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Here the autonomous planar system is $\dot{x}=v$, $\dot{v}=-\cos x$. Its equilibrium points areThe Jacobian matrix isFor even $n$, the eigenvalues are $1,-1$, so these points are unstable saddle equilibria. Writing $\xi =x-x_{\ast }$, their local unstable and stable directions are $v=\xi$ and $v=-\xi$ respectively. For odd $n$, the eigenvalues are $i,-i$.

To classify the latter points nonlinearly, differentiate the first integralAt $x_{\ast }=3\pi /2+2m\pi$ this energy has a strict local minimum. Its nearby level sets are closed curves, proving that the odd-$n$ points are stable center equilibria, with no attraction to the center. The phase portrait consists of closed oscillatory orbits for $-1<H<1$, heteroclinic orbits at $H=1$ joining consecutive saddle equilibria, and open rotating orbits for $H>1$. The separating curves haveAbove the horizontal axis $\dot{x}=v>0$, so flow is rightward; below it flow is leftward. The closed orbits are clockwise, and these same signs orient every separating and rotating orbit.

Introduce the velocity variable $v=\dot{x}$. The second-order differential equation becomes the autonomous planar systemA critical point of a dynamical system is a point where both components of the vector field vanish: here $v=0$ and $f(x,0)=0$. It is a stationary solution in the phase plane. The Jacobian matrix at each such point gives the local linear stability of a planar equilibrium. For the equations below, a conserved energy also determines the nonlinear phase portrait; this is important because purely imaginary eigenvalues alone do not generally prove that an equilibrium is a center equilibrium.

Write the linear differential equation as $\dot{\mathbf{x}}=M\mathbf{x}$, withIts characteristic polynomial is $(\lambda +4{)}^2+9$. Thus a choice of eigenvalues and eigenvectors isThe distinct eigenvalues give two linearly independent solutions, and the general complex solution isReal solutions have $a_2=\overline{a_1}$. Equivalently,They spiral counterclockwise towards the origin, which is a stable spiral.

In a forced linear differential equation, resonance in a differential equation occurs when exponential forcing excites a mode with the same exponent as its eigenvalue, producing a polynomial factor such as $t$ multiplying that exponential. Merely matching an exponent is not enough: its projection onto that mode must be nonzero. Here $t{e}^{{\lambda }_{j}t}$ still decays in absolute magnitude because the real part of ${\lambda }_j$ is negative; resonance refers to this extra factor, not necessarily to unbounded growth.

For each forcing vector, resolve it in the eigenvector basis:If $u_k$ is the coefficient of the $k$th eigenvector, its scalar equation is ${\dot{u}}_k={\lambda }_k{u}_k+{\sum }_j{c}_{kj}{e}^{{\lambda }_{j}t}$. Multiplying by $e^{-{\lambda }_{k}t}$ and integrating shows that the $j=k$ contribution is $c_{kk}t{e}^{{\lambda }_{k}t}$, whereas a $j\ne k$ contribution is $c_{kj}{e}^{{\lambda }_{j}t}/({\lambda }_j-{\lambda }_k)$. Therefore the exact conditions for no resonant response areUnder these conditions one particular solution iswhich contains no factor of $t$. For a real forcing, $p_2=\overline{p_1}$ and $q_2=\overline{q_1}$, and the two boxed conditions are conjugate to each other.

A real three-by-three matrix represents a proper rotation in the standard orthonormal basis precisely when $A^{T}A=I$ and $\det A=1$, that is, when $A\in \mathrm{SO}(3)$. For the supplied integer numerator $B=3A$, direct multiplication gives $B^{T}B=9I$ and $\det B=27$, verifying both conditions.

The rotation axis is the eigenspace for eigenvalue one. Solving $(A-I)n=0$ gives $n_3=0$ and $n_2=2n_1$, sois a unit axis vector; its negative describes the same axis. On its perpendicular plane the rotation matrix has eigenvalues $e^{i\theta }$ and $e^{-i\theta }$. The rotation trace formula therefore gives $\mathrm{tr}A=1+2\cos \theta$. Since the trace is $-1/3$,

Let $Az=\lambda z$ with $z\ne 0$, allowing complex eigenvectors. A real symmetric matrix is Hermitian, so $z^{\ast }Az$ is real and $\lambda =z^{\ast }Az/(z^{\ast }z)$ is real. For eigenvectors $u,v$ of distinct real eigenvalues $\lambda ,\mu$, self-adjointness gives $\lambda u^{\ast }v=u^{\ast }Av=\mu u^{\ast }v$, hence $u^{\ast }v=0$. Thus the distinct eigenspaces are orthogonal.

