Let \(F\) be a subfield of the complex numbers, and let \[ A = \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}. \] Find the monic generator of the ideal of all polynomials \(f\) in \(F[x]\) such that \(f(A) = 0\).
Solution
The matrix \(A\) has two distinct eigenvalues, \(1\) and \(3\). Indeed, \[ Ae_1=e_1, \qquad A\begin{bmatrix}1\\-1\end{bmatrix} =3\begin{bmatrix}1\\-1\end{bmatrix}. \] If \(f(A)=0\), applying \(f(A)\) to these two eigenvectors gives \[ f(1)=0, \qquad f(3)=0. \] Hence \((x-1)(x-3)\) divides \(f\).
Conversely, \[ (A-I)(A-3I)=0, \] so the polynomial \((x-1)(x-3)\) annihilates \(A\). Therefore the ideal of annihilating polynomials is \[ ((x-1)(x-3)), \] and its monic generator is \[ \boxed{(x-1)(x-3)}. \]