Let \(A\) be a \(2 \times 2\) matrix over a field \(F\), and suppose that \(A^2 = 0\). Show for each scalar \(c\) that \(\det(cI - A) = c^2\).
Solution
Write \[ A=\begin{bmatrix}a&b\\c&d\end{bmatrix}. \] The equation \(A^2=0\) gives \[ a^2+bc=0,\qquad b(a+d)=0,\qquad c(a+d)=0,\qquad d^2+bc=0. \] If either \(b\) or \(c\) is nonzero, then \(a+d=0\). If \(b=c=0\), then \(a^2=d^2=0\); since \(F\) is a field, \(a=d=0\). Thus in every case \[ a+d=0. \] Using \(d=-a\) and \(a^2+bc=0\), \[ \det A=ad-bc=-a^2-bc=0. \] Therefore, for every scalar \(c_0\in F\), \[ \det(c_0I-A) =c_0^2-c_0(a+d)+\det A =c_0^2. \] (The symbol \(c_0\) here denotes the scalar called \(c\) in the exercise, to distinguish it from the matrix entry \(A_{21}\).)