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A complex nilpotent matrix of size six is determined up to matrix similarity by its matrix rank and nilpotency index. Its Nilpotent Jordan block sizes form an integer partition of six. If there are $k$ blocks and the largest size is $m$, then its matrix rank is $6-k$ and its minimal polynomial is $t^m$. The eleven possible size lists areNo two have the same pair $(m,k)$, which proves the assertion by Jordan normal form. The dimension restriction matters: in size seven, block lists $(3,3,1)$ and $(3,2,2)$ both have matrix rank four and minimal polynomial $t^3$, but give different Jordan normal forms.

A stationary iterative method for a linear system converges to its specified fixed point for every initial vector if and only if its iteration matrix satisfies $\rho (H)<1$, where the spectral radius is the largest modulus of an eigenvalue.^[ref]

For a diagonalizable matrix, write $H=V\Lambda V^{-1}$. Then $H^k=V{\Lambda }^k{V}^{-1}$, so each eigenvector coefficient tends to zero exactly when its eigenvalue has modulus less than one. For a general matrix, the Jordan normal form gives the same result: a Jordan block $J=\lambda I+K$ with a nilpotent matrix $K$ of size $m$ satisfiesIf $0<|\lambda |<1$, every coefficient tends to zero because exponential decay beats any fixed polynomial in $k$; if $\lambda =0$, the block vanishes once $k\ge m$. Conversely, an eigenvector with $|\lambda |\ge 1$ provides an initial error that cannot tend to zero. For a real matrix with complex eigenvalues, the same obstruction appears in the real invariant subspace spanned by the real and imaginary parts of an eigenvector.

The words "every initial vector" matter. For example, $H=\mathrm{diag}(3/2,1/2)$ fails the criterion, but $e^{(0)}=(0,1{)}^T$ still tends to zero. Here a generic initial error has a nonzero first coordinate and diverges; the exceptional errors lie on the second coordinate axis. With $H=\mathrm{diag}(-1,1/2)$, generic errors oscillate rather than diverge in norm, while the same exceptional axis still converges. Both are valid matrix splittings with $M=I$ and $A=I-H$. In general the convergent errors form the direct sum of the generalized eigenspaces for $|\lambda |<1$; if $\rho (H)\ge 1$, this is a proper linear subspace.

A small spectral radius describes eventual convergence, but does not imply that the Euclidean norm of the error decreases on every step for a non-normal matrix. An operator norm bound $\Vert H\Vert <1$ does give the stepwise estimate $\Vert e^{(k)}\Vert \le \Vert H{\Vert }^k\Vert e^{(0)}\Vert$. For a real symmetric matrix, the matrix 2-norm equals its spectral radius, making this particularly direct.

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.

A cube root of a matrix $A$ is a square matrix $B$ of the same size over the specified field satisfying $B^3=A$. Every complex invertible matrix has one: put it in Jordan normal form, choose a scalar cube root of each nonzero eigenvalue, and apply the truncated binomial cube root of the identity plus a nilpotent matrix on each block. The Nilpotent Jordan block $J_3(0)$ has none, even though its square is nonzero. If $B^3=J_3(0)$, then $B^9=0$, making $B$ a nilpotent matrix of size three; the Cayley-Hamilton theorem then gives $B^3=0$, a contradiction.

The only diagonalizable nilpotent matrix is the zero matrix. Indeed, if $A^p=0$ and $Av=\lambda v$ with $v\ne 0$, then ${\lambda }^p=0$, so $\lambda =0$. A basis of eigenvectors would make every diagonal entry zero and therefore $A=0$. Nonzero nilpotent matrices require nontrivial Jordan blocks, even if their characteristic polynomial has only one distinct root.

When a square matrix satisfies $M^2=\kappa I$, its powers reduce to $M^{2n}={\kappa }^{n}I$ and $M^{2n+1}={\kappa }^{n}M$. Absolute convergence of the matrix exponential allows even and odd subseries to be separated. Thus $C(\kappa )={\sum }_{n\ge 0}{\kappa }^n/(2n)!$ and $S(\kappa )={\sum }_{n\ge 0}{\kappa }^n/(2n+1)!$. For real $\kappa >0$, these are $\cosh \sqrt{\kappa }$ and $\sinh \sqrt{\kappa }/\sqrt{\kappa }$. For $\kappa <0$, they are $\cos \sqrt{-\kappa }$ and $\sin \sqrt{-\kappa }/\sqrt{-\kappa }$. At $\kappa =0$, $e^M=I+M$, including nonzero nilpotent matrices of square zero. This also explains the limiting sine-over-argument coefficients without dividing by zero.