Here $A=3I-J$, where $J$ is the all-ones matrix. Since $J(1,1,1{)}^T=3(1,1,1{)}^T$ and $J$ vanishes on the plane of coordinate sum zero,Every nonzero vector in the indicated eigenspace is an eigenvector. The repeated eigenvalue three has a two-dimensional eigenspace.

A complex symmetric nilpotent matrix provides the requested counterexample:Its characteristic polynomial is ${\lambda }^2$, so its only eigenvalue is zero. It is not diagonalizable: a diagonalizable matrix with only zero eigenvalues would itself be zero. Complex symmetry $N^T=N$ is weaker than the Hermitian condition $N^{\ast }=N$.

Expanding the characteristic polynomial along the first column givesFor $s\ne 1$, solving the two nullspace equations givesThese are all the eigenvalues and eigenvectors, with nonzero vectors understood. The three displayed spanning vectors form an eigenbasis, so $A$ is diagonalizable.

At $s=1$, the sole eigenvalue is three, with algebraic multiplicity three but the same two-dimensional $E_3$. Thus $A$ is diagonalizable exactly when $s\ne 1$. Indeed, at the exceptional parameter, $A=3I+N$ with $N\ne 0$ and $N^2=0$, so its Jordan normal form has one block of size two and one of size one. This illustrates how an eigenvalue collision can destroy diagonalizability without changing the two-dimensional eigenspace.

The angular momentum commutation relations are $[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$. The commutator product rule givessince the bracketed expression is symmetric in $j,k$ and ${ϵ}_{ijk}$ is antisymmetric. Alsoproving $\boxed{L^2=L_{-}{L}_{+}+L_3^2+\hslash L_3}$.

For orbital angular momentum, $L_i=-i\hslash {ϵ}_{ijk}{x}_j{\partial }_k$. A differentiable radial function has ${\partial }_{k}f(r)=f^{\prime }(r)x_k/r$, so its contraction with the antisymmetric symbol vanishes: $Lf(r)=0$. Put $s=x_1+i{x}_2$. The coordinate differential operators give $L_{3}s=\hslash s$, whileannihilates every power of $s$, because $({\partial }_1+i{\partial }_2)s=0$. Since all $L_i$ annihilate the radial factor, the product rule yieldsSubstitution into the expression for $L^2$ gives the highest-weight complex-coordinate orbital wavefunction resultAs $[L^2,L_{-}]=0$, lowering preserves this eigenvalue whenever the lowered function is nonzero. Explicitly,For $n=0$ the lowered function is zero, so it is not an eigenfunction. For $n\ge 0$ the original nonzero function has the usual regular angular dependence. The PDF's “any integer” differential identities also hold locally for negative $n$ where $s\ne 0$, but those negative powers are singular on the axis and are not square-integrable regular angular eigenstates. The converted TeX incorrectly writes $L_4$ in place of the PDF's $L_{+}$.

Use coordinates $\mathbf{r}$ measured from the center of mass $G$, so $\int \mathbf{r}dm=0$. The displacement from $P$ is $\mathbf{r}-\mathbf{X}$. Expand the defining inertia tensor:All terms linear in $\mathbf{r}$ integrate to zero. The remaining constant terms give the tensor parallel-axis theorem,For the uniform cube, let $M=8\rho a^3$. Symmetry makes the off-diagonal inertia tensor entries zero. Direct integration gives $\int x_i^{2}dm=M{a}^2/3$ for each coordinate, and therefore $I(G)=(2M{a}^2/3)I_3$. With $\mathbf{X}=a(1,1,1{)}^T$, the tensor parallel-axis theorem givesThe unit eigenvector $u=(1,1,1{)}^T/\sqrt{3}$ has eigenvalue $2M{a}^2/3$. Every vector in the plane $x+y+z=0$ is annihilated by $11^T$ and hence has eigenvalue $11M{a}^2/3$. One orthonormal basis of that eigenspace isThe principal moments are $2M{a}^2/3$ along the body diagonal and $11M{a}^2/3$ twice in its perpendicular plane. Any orthonormal pair in that plane supplies the remaining principal axes.

