 Solution

Each entry of ${\mathrm{char}}_A(A)$ is a polynomial in the entries of $A$, so their simultaneous zero set $Z$ is Zariski closed. A matrix in $D(\Delta )$ has distinct eigenvalues and is diagonalizable. If $A=S\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)S^{-1}$, thenThus $D(\Delta )\subseteq Z$. Affine space is irreducible because its coordinate ring is an integral domain; by part (i), its nonempty open subset $D(\Delta )$ is dense. The closed set $Z$ containing it must be all of affine space. This proves the Cayley-Hamilton theorem: