= Solution
Translation $X\mapsto I+X$ is a <bijection> from <nilpotent matrices> to <unipotent matrices>. We can count the former without requiring a further partition identity. Let $a_n$ be their number, with $a_0=1$.
Every <linear operator> $T$ has a unique decomposition from the <Fitting lemma>
$$
V=\ker T^n\oplus\operatorname{im}T^n,
$$
on which it is respectively nilpotent and invertible. To justify the decomposition, kernels and images stabilize by step $n$; if $T^nv\in\ker T^n$, then $v\in\ker T^{2n}=\ker T^n$, so $T^nv=0$. The dimensions then show the sum is all of $V$, and stabilization makes $T$ invertible on the <image>. These two summands are determined by $T$, so there is no overcounting.
For a nilpotent summand of dimension $k$, the number of ordered complementary subspaces is $|GL_n(q)|/(|GL_k(q)||GL_{n-k}(q)|)$, by transporting two fixed bases. The restrictions of $T$ have $a_k$ and $|GL_{n-k}(q)|$ choices. Summing all $q^{n^2}$ <matrices> gives
$$
\frac{q^{n^2}}{|GL_n(q)|}=\sum_{k=0}^n\frac{a_k}{|GL_k(q)|}.
$$
Write $\phi_n(q^{-1})=\prod_{i=1}^n(1-q^{-i})$, so $|GL_n(q)|=q^{n^2}\phi_n(q^{-1})$. Subtract the same equation for $n-1$:
$$
\frac{a_n}{|GL_n(q)|}=\frac1{\phi_n(q^{-1})}-\frac1{\phi_{n-1}(q^{-1})}
=\frac{q^{-n}}{\phi_n(q^{-1})}.
$$
Thus <counting nilpotent matrices by the Fitting decomposition> proves
$$
\boxed{\#\{\text{unipotent elements of }GL_n(q)\}=a_n=q^{n^2-n}.}
$$
The supplied product identity is consistent with the resulting mass series: taking its variables to be $s=u$, $t=q^{-1}$ gives $\sum_na_nu^n/|GL_n(q)|=\prod_{i\ge1}(1-uq^{-i})^{-1}$.
Back to article page
