:: Answers to practise problems for scalar-triangular form

  1. Note that other answers are possible besides the ones presented here. In choosing our basis vectors to create transition matrices, we often chose parameter values other than 1 in order to clear fractions.
    1. \(P = \begin{bmatrix} 1 & -3 & 1 \\ -2 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \)      \( P^{-1} A P = \begin{bmatrix} 5 & 0 & 2 \\ 0 & 5 & -8 \\ 0 & 0 & 5 \end{bmatrix} \)
    2. \(P = \begin{bmatrix} -1 & -1 & 1 \\ 2 & 0 & 0 \\ 2 & 4 & 0 \end{bmatrix} \)      \( P^{-1} A P = \begin{bmatrix} 5 & 2 & -4 \\ 0 & 5 & -2 \\ 0 & 0 & 5 \end{bmatrix} \)
    3. \(P = \begin{bmatrix} 1 & -1 & -2 & 1 \\ 3 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 3 & 0 \end{bmatrix} \)      \( P^{-1} A P = \begin{bmatrix} 5 & 0 & 0 & -3 \\ 0 & 5 & 0 & 18 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \end{bmatrix} \)
    4. \(P = \begin{bmatrix} 1 & -1 & 1 & 0 \\ -3 & 6 & 0 & 1 \\ -3 & 12 & 0 & 0 \\ 3 & 0 & 0 & 0 \end{bmatrix} \)      \( P^{-1} A P = \begin{bmatrix} 5 & 3 & 1 & \frac{20}{3} \\ 0 & 5 & -2 & -\frac{2}{3} \\ 0 & 0 & 5 & -\frac{1}{3} \\ 0 & 0 & 0 & 5 \end{bmatrix} \)
    5. \(P = \begin{bmatrix} 1 & 1 & 1 & 0 \\ -1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \end{bmatrix} \)      \( P^{-1} A P = \begin{bmatrix} 5 & 0 & -6 & 16 \\ 0 & 5 & 5 & -13 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \end{bmatrix} \)
    6. \(P = \begin{bmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & 2 & 0 \\ 3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \end{bmatrix} \)      \( P^{-1} A P = \begin{bmatrix} 5 & 0 & -1 & 3 \\ 0 & 5 & 4 & -\frac{32}{3} \\ 0 & 0 & 5 & \frac{2}{3} \\ 0 & 0 & 0 & 5 \end{bmatrix} \)
      1. \( c_A(\lambda) = (\lambda + 1)^3 \)
      2. \( \mathcal{B}_{E_{-1}(A)} = \left\{ \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 4 \end{bmatrix} \right\} \)

        Geometric multiplicity is not equal to algebraic multiplicity for \( \lambda = -1 \).
      3. \( \mathcal{B}_{E^2_{-1}(A)} = \left\{ \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 4 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right\} \)

        \( G_{-1}(A) = E^2_{-1}(A) = \mathbb{R}^3 \)
      1. \( c_A(\lambda) = (\lambda + 1)^2 (x - 1)^2 \)
      2. \( \mathcal{B}_{E_{-1}(A)} = \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\} \)

        Geometric multiplicity is not equal to algebraic multiplicity for \( \lambda = -1 \).

        \( \mathcal{B}_{E_{1}(A)} = \left\{ \begin{bmatrix} 5 \\ 2 \\ 2 \\ 1 \end{bmatrix} \right\} \)

        Geometric multiplicity is not equal to algebraic multiplicity for \( \lambda = 1 \).
      3. \( \mathcal{B}_{E^2_{-1}(A)} = \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} \right\} \)

        \( G_{-1}(A) = E^2_{-1}(A) \)

        \( \mathcal{B}_{E_{1}(A)} = \left\{ \begin{bmatrix} 5 \\ 2 \\ 2 \\ 1 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} \right\} \)

        \( G_1(A) = E^2_1(A) \)