 Triangularization using dual spaces

A linear map on a finite-dimensional vector space over an algebraically closed field has both an upper triangular matrix and a lower triangular matrix in suitable bases. One proof uses invariant subspaces and annihilators to make the induction step entirely in the dual space.

Choose an eigenvector $v$ and let $U=\mathrm{span}\{v\}$. The dual map preserves the annihilator of a vector subspace $U^0$, of dimension one less. By induction, choose a basis of $U^0$ making the restriction upper triangular. Extend this to a basis of the whole dual space; its matrix $C$ remains upper triangular because $U^0$ is an invariant subspace. In the corresponding primal basis, the matrix is $C^T$, which is lower triangular. Reversing the primal basis makes it upper triangular. Dimension zero and one start the induction. This also allows the original eigenvector to be the first vector of the resulting upper triangular basis.