= Solution
The <Cayley-Hamilton theorem> states
$$
\boxed{\chi_T(T)=0.}
$$
Let $d=\dim V$. 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 $\lambda_1,\ldots,\lambda_d$ and put $V_i=\operatorname{span}(e_1,\ldots,e_i)$. Each $V_i$ is an <invariant subspace>, and
$$
(T-\lambda_iI)V_i\subseteq V_{i-1}.
$$
Apply $T-\lambda_dI$ first, then $T-\lambda_{d-1}I$, and continue. The resulting product sends $V_d$ into $V_0=\{0\}$. These factors are <commuting operators>, so
$$
\prod_{i=1}^d(T-\lambda_iI)=0.
$$
The <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.
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