Let \(A\) be the \(4 \times 4\) real matrix \[ A = \begin{bmatrix} 1 & 1 & 0 & 0 \\ -1 & -1 & 0 & 0 \\ -2 & -2 & 2 & 1 \\ 1 & 1 & -1 & 0 \end{bmatrix}. \] Show that the characteristic polynomial for \(A\) is \(x^2(x - 1)^2\) and that it is also the minimal polynomial.
Solution
Expanding \(\det(xI-A)\) gives \[ \chi_A(x)=x^4-2x^3+x^2=x^2(x-1)^2. \] Hence the minimal polynomial \(m_A\) divides \(x^2(x-1)^2\).
A direct multiplication gives \[ A^2(A-I)^2=0, \] so \(m_A\mid x^2(x-1)^2\), and also \[ A(A-I)\neq0,\qquad A^2(A-I)\neq0,\qquad A(A-I)^2\neq0. \] Since both \(0\) and \(1\) are roots of the characteristic polynomial, they are characteristic values of \(A\), and therefore both \(x\) and \(x-1\) divide \(m_A\). Among the divisors of \(x^2(x-1)^2\) divisible by \(x(x-1)\), the three proper possibilities are \[ x(x-1),\qquad x^2(x-1),\qquad x(x-1)^2. \] The three displayed nonvanishing products rule these out respectively. Therefore \[ m_A(x)=x^2(x-1)^2=\chi_A(x). \]