Solution: et us find the eigenvalues of the given matrix. \begin{align*} \det (A - \lambda I) = 0 & \implies \begin{vmatrix} -\lambda & 0 & 0 & -4 \\ 1 & -\lambda & 0 & 0 \\ 0 & 1 & -\lambda & 5 \\ 0 & 0 & 1 & -\lambda \\ \end{vmatrix} = 0 \\[1ex] & \implies -\lambda \begin{vmatrix} -\lambda & 0 & 0 \\ 1 & -\lambda & 5 \\ 0 & 1 & -\lambda \\ \end{vmatrix} + 4 \begin{vmatrix} 1 & -\lambda & 0 \\ 0 & 1 & -\lambda \\ 0 & 0 & 1 \\ \end{vmatrix} = 0 \\[1ex] & \implies \lambda ^2 \left( \lambda ^2 -5 \right) + 4 = 0 \\ & \implies \lambda ^4 - 5\lambda ^2 + 4 = 0 \\ & \implies (\lambda -1)(\lambda +1)(\lambda -2)(\lambda +2) = 0. \end{align*} Thus, the characteristic polynomial of $A$ will be \[ p(x) = (x - 1)(x + 1)(x - 2)(x + 2). \]

Since each of the factor only occurs once, the minimal polynomial will be same as the characteristic polynomial. Hence, the Jordan canonical form of $A$ will be \[ J = \begin{bmatrix} \begin{bmatrix} 1 \end{bmatrix} & 0 & 0 & 0 \\ 0 & \begin{bmatrix} -1 \end{bmatrix} & 0 & 0 \\ 0 & 0 & \begin{bmatrix} 2 \end{bmatrix} & 0 \\ 0 & 0 & 0 & \begin{bmatrix} -2 \end{bmatrix} \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & -2 \\ \end{bmatrix}. \]