The finite matrix geometric series telescopes as stated in the title. If its partial sums converge, then their successive differences show $D^{m+1}\to 0$. Passing to the limit gives $(I-D){\sum }_{j\ge 0}{D}^j=I$, so the sum is the matrix inverse $(I-D{)}^{-1}$. Conversely, existence of the inverse alone does not imply series convergence: the scalar matrix $D=2I$ is a counterexample. The Neumann series gives a sufficient condition from a matrix norm smaller than one.

For a nilpotent matrix with $D^r=0$, convergence is automatic and the inverse is the finite polynomial $I+D+\cdots +D^{r-1}$. Multiplying by $I-D$ directly verifies this inverse without computing a determinant or diagonalizing.

For $tv{t}^{-1}=Bv$, the commutator $[t,v]$ is $(B-I)v$ in additive notation. The lower central series from its second term is $(B-I){\mathbb{Z}}^r,(B-I{)}^2{\mathbb{Z}}^r,\dots$. It terminates exactly when $B-I$ is a nilpotent matrix. For rank two this means $(B-I{)}^2=0$, equivalently $\det B=1$ and $\mathrm{tr}B=2$. Nilpotence of a finite-index subgroup alone gives virtual nilpotence of the whole group.

The nilpotency index is the least positive power at which a nilpotent matrix becomes zero.

For conjugation of $n$ by $n$ complex matrices by the special linear group, the affine null cone consists exactly of the nilpotent matrices. If a conjugate tends to zero along an algebraic one-parameter subgroup, its constant characteristic polynomial must be $T^n$. Conversely, put a nilpotent matrix in strictly upper triangular form and conjugate it by $\mathrm{diag}(t^{n-1},t^{n-3},\dots ,t^{1-n})$. Every active matrix entry has positive weight and tends to zero. The Hilbert-Mumford criterion for projective stability therefore identifies the semistable locus of the projective conjugation representation as the complement of the projectivized nilpotent cone.

Write $f=m_T$. Part (ii) identifies the first superdiagonal block exactly:where $f^{\prime }$ is the formal derivative. Thus the candidate is $g=f^{\prime }$. If the irreducible polynomial $f$ has $f^{\prime }\ne 0$, its polynomial degree makes $f$ and $f^{\prime }$ coprime. Equivalently, $f$ is a separable polynomial. This includes every irreducible polynomial over a field of characteristic zero, and every irreducible polynomial over a perfect field.

Under this separability hypothesis, Bézout's identity for polynomials gives $a(x)f(x)+b(x)f^{\prime }(x)=1$. Evaluating at $A$ proves that $C=f^{\prime }(A)$ is an invertible matrix. Moreover, $f(B)$ has zero diagonal blocks, because $f(A)=0$. It is therefore a nilpotent matrix with $f(B{)}^N=0$. The only product of $N-1$ strictly upward block steps that reaches block $(1,N)$ uses the first superdiagonal every time, soFor $N=1$, the corresponding assertion is simply $f(B{)}^0=I\ne 0$. The block upper triangular matrix $B$ hasThe minimal polynomial divides this characteristic polynomial. Since $f$ is an irreducible polynomial, it must be a power of $f$; the exact nilpotency index just found forces

The arbitrary-field qualification is necessary. In the rational function field $F={\mathbb{F}}_2(t)$, take $f(x)=x^2-t$. This is an irreducible polynomial: a square of a rational function has even order of vanishing at $t=0$, while $t$ has odd order. Its formal derivative is zero. With $N=2$,The off-diagonal block in $f(B)$ is zero, and $m_U=f$, rather than $f^2$. No alternative polynomial $g$ coprime to $f$ could represent that block: $g(A)=0$ would imply $m_T=f\mid g$. Thus the printed coprimality claim fails as stated over a general field.

