Let \[ A=\left[\begin{array}{lll}{0} & {1} & {-2} \\ {1} & {1} & {-3} \\ {1} & {2} & {-4}\end{array}\right] \in M_{3}(\mathbb{C}) \]
Find the Jordan canonical form \(J\) of \(A\).
Find an invertible matrix \(P\) such that \(\inverseof{P} AP = J\).
You should not need to compute \(\inverseof{P}\).
Solution
(a) Characteristic polynomial and Jordan form:
Compute \(\det(\lambda I - A)\): \[ \det\begin{pmatrix} \lambda & -1 & 2 \\ -1 & \lambda-1 & 3 \\ -1 & -2 & \lambda+4 \end{pmatrix} = \lambda [(\lambda-1)(\lambda+4) + 6] + 1 [-(\lambda+4) + 3] + 2 [2 + (\lambda-1)] \] \[ = \lambda (\lambda^2 + 3\lambda + 2) + 1(-\lambda - 1) + 2(\lambda + 1) = (\lambda+1) [\lambda(\lambda+2) - 1 + 2] = (\lambda+1)(\lambda^2 + 2\lambda + 1) = (\lambda+1)^3. \] Thus the only eigenvalue is \(\lambda = -1\) with algebraic multiplicity 3.
Compute \(A - (-1)I = A + I\): \[ A + I = \begin{pmatrix} 1 & 1 & -2 \\ 1 & 2 & -3 \\ 1 & 2 & -3 \end{pmatrix}. \] Row-reducing \(A+I\): \[ \begin{pmatrix} 1 & 1 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \implies \rank(A+I) = 2. \] The geometric multiplicity of \(\lambda = -1\) is \(\nullity(A+I) = 3 - 2 = 1\). Since there is only 1 Jordan block for \(\lambda = -1\), the Jordan canonical form is a single \(3 \times 3\) block: \[ J = \begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{pmatrix}. \]
(b) Finding the change-of-basis matrix \(P\):
We seek a generalized eigenvector chain \((v_1, v_2, v_3)\) such that: \[ (A+I)v_3 = v_2, \quad (A+I)v_2 = v_1, \quad (A+I)v_1 = 0 \quad (\text{with } v_1 \neq 0). \] Compute \((A+I)^2\): \[ (A+I)^2 = \begin{pmatrix} 1 & 1 & -2 \\ 1 & 2 & -3 \\ 1 & 2 & -3 \end{pmatrix}\begin{pmatrix} 1 & 1 & -2 \\ 1 & 2 & -3 \\ 1 & 2 & -3 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ 0 & -1 & 1 \\ 0 & -1 & 1 \end{pmatrix}. \] We choose \(v_3\) such that \((A+I)^2 v_3 \neq 0\). Choose \(v_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\). Then: \[ v_2 = (A+I)v_3 = \begin{pmatrix} -2 \\ -3 \\ -3 \end{pmatrix}, \] \[ v_1 = (A+I)v_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}. \] Check: \((A+I)v_1 = \begin{pmatrix} 1 & 1 & -2 \\ 1 & 2 & -3 \\ 1 & 2 & -3 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\).
Thus the columns of \(P = [v_1 \mid v_2 \mid v_3]\) give: \[ P = \begin{pmatrix} 1 & -2 & 0 \\ 1 & -3 & 0 \\ 1 & -3 & 1 \end{pmatrix}. \] Since \(\det(P) = 1(-3) - (-2)(1) = -1 \neq 0\), \(P\) is invertible and satisfies \(P^{-1}AP = J\).