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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 nilpotent matrix $T$, define the polynomial $E(z)={\sum }_{k=0}^{m-1}{z}^k{T}^k/k!$. Because $T^m=0$, differentiating gives $E^{\prime }(z)=TE(z)=E(z)T$. The derivative of the preserved bilinear form matrix is thereforeIt is a constant matrix polynomial, equal to $A$ at $z=0$. Evaluation at one provesAlso $E(-1)$ is the inverse of $E(1)$ by the finite exponential product identity, so this really is an invertible isometry, even when $A$ is degenerate.

Put $S=u^2+v^2$. At a nonzero equilibrium point the two equations form the homogeneous linear systemIts determinant must vanish, giving $S^2/2-(\mu +\sigma )S+{\mu }^2+{\sigma }^2-1=0$. ThusEach positive real polynomial root gives an antipodal equilibrium point pair; a negative polynomial root does not describe a real equilibrium point. A direction is specified by $\sin 2\theta =S/2-\mu$, $\cos 2\theta =S/2-\sigma$, where $u=\sqrt{S}\cos \theta$, $v=\sqrt{S}\sin \theta$. This is the polar form of the cubic confinement model. The origin is always an equilibrium point.

The origin's eigenvalues are ${\lambda }_{\pm }=\mu \pm \sqrt{1-{\sigma }^2}$. A simple zero eigenvalue occurs on ${\mu }^2+{\sigma }^2=1$, except at $\mu =0$. The symmetry of an odd function $(u,v)\mapsto (-u,-v)$ makes the generic steady bifurcation a pitchfork bifurcation. At a point $(\sigma ,{\mu }_0)$ on this circle with ${\mu }_0\ne 0$, vary $\mu ={\mu }_0+\delta$. Expanding the radius equation gives $S\sim 2{\mu }_0\delta /({\mu }_0+\sigma )$. In a signed, unit-eigenvector centre manifold coordinate, the cubic coefficient therefore has sign $-({\mu }_0+\sigma )/(2{\mu }_0)$.

On the lower semicircle ${\mu }_0=-\sqrt{1-{\sigma }^2}$ the transverse eigenvalue is negative: the pitchfork bifurcation is supercritical for $\sigma <1/\sqrt{2}$ and subcritical for $\sigma >1/\sqrt{2}$. On the upper semicircle it is supercritical for $\sigma >-1/\sqrt{2}$ and subcritical for $\sigma <-1/\sqrt{2}$ in the centre manifold direction, but its transverse eigenvalue is positive, so the bifurcating equilibrium points are not fully attracting. At $(\sigma ,{\mu }_0)=(\pm 1/\sqrt{2},\mp 1/\sqrt{2})$ the cubic vanishes. The radius equation instead gives $\delta =-S^2/(4{\mu }_0)+o(S^2)$, a nonzero quintic degeneracy with amplitude proportional to $|\delta {|}^{1/4}$: these are degenerate pitchforks in the cubic confinement model.

A Hopf bifurcation occurs at $\mu =0$, $|\sigma |>1$, with frequency $\sqrt{{\sigma }^2-1}$. To determine its direction, first note that the divergence is $2\mu -2S$. The Bendixson-Dulac criterion excludes every nonconstant periodic orbit for $\mu \le 0$, since this divergence is negative except possibly at the origin. We can also calculate a nonzero saturation coefficient. Set $a=|\sigma +1|$, $b=|\sigma -1|$ and $V=a{u}^2+b{v}^2$. At $\mu =0$ the linear orbits are ellipses of constant $V$. Direct differentiation givesOn a linear ellipse, write $u=\sqrt{V/a}\cos \varphi$, $v=\sqrt{V/b}\sin \varphi$. The angular speed is constant; the averages of $u^{3}v$ and $u{v}^3$ vanish, while $⟨S⟩=V|\sigma |/({\sigma }^2-1)$. Averaging the cubic terms, or removing their oscillatory parts by a periodic change of coordinates in its normal form, gives for $\varrho =\sqrt{V}$The cubic coefficient is strictly negative. Hence a small attracting limit cycle exists for $\mu >0$, andFinally $(\sigma ,\mu )=(\pm 1,0)$ have a nonzero nilpotent matrix as their linearization with two zero eigenvalues. They are reflection-symmetric double-zero bifurcations where the Hopf bifurcation and steady thresholds meet; an ordinary nonzero-frequency Hopf bifurcation calculation does not apply there. These, the two degenerate pitchfork bifurcation points, and the generic pitchfork bifurcation/Hopf bifurcation loci above exhaust the origin's local loss-of-hyperbolicity possibilities.

