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Fix an irreducible polynomial f of degree d with nonzero constant term. In $G{L}_{dm}(q)$, the number of matrices whose minimal polynomial is a power of f is $|G{L}_{dm}(q)|q^{d(m^2-m)}/|G{L}_m(q^d)|$. The primary matrix centralizer formula is identical to the unipotent centralizer formula over $F_{q^d}$, and summing reciprocal centralizer orders reduces to counting nilpotent matrices by the Fitting decomposition.

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 lemmaon which it is respectively nilpotent and invertible. To justify the decomposition, kernels and images stabilize by step $n$; if $T^{n}v\in \ker T^n$, then $v\in \ker T^{2n}=\ker T^n$, so $T^{n}v=0$. 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 $|G{L}_n(q)|/(|G{L}_k(q)||G{L}_{n-k}(q)|)$, by transporting two fixed bases. The restrictions of $T$ have $a_k$ and $|G{L}_{n-k}(q)|$ choices. Summing all $q^{n^2}$ matrices givesWrite ${\varphi }_n(q^{-1})={\prod }_{i=1}^n(1-q^{-i})$, so $|G{L}_n(q)|=q^{n^2}{\varphi }_n(q^{-1})$. Subtract the same equation for $n-1$:Thus counting nilpotent matrices by the Fitting decomposition provesThe supplied product identity is consistent with the resulting mass series: taking its variables to be $s=u$, $t=q^{-1}$ gives ${\sum }_n{a}_n{u}^n/|G{L}_n(q)|={\prod }_{i\ge 1}(1-u{q}^{-i}{)}^{-1}$.