Let \(A\) be a \(3 \times 3\) matrix over the field of complex numbers. We form the matrix \(xI - A\) with polynomial entries, the \(i\), \(j\) entry of this matrix being the polynomial \(\delta_{ij}x - A_{ij}\). If \(f = \det(xI - A)\), show that \(f\) is a monic polynomial of degree 3. If we write \[ f = (x - c_1)(x - c_2)(x - c_3) \] with complex numbers \(c_1\), \(c_2\), and \(c_3\), prove that \[ c_1 + c_2 + c_3 = \operatorname{trace}(A) \quad \text{and} \quad c_1 c_2 c_3 = \det A. \]
Solution
In the determinant expansion of \(xI-A\), the identity permutation contributes \[ (x-A_{11})(x-A_{22})(x-A_{33}), \] whose leading term is \(x^3\). Every other permutation uses at least two off-diagonal entries and therefore has degree at most \(1\) in \(x\). Thus \[ f(x)=\det(xI-A) \] is monic of degree \(3\).
The coefficient of \(x^2\) comes only from the identity-permutation term, and is \[ -(A_{11}+A_{22}+A_{33})=-\operatorname{trace}(A). \] On the other hand, from \[ f(x)=(x-c_1)(x-c_2)(x-c_3) \] the coefficient of \(x^2\) is \(-(c_1+c_2+c_3)\). Hence \[ c_1+c_2+c_3=\operatorname{trace}(A). \]
Finally, \[ f(0)=\det(-A)=(-1)^3\det A=-\det A, \] while the factorization gives \[ f(0)=(-c_1)(-c_2)(-c_3)=-c_1c_2c_3. \] Therefore \[ c_1c_2c_3=\det A. \]