There is also a precise general replacement. In characteristic $p>0$, let $q=p^r$ be the largest power for which $f(x)=h(x^q)$, so that $h^{\prime }\ne 0$; set $q=1$ in characteristic zero. Irreducibility of $f$ implies that $h$ is an irreducible polynomial, so $h$ and $h^{\prime }$ are coprime. The binomial theorem in characteristic $p$ gives $B^q=D^q+K^q$. Expanding $h(D^q+K^q)$, with the coefficients equivalently supplied by the Hasse derivatives, giveswhere $H$ commutes with $K$ and is a block upper triangular matrix whose diagonal blocks are $h^{\prime }(A^q)$. These blocks are invertible by Bézout's identity for polynomials, since $h(A^q)=f(A)=0$. Thus $H$ is invertible, andThe same characteristic polynomial and divisibility argument now prove the corrected arbitrary-field answer:This is the minimal polynomial of a companion block chain; $q=1$ recovers the intended separable case.

Subtract the identity matrix and putDirect matrix multiplication givesFor the last equality, each column of $N^2$ is a multiple of $(1,-1,0{)}^T$, and $N(1,-1,0{)}^T=0$. Thus $N$ is a nilpotent matrix. The minimal polynomial of $A_{\lambda ,\mu }$ divides $(t-1{)}^3$ and equals $(t-1{)}^s$, where $s$ is the nilpotency index of $N$.

The first row of $N$ forces $\lambda =0$ if $N=0$, and its third row then forces $\mu =0$. Conversely $\lambda =\mu =0$ gives $N=0$. If exactly one parameter is nonzero, then $N\ne 0$ but $N^2=0$. If both are nonzero, then $N^2\ne 0$ while $N^3=0$. Consequently the complete answer isNo lower power can annihilate the matrix in any of these cases, as the displayed nonvanishing matrix powers show.

By Jordan normal form, each complex nilpotent matrix of size six can be put by matrix similarity into a block diagonal matrix whose Nilpotent Jordan blocks have sizesA Nilpotent Jordan block of size $s$ has matrix rank $s-1$ and minimal polynomial $t^s$. Therefore, for the whole matrix,Thus the given matrix rank determines the number $k$ of Jordan blocks, and the given minimal polynomial determines the largest size $s_1$. For size six these two data determine the entire integer partition. Here is the complete list, arranged by the largest size:To see that the list is exhaustive, choose the largest part $s_1$ between one and six and partition the remainder $6-s_1$ into parts at most $s_1$. The resulting rows all have distinct pairs $(s_1,k)$. Hence $A$ and $B$ have the same Jordan normal form $J$.

Explicitly, choose invertible matrices $X,Y$ such that $A=XJ{X}^{-1}$ and $B=YJ{Y}^{-1}$. ThenThis proves matrix similarity and illustrates the classification of six-dimensional nilpotent matrices by rank and nilpotency index.

Take the Nilpotent Jordan blockThenSuppose that a complex matrix $B$ of size three satisfied $B^3=A$. We would have $B^9=A^3=0$, so $B$ would be a nilpotent matrix. Every eigenvalue of $B$ must then be zero: if $Bv=\theta v$ with $v\ne 0$, applying $B^9$ gives ${\theta }^{9}v=0$. Its characteristic polynomial is therefore $t^3$, and the Cayley-Hamilton theorem gives $B^3=0$. This contradicts $B^3=A\ne 0$. HenceThe obstruction is that a nilpotent matrix of size three cannot have nilpotency index greater than three.

The increments ${\Delta }_i=d_i-d_{i-1}$ are positive integers by part (a)(i), are nonincreasing by part (a)(ii), and sum to $d_m=4$. Thus they form an integer partition of $4$. The complete list is $(4)$, $(3,1)$, $(2,2)$, $(2,1,1)$, and $(1,1,1,1)$.

To prove all five occur over every field, let $J_s(0)$ be a Nilpotent Jordan block of size $s$. It sends each standard basis vector to its predecessor, with the first sent to zero, soFor a block diagonal matrix, the nullities add. The following explicit nilpotent matrices therefore realize every possible sequence:In each row the largest Jordan block size is exactly the nilpotency index $m$, so the displayed sequence stops at the correct place.

