Let \[ A=\left(\begin{array}{ccc} -3 & 3 & -2 \\ -7 & 6 & -3 \\ 1 & -1 & 2 \end{array}\right) \in M_{3}(\mathrm{C}) .\]
Find the Jordan canonical form \(J\) of \(A\).
Find an invertible matrix \(P\) such that \(\inverseof{P} A P = J\). You do not need to compute \(\inverseof{P}\).
Solution
The characteristic polynomial is \[ \chi_A(x)=(x-1)(x-2)^2. \] For the eigenvalue \(2\), the eigenspace is one-dimensional, so there is a single Jordan block of size \(2\). Thus \[ J=\begin{bmatrix} 1&0&0\\ 0&2&1\\ 0&0&2 \end{bmatrix}. \]
Take \[ P=\begin{bmatrix} 1&\tfrac12&\tfrac12\\ 2&\tfrac12&1\\ 1&-\tfrac12&0 \end{bmatrix}. \] Its determinant is \(1/4\), so \(P\) is invertible. Its first column is an eigenvector for eigenvalue \(1\); if \(v_2,v_3\) are the second and third columns, then \[ (A-2I)v_2=0, \qquad (A-2I)v_3=v_2. \] Therefore the columns of \(P\) form a Jordan basis and \[ P^{-1}AP=J. \]