 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 $U=⟨v⟩$ is an invariant subspace.

By parts (ii) and (iii), $U^0$ has dimension $n-1$ and is an invariant subspace of the dual map $T^{\prime }$. Apply the induction hypothesis to $T^{\prime }{|}_{U^0}$ to choose a basis ${\phi }_1,\dots ,{\phi }_{n-1}$ with an upper triangular matrix. Extend it by a linear functional ${\phi }_n$ satisfying ${\phi }_n(v)=1$. Such a functional exists by basis extension, and cannot belong to $U^0$. In this basis of $V^{\prime }$, the matrix of $T^{\prime }$ has the formwhere $B$ is an upper triangular matrix; here the last diagonal entry follows by evaluating $T^{\prime }{\phi }_n$ at $v$, where $Tv=\lambda v$.

There is a basis $e_1,\dots ,e_n$ of $V$ with ${\phi }_i(e_j)={\delta }_{ij}$, 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 givesa 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 provedThis completes the induction by triangularization using dual spaces.