= Solution
Proceed by induction on $n=\dim V$. The zero-dimensional case is immediate. Over $\mathbb C$, the <fundamental theorem of algebra> supplies a root $\lambda$ of the <characteristic polynomial>. Since $\det(T-\lambda I)=0$, there is a nonzero vector $v_1$ with $Tv_1=\lambda v_1$. The line $W=\mathbb Cv_1$ is invariant, so $T$ induces a <linear map> $\overline T$ on $V/W$. By induction, choose an upper triangular <basis> for that quotient and lift its vectors to $v_2,\ldots,v_n$. Then $Tv_j\in\operatorname{span}(v_1,\ldots,v_j)$ for every $j$. Thus $v_1,\ldots,v_n$ is the required <basis> and $\boxed{T\text{ is triangularizable}}$.
The same conclusion holds over $F$ when the <characteristic polynomial> splits over $F$. Indeed an <eigenvalue> then lies in $F$, and the <block matrix> arising from the invariant line gives $\chi_T(X)=(X-\lambda)\chi_{\overline T}(X)$. The quotient <characteristic polynomial> therefore also splits, so the induction works over $F$ without requiring <algebraic closure>.
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