To the Jordan canonical form by the factor space (Q1383666)

It is shown that each linear mapping \(f\) of an \(n\)-dimensional vector space \(V\) over an algebraically closed field has a Jordan basis \(B\), i.e., the matrix of \(f\) with respect to \(B\) is Jordan. The result is obvious for \(n=1\). For \(n>1\) the author uses induction as follows: There is a (non-zero) eigenvector \(v\) of \(f\), whence \(f\) induces a linear mapping \(f'\) on the factor space \(V/\text{span} (v)\). So \(f'\) has a Jordan basis \(B'\). Then each element of \(B'\) can be represented by such a vector \(d\) of \(V\) that \(v\) together with all \(d\)'s yields a Jordan basis of \(f\). Reviewer's remark: A similar proof can be found in the book of \textit{A. I. Kostrikin} and \textit{Y. I. Manin} [Linear algebra and geometry, (1989; Zbl 0755.15002) pp. 61-62].