An $R$-point means a unital algebra homomorphism over a field $\alpha :k_q[x,y]\to R$, as in points of a noncommutative algebra. It is specified by $X=\alpha (x)$ and $Y=\alpha (y)$ satisfying $YX=qXY$. For $R=\mathbb{C}$ and $q$ not a root of unity, $q\ne 1$, so scalar commutativity gives $(1-q)XY=0$. Hence the scalar points are the union of the two coordinate axes:

For matrix-valued points of a noncommutative algebra, put $H={\mathbb{C}}^n$ and let $H_{\lambda }(X)=\ker (X-\lambda I{)}^n$ be a generalized eigenspace. From $XY=q^{-1}YX$,Therefore $Y{H}_{\lambda }(X)\subseteq H_{q^{-1}\lambda }(X)$, interpreting a missing eigenvalue as the zero space. In particular the linear subspaces $V_X={⨁}_{\lambda \ne 0}{H}_{\lambda }(X)$ and $H_0(X)$ are invariant under both matrices. If $\lambda \ne 0$, the numbers $\lambda ,q^{-1}\lambda ,\dots ,q^{-n}\lambda$ are distinct. A nonzero vector $Y^{n}v$ with $v\in H_{\lambda }(X)$ would force all $n+1$ intervening generalized eigenspaces to be nonzero, impossible in dimension $n$. Thus $Y^n{V}_X=0$: $Y$ is nilpotent on $V_X$, although $X$ is invertible there.

Apply the same argument with $X,Y$ exchanged, using $XY=q^{-1}YX$. The sum $V_Y$ of the nonzero $Y$ generalized eigenspaces is invariant under both matrices, and $X$ is nilpotent on it. Consequently $V_Y\subseteq H_0(X)$. Conversely, since $Y$ is nilpotent on $V_X$, its nonzero generalized eigenspaces all lie in $H_0(X)$. Decompose the restriction of $Y$ to that invariant space into its nonzero and zero generalized eigenspaces:The zero generalized eigenspace of $Y$ is $X$-invariant by the scaling identity, so $U$ is invariant under both matrices. On $U$, both matrices have only zero eigenvalues, hence are nilpotent matrices. This proves the spectral decomposition of a quantum-plane point:Here $V_X,V_Y$ mean sums of generalized eigenspaces, not merely ordinary eigenspaces; that distinction is necessary for non-diagonalizable points.

Put $p=\dot{u}$ and extend the system by $\dot{\lambda }=\dot{\kappa }=0$. At the zero equilibrium with zero parameters, the $(u,p)$ block is the nilpotent matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, while the $v,w$ eigenvalues are $-1,-\tau$. Since $\tau >0$, these are stable. The extended centre manifold is a graph $v=h(u,p,\lambda ,\kappa )$, $w=k(u,p,\lambda ,\kappa )$ tangent to the four-dimensional centre space containing the two parameter directions. Its invariance equations areFor the quadratic jets in the ordinary total degree, the coefficient multiplying $h_p,k_p$ has no linear part. Hence $p{\partial }_u{h}_2=-h_2+u^2$ and $p{\partial }_u{k}_2=-\tau k_2+u^2/\tau$. With $h_2=a{u}^2+bup+c{p}^2$, comparison gives $a=1$, $b=-2$, $c=2$. The second graph is obtained in the same way. Terms involving parameters at quadratic order have zero forcing and vanish. Thus the quadratic centre manifold for a pair of linear filters isSubstitution into the centre equations gives the cubic termsThe last displayed cubic term is absent from the proposed second-order approximation, so its omission needs the weighted scaling rather than an ordinary-degree truncation.

