Let \(V\) be a two-dimensional vector space over the field \(F\), and let \(\mathfrak{B}\) be an ordered basis for \(V\). If \(T\) is a linear operator on \(V\) and \[ [T]_{\mathfrak{B}} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] prove that \(T^2 - (a + d)T + (ad - bc)I = 0\).
Solution
Let \[ A=[T]_{\mathfrak B}=\begin{bmatrix}a&b\\c&d\end{bmatrix}. \] A direct multiplication gives \[ A^2-(a+d)A+(ad-bc)I_2=0. \] Indeed, the four entries cancel separately. Matrix representation in a fixed basis respects sums, scalar multiples, products, and the identity, so \[ [T^2-(a+d)T+(ad-bc)I]_{\mathfrak B}=0. \] The zero matrix represents only the zero operator. Hence \[ T^2-(a+d)T+(ad-bc)I=0. \]