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There are $q^{n^2-n}$ nilpotent matrices of size n over $F_q$. Every matrix has a unique invariant direct sum on which it is nilpotent and invertible. Counting those ordered decompositions gives $q^{n^2}/|G{L}_n(q)|={\sum }_{k=0}^n{a}_k/|G{L}_k(q)|$. Subtraction at n and n−1 yields $a_n=q^{n^2-n}$. Translation by the identity counts unipotent matrices.

The upper and lower unipotent matrices generate SL2R. For $a\ne 0$, let $w(a)=U(a)L(-a^{-1})U(a)$; multiplication gives $w(a)=\left(\begin{array}{cc}0& a\\ -a^{-1}& 0\end{array}\right)$ and $w(a)w(-1)=\mathrm{diag}(a,a^{-1})$. A determinant-one matrix with upper-left entry $a\ne 0$ factors as $L(c/a)\mathrm{diag}(a,a^{-1})U(b/a)$. If that entry vanishes, first add the second row to the first by $U(1)$. Lower unipotents are conjugates of upper unipotents, so the latter also generate the whole group normally.

Let $E_{12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, so $A(t)=I+t{E}_{12}$ and $E_{12}^2=0$. Multiplication givesThe nilpotence also gives $(I+t{E}_{12}{)}^k=I+kt{E}_{12}$ for every nonnegative integer $k$, since all higher binomial terms vanish. Hence, for each positive integer $m$,These are unipotent matrices in SL2R. The integer is positive, since the displayed diagonal matrix is undefined at $m=0$.

If $U$ is a unipotent matrix, $N=U-I$ is a nilpotent matrix. If $Uv=\lambda v$ for a nonzero eigenvector, then $N^{m}v=(\lambda -1{)}^{m}v=0$ for some $m$, forcing $\lambda =1$.

Conversely, suppose the only eigenvalue is $1$. Over $\mathbb{C}$, the characteristic polynomial is then $(t-1{)}^n$. The Cayley-Hamilton theorem gives $(U-I{)}^n=0$, so $U-I$ is nilpotent. Thus $U$ is unipotent if and only if its only eigenvalue is one.

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 ${\sum }_n{a}_n{u}^n/|G{L}_n(q)|={\prod }_{i\ge 1}(1-u{q}^{-i}{)}^{-1}$.

For the fixed degree-two irreducible polynomial $f$, a matrix with minimal polynomial a power of $f$ has only one primary component. Its conjugacy classes correspond to integer partitions $\lambda$ of $m$. By the primary matrix centralizer formula, its centralizer has exactly the same order $c_{\lambda }(q^2)$ as the centralizer of the corresponding unipotent matrix in $G{L}_m(q^2)$.

Sum the conjugacy class sizes:The same sum times $|G{L}_m(q^2)|$ is the number of unipotent matrices of $G{L}_m(q^2)$, which 5(c) shows is $(q^2{)}^{m^2-m}$. Hence counting matrices with a fixed primary polynomial givesIn particular the exponent $a$ of the minimal polynomial is not fixed in advance: summing all integer partitions includes every possible largest block size. The usual statement takes $m\ge 1$; dimension zero can instead be included with its identity convention.

A unipotent algebraic group admits a faithful linear representation in which every group element is a unipotent matrix. For $E={\mathrm{End}}_Q(X)$, invertibility is the nonvanishing condition ${\prod }_i\det h_i\ne 0$. Therefore ${\mathrm{Aut}}_Q(X)=E^{\times }$ is a nonempty Zariski-open subset of the vector space $E$.

Take a Krull-Schmidt decomposition $X\cong {⨁}_{a=1}^r{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposable modules. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local endomorphism ring. Its residue division algebra is $k$: over an algebraically closed field, every element of a finite-dimensional division algebra has an eigenvalue and hence must be scalar. The semisimple quotient of a module endomorphism algebra consequently gives, for the Jacobson radical $J=J(E)$,is surjective with kernel $1+J$. The nilpotence of $J$ makes $(1+j{)}^{-1}=1-j+j^2-\cdots$ finite and each $1+j$ unipotent. The kernel is closed and normal. Acting on the multiplicity spaces embeds the product of general linear groups back into $E^{\times }$ and splits this quotient. This proves the Levi decomposition of a quiver automorphism groupSince $1+J$ is the unipotent radical, a nonzero $X$ is indecomposable exactly when ${\mathrm{Aut}}_Q(X)/R_u({\mathrm{Aut}}_Q(X))\cong {\mathbb{G}}_m$: the product has a single factor of size one.

For the base change action on quiver representations, the orbit map is $g\mapsto (g_j{x}_{\rho }{g}_i^{-1}{)}_{\rho :i\to j}$. Substituting $g_i=I+ϵu_i$, with ${ϵ}^2=0$, shows its differential isIts kernel is ${\mathrm{End}}_Q(X)$. The stabilizer is smooth because it is open in that vector space. Hence the differential has rank $\dim \mathrm{GL}(\mathbf{n})-\dim {\mathrm{Aut}}_Q(X)=\dim {\mathcal{O}}_X$, and its image is the Zariski tangent space $T_x{\mathcal{O}}_X$. The normal space to a quiver orbit is thereforeThe ambient quiver representation space is an irreducible affine space, and orbits are locally closed. An orbit is open exactly when its dimension equals that ambient dimension, equivalently when ${\mathrm{Ext}}_Q^1(X,X)=0$. This proves that rigid quiver representations have open orbits. Such an orbit is dense and unique, since two nonempty open subsets of an irreducible space intersect.

For a finite field of odd characteristic and nonzero $a,b$, let $U_a=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)$. An arbitrary determinant-one matrix $P=\left(\begin{array}{cc}r& s\\ t& u\end{array}\right)$ satisfies $P{U}_a=U_{b}P$ only if $t=0$ and $ar=bu$. The determinant condition $ru=1$ then gives $b/a=r^2$. Conversely $\mathrm{diag}(r,r^{-1})U_a\mathrm{diag}(r,r^{-1}{)}^{-1}=U_{a{r}^2}$. Thus nonzero upper unipotent matrices fall into two conjugacy classes under this square-class criterion. For example, $U_1$ and $U_3$ are conjugate over ${\mathbb{F}}_{11}$ because $5^2=3$, but not over ${\mathbb{F}}_5$. The criterion compares these particular upper unipotent representatives; it does not classify all determinant-one matrices.