Under the hinted scaling $u=\epsilon \tilde{u}$, $\kappa =\epsilon \tilde{\kappa }$, $\lambda ={\epsilon }^2\tilde{\lambda }$, $d/dt=\epsilon d/d\tilde{t}$, one has $p=O({\epsilon }^2)$ and $\dot{p}=O({\epsilon }^3)$. The invariance equations at weighted orders two and three give $v=u^2-2up+O({\epsilon }^4)$ and $w=u^2/{\tau }^2-2up/{\tau }^3+O({\epsilon }^4)$. In particular no additional parameter-dependent weighted-order-three graph terms appear. Therefore the $u{\dot{u}}^2$ term and the omitted manifold corrections affect the reduced acceleration only at order ${\epsilon }^5$, whereas $u^3$ has order ${\epsilon }^3$ and $u^2\dot{u}$ order ${\epsilon }^4$. After division by ${\epsilon }^3$, the slow equation isThis proves why the stated approximation retains the leading nonlinear restoring force and its first nonlinear damping correction. The centre manifold theorem removes the exponentially decaying transverse directions, while the retained generic coefficients determine the nearby reduced bifurcations. It is a local asymptotic description, not an exact global reduction. If either relevant coefficient vanishes, higher-order terms may be needed to resolve that degeneracy; the two extreme $s$ cases in part (b), with $\tau =1$, avoid it.

Set $T=ϵt$ and use the method of multiple scales:Since ${\omega }^2=1/4+kϵ+O({ϵ}^2)$, the first correction satisfiesThe resonant cosine and sine coefficients must vanish to avoid secular terms. Using the product-to-sum formulas givesThe remaining forcing has frequency $3/2$ and permits the bounded particular correction $y_1=(a/4)\cos (3t/2)+(b/4)\sin (3t/2)$. The two slow amplitudes carry the two arbitrary constants of the original equation.

For an explicit general leading solution, putThe matrix exponential givesThe error is uniform on bounded intervals of $T$, for bounded initial data. When $\lambda$ is imaginary this expression uses ordinary sine and cosine; at $\lambda =0$ its continuous limit is $I+TM$.

The primary instability tongue of a weak Mathieu oscillator has a positive slow eigenvalue precisely whenThese are the leading edges, rather than an exact finite-$ϵ$ interval. At the leading endpoints the amplitude matrix is a nilpotent matrix, so the leading solution can grow linearly in $T$ and does not decide the exact exponential instability. For completeness, set $\omega =1/2+\beta ϵ+\gamma {ϵ}^2$. The periodic or antiperiodic edge calculation uses a pure cosine when $\beta =-1/2$ and a pure sine when $\beta =1/2$, with the third-harmonic correction just found. The next resonant coefficient is $-(\gamma +3/8)y_0$, soThus the exact growing interval is between these shifted edges. In particular, for positive small $ϵ$ the fixed leading value $k=-1/2$ lies just inside it, whereas $k=1/2$ lies just outside. Interior detunings bounded away from the edges are already resolved by the leading method of multiple scales.

If $A$ is a strictly upper triangular matrix, then $A_{ij}=0$ unless $j-i\ge 1$. We prove by mathematical induction thatThe assertion for $k=1$ is the definition. In the matrix multiplication formulaa nonzero summand would require $\ell -i\ge k$ and $j-\ell \ge 1$, hence $j-i\ge k+1$. This proves the induction step. Since $j-i\le n-1$ for all entries, $\boxed{A^n=0}$. In particular, every strictly upper triangular matrix is a nilpotent matrix.

Let $S_m=I+D+\cdots +D^m$. The finite matrix geometric series telescopes to $(I-D)S_m=I-D^{m+1}$. By the assumed convergence, its terms $D^{m+1}$ tend to zero, so the limit $S$ satisfies $(I-D)S=I$. Consequently $S=(I-D{)}^{-1}$. This is the finite-dimensional Neumann series argument; invertibility by itself would not imply convergence.

For the stated nilpotent matrix, $D^3=0$ and only the first three terms remain:

Nilpotence and eigenvalues. If $Av=\lambda v$ with $v\ne 0$, then $A^{n}v={\lambda }^{n}v$. Thus $A^n=0$ forces $\lambda =0$ for every eigenvalue. Conversely, if all $n$ complex eigenvalues, counted with algebraic multiplicity, vanish, the monic characteristic polynomial is ${\chi }_A(t)=t^n$. The Cayley-Hamilton theorem then gives $A^n=0$. HenceThis is the criterion for a nilpotent matrix.

Symmetric case. Here a symmetric matrix is understood to be real. The spectral theorem gives $A=QD{Q}^T$ with $Q$ orthogonal and $D$ real diagonal. The previous result makes every diagonal entry zero, so $A=0$. Alternatively, choose $2^r\ge n$, so $A^{2^r}=0$. For every real vector $x$,It follows that $A^{2^{r-1}}=0$. Repeat until $A=0$. This uses positivity of the quadratic form $x^T{M}^{T}Mx$. The restriction to real symmetric, or complex Hermitian matrices, matters: a complex matrix satisfying only $A^T=A$ need not have this positivity.

