f^{3}\setminus\ker f^{2},\exists v_{2}\in\ker\left(f-1\right )\setminus\left\lbrace0\right\rbrace,\exists v_{3}\in\ker\left(f+2\right)\setminus \left\lbrace0\right\rbrace\)​

    Since \(\R^5=W_1\oplus W_2\oplus W_3\), then \(v=u_1+u_2+u_3\), \(\forall u_i\in W_i\)​

    Then \(v=v_{1}+v_{2}+v_{3}\) where \(v\) is an cyclic vector

    Moreover, \(v=(-1,1,0,0,0)+(0,0,0,0,1)+(0,1,0,0,1)=(-1,2,0,0,2)\)​

Problem 3

Let \(f\) be a linear operator on \(\mathbb{R}^8\) whose matrix representation in the standard basis is \(\begin{bmatrix} 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \end{bmatrix}\)

Find a vector \(v \in \mathbb{R}^8\) such that the minimal polynomial of \(v\) with respect to \(f\) coincides with the minimal polynomial of \(f\), i.e., \(m_{f,v}(x) = m_f(x)\).

Let's compute the \(\chi_{f}(x)=\det \begin{bmatrix} x & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ -1 & x & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & x & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & x & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & x & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & x & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & x-1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & x-1 \end{bmatrix}=\left(x-1\right)\det \begin{bmatrix} x & 0 & 0 & 1 & 0 & 0 & 0 \\ -1 & x & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & x & 2 & 0 & 0 & 0 \\ 0 & 0 & -1 & x & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & x & 1 & 0 \\ 0 & 0 & 0 & 0 & -1 & x & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & x-1 \end{bmatrix}\)

\(=\left(x-1\right)^{2}\det \begin{bmatrix} x & 0 & 0 & 1 & 0 & 0 \\ -1 & x & 0 & 0 & 0 & 0 \\ 0 & -1 & x & 2 & 0 & 0 \\ 0 & 0 & -1 & x & 0 & 0 \\ 0 & 0 & 0 & 0 & x & 1 \\ 0 & 0 & 0 & 0 & -1 & x \end{bmatrix}=\left(x-1\right)^{2}\left\lbrack x\det \begin{bmatrix} x & 0 & 0 & 0 & 0 \\ -1 & x & 2 & 0 & 0 \\ 0 & -1 & x & 0 & 0 \\ 0 & 0 & 0 & x & 1 \\ 0 & 0 & 0 & -1 & x \end{bmatrix}+\left(-1\right)\cdot\left(-1\right)^{3}\det \begin{bmatrix} 0 & 0 & 1 & 0 & 0 \\ -1 & x & 2 & 0 & 0 \\ 0 & -1 & x & 0 & 0 \\ 0 & 0 & 0 & x & 1 \\ 0 & 0 & 0 & -1 & x \end{bmatrix}\right\rbrack\)

\(=\left(x-1\right)^{2}\left\lbrack x^{2}\det \begin{bmatrix} x & 2 & 0 & 0 \\ -1 & x & 0 & 0 \\ 0 & 0 & x & 1 \\ 0 & 0 & -1 & x \end{bmatrix}+\det \begin{bmatrix} -1 & x & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & x & 1 \\ 0 & 0 & -1 & x \end{bmatrix}\right\rbrack=\left(x-1\right)^{2}\left\lbrack x^{2}\left(\left(x^{2} +1\right)\left(x^{2}+2\right)\right)+x^{2}+1\right\rbrack\)

\(=\left(x-1\right)^{2}\left(x^{2}\left(x^{4}+3x^{2}+2\right)+x^{2}+1\right)=\left (x-1\right)^{2}\left(x^{6}+3x^{4}+3x^{2}+1\right)=\left(x-1\right)^{2}\left(x^{2} +1\right)^{3}=0\)

Then \(x=1\) or \(x^2=-1\)

\(E_{1}=\ker \begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \end{bmatrix}=\ker \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}=\text{span}\{(0,0,0,0,0,0,0,1)\}\)

And \(\dim E_{1}<mult(1,\chi_{f}(x))\), thus this is impossible

\(E_{1}^{2}=\ker \begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \end{bmatrix}^{2}=\text{span}\{(0,0,0,0,0,0,0,1),(0,0,0,0,0,0,1,0)\}\)

\(E_{x^2+1}=\ker \begin{bmatrix} 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \end{bmatrix}^{2}+Id=\ker \begin{bmatrix} 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0 \\ 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 2 \end{bmatrix}=\text{span}\{(x_1,x_2,x_1,x_2,x_5,x_6,0,0)\}\)

And \(\dim E_{x^2+1}<mult\), thus this is impossible

\(E_{\left(x^2+1\right)^2}=\ker\left( \begin{bmatrix} 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \end{bmatrix}^{2}+Id\right)^{2}=\ker \begin{bmatrix} 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0 \\ 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 2 \end{bmatrix}^{2}=\text{span}\{(x_1,x_2,x_3,x_4,x_5,x_6,0,0)\}\)

This is 6 dimension. Thus \(m_{f}(x)=(x-1)^{2}\left(x^{2}+1\right)^{2}\).

Let \(v_{1}\in\ker(f-1)^{2}\setminus\ker(f-1)\), then \(v_{2},v_{3}\in\ker\left(f^{2}+1\right)^{2}\setminus\ker\left(f^{2}+1\right)\) where \(v_1=e_7,v_2=e_3,v_3=e_4\)

Then \(v=v_{1}+v_{2}+v_{3}=e_{3}+e_{4}+e_{7}\) can satisfy \(m_{f,v}(x) = m_f(x)\).

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