Let \(A\) be a \(2 \times 2\) matrix over a field \(F\). Then the set of all matrices of the form \(f(A)\), where \(f\) is a polynomial over \(F\), is a commutative ring \(K\) with identity. If \(B\) is a \(2 \times 2\) matrix over \(K\), the determinant of \(B\) is then a \(2 \times 2\) matrix over \(F\), of the form \(f(A)\). Suppose \(I\) is the \(2 \times 2\) identity matrix over \(F\) and that \(B\) is the \(2 \times 2\) matrix over \(K\) \[ B = \begin{bmatrix} A - A_{11}I & -A_{12}I \\ -A_{21}I & A - A_{22}I \end{bmatrix}. \] Show that \(\det B = f(A)\), where \(f = x^2 - (A_{11} + A_{22})x + \det A\), and also that \(f(A) = 0\).
Solution
Put \(a=A_{11}\), \(b=A_{12}\), \(c=A_{21}\), and \(d=A_{22}\). Since every element of \(K\) is a polynomial in \(A\), the entries of \(B\) commute. Therefore the \(2\times2\) determinant over \(K\) gives \[ \begin{aligned} \det B &=(A-aI)(A-dI)-(-bI)(-cI)\\ &=A^2-(a+d)A+(ad-bc)I\\ &=f(A), \end{aligned} \] where \[ f(x)=x^2-(a+d)x+(ad-bc). \] Since \(ad-bc=\det A\), this is the polynomial asserted in the exercise.
It remains to show that \(f(A)=0\). Direct multiplication gives \[ A^2= \begin{bmatrix} a^2+bc&ab+bd\\ ac+cd&bc+d^2 \end{bmatrix}. \] Hence \[ A^2-(a+d)A = \begin{bmatrix} bc-ad&0\\0&bc-ad \end{bmatrix} =-(ad-bc)I. \] Consequently \[ f(A)=A^2-(a+d)A+(ad-bc)I=0. \]