= Triangularization using dual spaces
{title2=$[T']_{\mathcal B'}=[T]_{\mathcal B}^{T}$}
= Dual triangularization
{synonym}
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 $U=\operatorname{span}\{v\}$. 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>.
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