Let \(L \in M_5(\mathbb{R})\) be a \(5 \times 5\) real matrix with eigenvalues \(0, 1 + i, 1 + 2i\). Find the characteristic polynomial \(\chi_L(x)\) and the minimal polynomial \(\mu_L(x)\) of \(L\) over \(\mathbb{R}\).
Solution
Because \(L\) has real entries, nonreal eigenvalues occur in complex-conjugate pairs. Thus from the given eigenvalues \(0\), \(1+i\), and \(1+2i\), we also obtain \(1-i\) and \(1-2i\). These are five distinct complex eigenvalues of a \(5\times5\) matrix, so each has algebraic multiplicity \(1\).
Hence \[ \begin{aligned} \chi_L(x) &=x(x-(1+i))(x-(1-i))(x-(1+2i))(x-(1-2i))\\ &=x\bigl((x-1)^2+1\bigr)\bigl((x-1)^2+4\bigr)\\ &=x(x^2-2x+2)(x^2-2x+5). \end{aligned} \] Expanding, \[ \chi_L(x)=x^5-4x^4+11x^3-14x^2+10x. \]
Every eigenvalue is a root of the minimal polynomial, so \(\mu_L\) has at least these five distinct roots. Therefore \(\deg\mu_L\ge5\). Since \(\mu_L\mid\chi_L\) by Cayley–Hamilton and \(\deg\chi_L=5\), we obtain \[ \mu_L(x)=\chi_L(x)=x(x^2-2x+2)(x^2-2x+5). \]