= Solution
We prove by <mathematical induction> on $n=\dim V$ that every <linear map> on an $n$-dimensional <complex vector space> admits both triangular forms. For $n=0$ or $1$ this is immediate. When $n>1$, the <fundamental theorem of algebra> supplies an <eigenvector> $v$; its <span> $U=\langle v\rangle$ is an <invariant subspace>.
By parts (ii) and (iii), $U^0$ has <dimension> $n-1$ and is an <invariant subspace> of the <dual map> $T'$. Apply the induction hypothesis to $T'|_{U^0}$ to choose a <basis> $\varphi_1,\ldots,\varphi_{n-1}$ with an <upper triangular matrix>. Extend it by a <linear functional> $\varphi_n$ satisfying $\varphi_n(v)=1$. Such a functional exists by <basis extension>, and cannot belong to $U^0$. In this <basis> of $V'$, the <matrix> of $T'$ has the form
$$
C=\begin{pmatrix}B&b\\0&\lambda\end{pmatrix},
$$
where $B$ is an <upper triangular matrix>; here the last diagonal entry follows by evaluating $T'\varphi_n$ at $v$, where $Tv=\lambda v$.
There is a <basis> $e_1,\ldots,e_n$ of $V$ with $\varphi_i(e_j)=\delta_{ij}$, since the <dual basis> construction gives a finite-dimensional <bidual evaluation map> isomorphism. In particular $e_n=v$. The <matrix transpose> relationship for a <dual map> gives
$$
[T]_{(e_1,\ldots,e_n)}=C^T,
$$
a <lower triangular matrix>. Reversing this <basis> turns a <lower triangular matrix> into an <upper triangular matrix>. If $R$ is the <permutation matrix> reversing the order, we have proved
$$
\boxed{[T]_{(e_1,\ldots,e_n)}=C^T\text{ lower triangular},\qquad
[T]_{(e_n,\ldots,e_1)}=RC^TR\text{ upper triangular}}.
$$
This completes the induction by <triangularization using dual spaces>.
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