:: Answers to practise problems for consequences of triangular block form

      1. \( c_A(\lambda) = (\lambda + 1)^3 \)
      2. \( A + I = \begin{bmatrix} -4 & 3 & 1 \\ -4 & 3 & 1 \\ -4 & 3 & 1 \end{bmatrix} \)

        \( (A + I)^2 = \mathbf{0} \implies (A + I)^3 = \mathbf{0} \)
      1. \( c_A(\lambda) = (\lambda + 1)^2 (\lambda - 1)^2 \)
      2. \( A + I = \begin{bmatrix} -2 & 6 & 3 & 2 \\ -2 & 4 & 2 & 2 \\ -1 & 3 & 1 & 1 \\ -1 & 1 & 2 & 1 \end{bmatrix} \)      \( A - I = \begin{bmatrix} -4 & 6 & 3 & 2 \\ -2 & 2 & 2 & 2 \\ -1 & 3 & -1 & 1 \\ -1 & 1 & 2 & -1 \end{bmatrix} \)

        \( (A + I)^2 = \begin{bmatrix} -13 & 23 & 13 & 13 \\ -8 & 12 & 8 & 8 \\ -6 & 10 & 6 & 6 \\ -3 & 5 & 3 & 3 \end{bmatrix} \)      \( (A - I)^2 = \begin{bmatrix} -1 & -1 & 1 & 5 \\ 0 & 0 & 0 & 0 \\ -2 & -2 & 6 & 2 \\ 1 & 1 & -5 & 3 \end{bmatrix} \)

        \( (A + I)^2 (A - I)^2 = \mathbf{0} \)
  2. \( C_p = \begin{bmatrix} 0 & 0 & 0 & 2 \\ 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & -1 \end{bmatrix} \)

    \( c_{C_p}(\lambda) = \lambda^4 + \lambda^3 + 4 \lambda^2 - 5 \lambda - 2 \)