Here is yet another definition in mathematics where it seems that conventions vary (almost like the orientation of titles on the spines of books): is a Jordan block an upper-triangular or lower-triangular matrix? In other words, which of the matrices
$latex A_1=\begin{pmatrix}\alpha & 1\\0&\alpha\end{pmatrix},\quad\quad A_2=\begin{pmatrix} \alpha & 0\\1&\alpha\end{pmatrix}$
is a Jordan block of size 2 with respect to the eigenvalue $latex \alpha$?
I have the vague impression that most elementary textbooks in Germany (I taught linear algebra last year…) use $latex A_1$, but for instance Bourbaki (Algèbre, chapitre VII, page 34, définition 3, in the French edition) uses $latex A_2$, and so does Lang’s “Algebra”. Is it then a cultural dichotomy (again, like spines of books)?
I have to admit that I tend towards $latex A_2$ myself, because I find it much easier to remember a good model for a Jordan block: over the field $latex K$, take the vector space $latex V=K[X]/X^nK[X]$, and consider the linear map $latex u\colon V\to V$ defined by $latex u(P)=\alpha P+XP$. Then the matrix of $latex u$ with respect to the basis $latex (1,X,\ldots,X^{n-1})$ is the Jordan block in its lower-triangular incarnation. The point here (for me) is that passing from $latex n$ to $latex n+1$ is nicely “inductive”: the formula for the linear map $latex u$ is “independent” of $latex n$, and the bases for different $latex n$ are also nicely meshed. (In other words, if one finds the Jordan normal form using the classification of modules over principal ideal domains, one is likely to prefer the lower-triangular version that “comes out” more naturally…)
