Jordan block

A Jordan block has one eigenvalue on its diagonal and ones on its superdiagonal.

For $T=\lambda I+N$ with $N^2=0\ne N$ and a polynomial $p$, the binomial formula gives $p(T)=p(\lambda )I+p^{\prime }(\lambda )N$. Hence its module on the two-dimensional space is indecomposable exactly when $p^{\prime }(\lambda )\ne 0$: in that case the nonzero nilpotent part is one Jordan block; otherwise the operator is scalar and every line is a submodule. For $p(X)=X^2$ over a field of characteristic different from two, the criterion is $\lambda \ne 0$.

The model matrix $B=\left(\begin{array}{cc}0& 1\\ ϵ& d\end{array}\right)$ has a defective double eigenvalue when $d=ϵ=0$. Its characteristic polynomial is ${\lambda }^2-d\lambda -ϵ$, giving the displayed roots. For fixed nonzero $d$ the small root is $-ϵ/d+O({ϵ}^2)$ and the other is $d+ϵ/d+O({ϵ}^2)$. These expansions cease to be ordered at $d=O(\sqrt{ϵ})$, the distinguished limit in which both roots split on a square-root scale. This is the generic leading behavior when a perturbation closes a two-dimensional Jordan block through a nonzero lower-left entry; it need not occur for a perturbation that leaves that entry zero.

A nilpotent Jordan block has zeros on its diagonal and ones on its superdiagonal. Its $k$th power has ones on the $k$th superdiagonal, and a block of size $d$ satisfies $J_d(0{)}^d=0$.