The parameters and powers. The first two columns of $T$ coincide, so $\det T=0$ for all $\alpha ,\beta$. The sum of its principal two-by-two determinants, each a minor of a matrix, isTogether with $\mathrm{tr}T=2+\beta$, this givesFor every eigenvalue to vanish both remaining coefficients must be zero. ThereforeDirect multiplication at these values givesIndeed the first row of $T^2$ multiplied by $T$ is $3{\mathrm{row}}_1(T)+3{\mathrm{row}}_2(T)-6{\mathrm{row}}_3(T)=0$, and the second row is its negative.

Geometry of repeated application. For any parameters the image of a linear map lies in the span of $a=(1,1,1{)}^T$ and $b=(1,\alpha ,\beta {)}^T$, since $Tx=(x_1+x_2)a+x_{3}b$. Thus it lies in a fixed plane through the origin, independently of the input. When $a,b$ are independent its normal is $a\times b=(\beta -\alpha ,1-\beta ,\alpha -1)$; if they coincide, the image is a line and is still contained in a plane.

For $\alpha =-5,\beta =-2$, put $s=x_1+x_2$ and $t=x_3$. ThenThe columns span this entire plane, so $T$ has rank two. The next two steps areThus the image of a linear map contracts from a plane to a line and then to the origin under successive applications. The line from $T^2$ is precisely the kernel of $T$.

Taking determinants of $A^p=0$ gives $(\det A{)}^p=0$, so $A$ is singular and has the eigenvalue zero. More strongly, if $Av=\lambda v$ with $v\ne 0$, then $0=A^{p}v={\lambda }^{p}v$, so every eigenvalue is zero.

If the nilpotent matrix were diagonalizable, write $A=PD{P}^{-1}$. Then $0=A^p=P{D}^p{P}^{-1}$, hence each diagonal entry of $D$ has zero $p$th power and is zero. Thus $D=0$ and $A=0$. Conversely the zero matrix is diagonal. ThereforeThe conclusion is valid over $\mathbb{C}$ and also over $\mathbb{R}$.

Use the full $3\times 3$ matrix $B$ in the original PDF; the transcription has dropped its middle column. For either matrix, the three-dimensional characteristic polynomial is ${\chi }_M(t)=t^3-(\mathrm{tr}M)t^2+[(\mathrm{tr}M{)}^2-\mathrm{tr}(M^2)]t/2-\det M$. Both have zero trace and zero determinant, while $\mathrm{tr}(B^2)=0$ and $\mathrm{tr}(C^2)=72$. ConsequentlyFor $B$, a useful check isfor every value of $\alpha$. Thus $B$ is a nonzero nilpotent matrix for all $\alpha$; for example its second column never vanishes. By the preceding result, $\boxed{B\text{ is not diagonalizable for any }\alpha }$.

For $C$, the three eigenvalues $-6,0,6$ are distinct and real, so their eigenvectors are linearly independent. An explicit eigenbasis isWith $P=(v_{-},v_0,v_{+})$, $\det P=18\ne 0$ and $P^{-1}CP=\mathrm{diag}(-6,0,6)$. Hence $\boxed{C\text{ is diagonalizable over }\mathbb{R}\text{ and }\mathbb{C}}$.

A repeated root of the characteristic polynomial alone would not settle diagonalizability; for $B$ it is the nonzero nilpotence that completes the argument.

For every vector $v$, the Hermitian inner product givesThe norm vanishes only on the zero vector, so $Cv=0$ for every $v$ and $\boxed{C=0}$. In contrast,is a nonzero nilpotent matrix. Multiplying a matrix by its conjugate transpose detects column norms; multiplying it by itself has no corresponding positivity property.

The order-${ϵ}^2$ equation isUsing $\cos t\cos 2t=(\cos 3t+\cos t)/2$ and $\cos t\sin 2t=(\sin 3t+\sin t)/2$, its resonance coefficients aresecular term are removed by setting these to zero:The eigenvalues of this two-by-two system satisfyThey are real with opposite signs for $-1/12<a_2<5/12$. Generic initial amplitudes then contain an exponentially growing slow-time component. At either leading endpoint the matrix is nonzero nilpotent matrix, so suitable initial amplitudes give linear slow-time growth; higher-order detuning decides the precise corrected boundary. Outside this interval the leading eigenvalues are purely imaginary and the slow amplitudes remain oscillatory.

