Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}4&2&-5\\-3&2&6\\0&0&4\end{matrix} \right) $$The characteristic polynomial for matrix A is:
$$ p(\lambda) = -\lambda^3+10\lambda^2-38\lambda+56 $$The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(x) &= det(A - x I) = det \left( \left[ \begin{matrix}4&2&-5\\-3&2&6\\0&0&4\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}4 - \lambda &2&-5\\-3&2 - \lambda &6\\0&0&4 - \lambda \end{matrix} \right| = \\ &= (-\lambda+4) (-\lambda+2) (-\lambda+4) + 2 \cdot 6 \cdot 0 + (-5) \cdot (-3) \cdot 0 - (-\lambda+4) \cdot 6 \cdot 0 - (-3) \cdot 2 \cdot (-\lambda+4) - 0 \cdot (-\lambda+2) \cdot (-5) = \\ & = -\lambda^3+10\lambda^2-32\lambda+32 + 0 + 0 - 0 - (6\lambda-24) - 0 = \\ & = -\lambda^3+10\lambda^2-38\lambda+56 \end{aligned} $$