Let \[ A = \begin{bmatrix} 0 & 1 & 0 \\ 2 & -2 & 2 \\ 2 & -3 & 2 \end{bmatrix}. \] Is \(A\) similar over the field of real numbers to a triangular matrix? If so, find such a triangular matrix.
Solution
Let \(v=e_1=(1,0,0)^t\). Direct calculation gives \[ Av=(0,2,2)^t,\qquad A^2v=(2,0,-2)^t, \] and \[ A^3=0. \] Moreover \[ \det\begin{bmatrix}A^2v&Av&v\end{bmatrix}=4\ne0, \] so \[ \mathcal B=(A^2v,Av,v) \] is a basis of \(\mathbb R^3\). Relative to this basis, \[ A(A^2v)=0,\qquad A(Av)=A^2v,\qquad Av=Av, \] so the matrix of the operator is \[ [A]_{\mathcal B}= \begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix}. \] Hence \(A\) is similar over \(\mathbb R\) to this upper triangular matrix.