Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}1&1&0\\0&2&0\\0&0&1\end{matrix} \right) $$The characteristic polynomial for matrix A is:
$$ p(\lambda) = -\lambda^3+4\lambda^2-5\lambda+2 $$The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(x) &= det(A - x I) = det \left( \left[ \begin{matrix}1&1&0\\0&2&0\\0&0&1\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}1 - \lambda &1&0\\0&2 - \lambda &0\\0&0&1 - \lambda \end{matrix} \right| = \\ &= (-\lambda+1) (-\lambda+2) (-\lambda+1) + 1 \cdot 0 \cdot 0 + 0 \cdot 0 \cdot 0 - (-\lambda+1) \cdot 0 \cdot 0 - 0 \cdot 1 \cdot (-\lambda+1) - 0 \cdot (-\lambda+2) \cdot 0 = \\ & = -\lambda^3+4\lambda^2-5\lambda+2 + 0 + 0 - 0 - 0 - 0 = \\ & = -\lambda^3+4\lambda^2-5\lambda+2 \end{aligned} $$