Generate independent Bernoulli random variables above the diagonal and set everything else to zero. Taking the strict upper triangle of a random square array gives each allowed entry probability $1/2$ of being one:

import numpy as np
rng = np.random.default_rng(2024)


def rand_triu(n):
    return np.triu(rng.integers(0, 2, size=(n, n)), k=1)


example = rand_triu(10)
print(example)

One reproducible strictly upper triangular matrix is:

0 1 0 0 0 0 1 1 1 1
0 0 1 0 0 0 1 0 0 0
0 0 0 1 1 0 0 1 1 0
0 0 0 0 1 1 1 1 0 0
0 0 0 0 0 0 1 1 1 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0

Every matrix returned in this way is a nilpotent matrix; randomness affects its nilpotency index, not the fact of the nilpotent property.

Put $t=\mathrm{tr}A=a+d$ and $\delta =\det A=ad-bc$. Direct matrix multiplication gives the two-dimensional Cayley-Hamilton theorem identity $A^2-tA+\delta I=0$. If $A^2=0$, determinant multiplicativity gives ${\delta }^2=0$, hence $\delta =0$. The identity then says $tA=0$. If $A=0$, its trace is zero; otherwise a nonzero entry forces $t=0$. Thus $d=-a$ and $bc=ad=-a^2$.

Conversely, those two scalar relations give $t=\delta =0$, so the same Cayley-Hamilton theorem identity implies $A^2=0$. ThereforeThis is the square-zero criterion for a two-by-two matrix. If $A^3=0$, determinant multiplicativity first gives ${\delta }^3=0$, hence $\delta =0$. Now $A^2=tA$ and $A^3=t^{2}A$. Either $A=0$, or a nonzero entry gives $t^2=0$ and hence $t=0$; in both cases $A^2=0$. The converse follows by multiplying $A^2=0$ by $A$. Consequently $A^3=0⟺A^2=0$. A nonzero nilpotent matrix of this size therefore has nilpotency index two.

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.

For $D=P^{-1}{D}_{0}P$, conjugating $U=D^{-1}T$ givesBecause $D_0^{-1}$ commutes with $N$, and $N^n=0$,Thus $U-I$ is similar to a nilpotent matrix, and is itself nilpotent. Therefore $U$ is a unipotent matrix.

Use the natural linearization of a line bundle on ${\mathcal{O}}_X(1)$ induced by the conjugation representation on $M_n\mathbb{C}$. The notation involving the dual does not change this matrix description: the nondegenerate pairing $(A,B)\mapsto \mathrm{tr}(AB)$ identifies the representation with its dual equivariantly. Let $[A]$ be a projective point, so $A\ne 0$.

The Hilbert-Mumford criterion for projective stability says that $[A]$ is not a semistable projective point in geometric invariant theory precisely when some algebraic one-parameter subgroup $\lambda :{\mathbb{G}}_m\to S{L}_n$ sends its affine representative to zero:Equivalently, all active weights for that subgroup are strictly positive. With the convention $\mu ([A],\lambda )=-\min \{\text{active weights}\}$, semistability requires $\mu \ge 0$ for every subgroup, including conjugates of diagonal ones.

Suppose such a zero limit exists. The coefficients $c_j(A)$ inare invariant under conjugation. Continuity along the zero limit gives $c_j(A)=c_j(0)=0$ for every $j$. By the Cayley-Hamilton theorem, $A^n=0$: the matrix is nilpotent.

Conversely, if $A$ is a nonzero nilpotent matrix, choose a basis in which it is strictly upper triangular, for example a Jordan normal form. The change of basis can be chosen in $S{L}_n$: multiply any invertible change-of-basis matrix by a scalar with the required $n$th power, which does not affect its conjugation. In this basis takeThe exponents sum to zero, so this is an algebraic one-parameter subgroup of the special linear group. The $(i,j)$ matrix entry has weightEvery potentially nonzero entry has $i<j$ and therefore positive weight. Thus the conjugated matrix tends to zero. Conjugate this subgroup back to the original basis to obtain the required destabilizing subgroup for $A$.

It follows that the nilpotent cone for conjugation of matrices is exactly the affine null cone, andEach $c_j$ is a homogeneous invariant of degree $j$, so the second description also gives invariant open sets witnessing semistability. Equivalently, a representative has at least one nonzero eigenvalue. Nonzero singular matrices can be semistable; invertibility is not the criterion. For example, a diagonal matrix with diagonal entries $1,0,\dots ,0$ is semistable, while a nonzero strictly upper triangular matrix is unstable.