 Solution

The Cayley-Hamilton theorem statesLet $d=\dim V$. Extend the field to an algebraically closed field and choose a basis in which the matrix of $T$ is an upper triangular matrix, as permitted. Write its diagonal entries as ${\lambda }_1,\dots ,{\lambda }_d$ and put $V_i=\mathrm{span}(e_1,\dots ,e_i)$. Each $V_i$ is an invariant subspace, andApply $T-{\lambda }_{d}I$ first, then $T-{\lambda }_{d-1}I$, and continue. The resulting product sends $V_d$ into $V_0=\{0\}$. These factors are commuting operators, soThe determinant of an upper triangular matrix gives ${\chi }_T(x)=(-1{)}^d{\prod }_i(x-{\lambda }_i)$. Thus ${\chi }_T(T)=0$ after the field extension. Its entries were in the original field, whose embedding is injective, so they were already zero there. The zero-dimensional case is immediate as well.