Substitution into $a=1+a_2{ϵ}^2+\cdots$ gives the requested leading instability interval“Can become unbounded” permits special initial amplitudes without growth. The calculation identifies the second instability tongue of a weak Mathieu oscillator and its slow amplitudes for the second Mathieu instability tongue; it is not an assertion that every solution grows throughout the closed interval.

Take the linear mapIts kernel of a linear map consists of all $(x,0)$, and its image of a linear map is the same line:Their set intersection is nonzero and their sum is only one-dimensional. In particular, $(0,1)$ cannot be written as a sum of one vector from each of these vector subspaces. The nonzero set intersection also prevents an internal direct sum of vector spaces. ThusThe matrix is a nonzero nilpotent matrix; its failure of diagonalizability explains why the preceding result does not apply.

For $n=2$, takeThey are commuting operators because every matrix commutes with the identity matrix. However $A$ has only the eigenvalue zero, and $\ker A=\mathrm{span}\{(1,0{)}^T\}$ has dimension one. It has no eigenbasis, so it cannot appear in any simultaneous diagonalization. Equivalently, a diagonal matrix similar to this nonzero nilpotent matrix would have to be zero, an impossibility. For any prescribed $n\ge 2$, use $A=E_{12}$ and $B=I_n$; its zero eigenspace has dimension $n-1$. For $n=1$ no such example exists.

If $B^k=0$ with $k\ge 1$, put $S=I+B+\cdots +B^{k-1}$. The matrix geometric series telescopes on both sides:Thus $I-B$ is nonsingular, andThis is a finite polynomial identity for a nilpotent matrix; no convergence argument or bound $\Vert B\Vert <1$ is required.

The binomial theorem for the commuting matrices $I,N$ proves the formula for nonnegative integers: every term containing $N^2$ vanishes. Since $(I+N)(I-N)=I$, the matrix inverse is $I-N$; applying the same argument to its powers gives the negative integers as well. The same identity shows that $I+N$ is invertible even when the nilpotent matrix $N$ is nonzero. In dimension two, the example $N=\left(\begin{array}{cc}0& p\\ 0& 0\end{array}\right)$ describes repeated shear mappings, whose displacement parameters add under composition.

Take a different real matrix,Its only eigenvalue is $2$, with algebraic multiplicity two. However, the polynomial $q(t)=t-2$, which vanishes at that eigenvalue, gives $q(B)=N\ne 0$. Merely checking values at eigenvalues cannot prove a matrix identity. The full characteristic polynomial does work:The matrix is a Jordan block, and its minimal polynomial is also $(t-2{)}^2$.

The binomial theorem gives, for positive integers $n$,Terms containing $N^2$ or a higher matrix power vanish. After removing the common growth factor $2^n$, the remaining linear map is the shear mapping $I+(n/2)N$. It moves $(x,y)$ to $(x+ny/2,y)$. The single eigenvalue controls the exponential scale, while the nilpotent matrix adds a factor growing linearly in $n$.

In terms of polynomial interpolation, reduction modulo $(t-2{)}^2$ retains both a value and a derivative:Indeed, polynomial division gives $f(t)=f(2)+f^{\prime }(2)(t-2)+(t-2{)}^{2}h(t)$. This is a simple Hermite interpolation condition. For $f(t)=t^n$, it gives exactly the displayed matrix power. Repeated eigenvalues retain information about how a polynomial changes, not just where it vanishes.

On a finite-dimensional positive inner-product space, a skew-adjoint nilpotent operator is zero. In an orthonormal basis, $\Vert A{\Vert }_F^2=\mathrm{tr}(A^{\ast }A)=-\mathrm{tr}(A^2)=0$, since a nilpotent matrix has zero trace and so does its square. The Frobenius norm therefore vanishes. In particular, a nonzero nilpotent adjoint map rules out a compact Lie algebra.

For a real or complex two-by-two matrix $A$, the Cayley-Hamilton theorem reads $A^2-(\mathrm{tr}A)A+(\det A)I=0$. Hence $A^2=0$ exactly when $\mathrm{tr}A=\det A=0$. Conversely, if $A^m=0$ for any positive integer $m$, every eigenvalue is zero and the same identity gives $A^2=0$. Thus a nonzero nilpotent matrix of this size has nilpotency index two.