Truncated binomial cube root of the identity plus a nilpotent matrix (${\left({\sum }_{j=0}^{m-1}\left(\genfrac{}{}{0ex}{}{1/3}{j}\right)N^j\right)}^3=I+N,N^m=0$)

Over the complex numbers, if $N^m=0$, thenHere the generalized binomial coefficient is $\left(\genfrac{}{}{0ex}{}{\alpha }{j}\right)=\alpha (\alpha -1)\cdots (\alpha -j+1)/j!$, with $\left(\genfrac{}{}{0ex}{}{\alpha }{0}\right)=1$. This is an exact finite identity, even when an infinite binomial series would not converge.

One can construct the required polynomial $q(z)=1+{\sum }_{j=1}^{m-1}{c}_j{z}^j$ without any infinite series. At polynomial degree $j$, the coefficient of $q(z{)}^3$ is $3c_j$ plus terms involving only earlier coefficients. Recursively choose it to agree with $1+z$ through polynomial degree $m-1$; the first coefficients are $1/3,-1/9,5/81$. The stated generalized binomial coefficients solve this recursion: near $z=0$, the binomial series for $(1+z{)}^{1/3}$ cubes to $1+z$, and truncating it changes no coefficient below polynomial degree $m$. Thus $q(z{)}^3-(1+z)$ is divisible by $z^m$ as a polynomial identity. Substituting the nilpotent matrix $N$ kills the remaining terms, all divisible by $N^m$. For a Jordan block $\lambda I+N$ with $\lambda \ne 0$, replace $N$ by $N/\lambda$ and multiply $Q$ by a scalar cube root of $\lambda$.