Take a size-three Jordan block at one and a scalar block at three:Triangularity gives ${\chi }_A(x)=(x-1{)}^3(x-3)$. On the first block, $(A-I{)}^3$ is zero but $(A-I{)}^2$ is not; on the last component the eigenvalue is three. Thus both factors with exactly those powers are needed, and their product annihilates $A$. Its minimal polynomial is therefore also $(x-1{)}^3(x-3)$.

Let $m_{\alpha }(x)$ be the monic minimal polynomial. The Cayley-Hamilton theorem says ${\chi }_{\alpha }(\alpha )=0$. Divide ${\chi }_{\alpha }$ by $m_{\alpha }$; its remainder also annihilates $\alpha$, and a nonzero remainder of smaller degree would contradict minimality. Thus $m_{\alpha }$ divides ${\chi }_{\alpha }$.

Every characteristic root ${\lambda }_i$ is an eigenvalue, with a nonzero eigenvector $v_i$. Since $m_{\alpha }(\alpha )v_i=m_{\alpha }({\lambda }_i)v_i=0$, it must be a root of the minimal polynomial as well. This proves there are no possibilities beyondEvery such choice really occurs. For each $i$, take one Jordan block of size $r_i$ at ${\lambda }_i$, together with $n_i-r_i$ scalar blocks at that eigenvalue, and form their direct sum. The characteristic polynomial has the prescribed multiplicities. The nilpotent part of a block of size $r_i$ has its $r_i$th power zero and its $(r_i-1)$st power nonzero, so this component's minimal exponent is exactly $r_i$. A polynomial annihilates the direct sum precisely when it annihilates every component, establishing the displayed minimal polynomial without any extra restrictions.

For a Hermitian operator $A$, every eigenvalue is real. Indeed, if $Av=\lambda v$ with $v\ne 0$, self-adjointness gives $\lambda ⟨v,v⟩=⟨Av,v⟩=⟨v,Av⟩=\overline{\lambda }⟨v,v⟩$. To prove the finite-dimensional spectral theorem, proceed by induction on the dimension. The fundamental theorem of algebra supplies a root of the characteristic polynomial, hence a unit eigenvector $v$. Its orthogonal complement is invariant: for $w\perp v$, $⟨Aw,v⟩=⟨w,Av⟩=0$. The restriction of $A$ to that complement is again Hermitian. Induction supplies an orthonormal basis of eigenvectors there; adjoining $v$ gives an orthonormal eigenbasis of the whole space. The zero-dimensional case is immediate. Consequently a Hermitian operator is diagonal in an orthonormal basis, with real diagonal entries.

Let $P$ be the transition matrix of a finite irreducible Markov chain with at least two states, reversible with stationary distribution $\pi$. Its Dirichlet form of a Markov chain isThe Discrete Poincare inequality states that ${\mathrm{Var}}_{\pi }f\le C\mathcal{E}(f,f)$ for every real $f$. Its optimal constant is $C=(1-{\lambda }_2{)}^{-1}$, where ${\lambda }_2$ is the second largest eigenvalue of $P$. Indeed, detailed balance makes $P$ self-adjoint on $L^2(\pi )$. Expand $f-\pi (f)={\sum }_{j\ge 2}{a}_j{v}_j$ in an orthonormal eigenbasis. Then ${\mathrm{Var}}_{\pi }f=\sum a_j^2$ and $\mathcal{E}(f,f)=\sum (1-{\lambda }_j)a_j^2$, proving the inequality and its sharpness.

The useful constructive version is the canonical paths Poincare bound. Choose transition graph paths ${\gamma }_{xy}$ between each ordered pair, write $Q(u,v)=\pi (u)P(u,v)$, and definewhere $N_{uv}$ counts occurrences of the directed transition. The Cauchy-Schwarz inequality along each path in a graph gives $(f(x)-f(y){)}^2\le |{\gamma }_{xy}|{\sum }_{{\gamma }_{xy}}(f(u)-f(v){)}^2$. Multiply by $\pi (x)\pi (y)/2$ and sum. The identity ${\mathrm{Var}}_{\pi }f=\frac{1}{2}{\sum }_{x,y}\pi (x)\pi (y)(f(x)-f(y){)}^2$ proves ${\mathrm{Var}}_{\pi }f\le \rho \mathcal{E}(f,f)$.

For a lazy Markov chain this also proves the mixing time estimateTo see this, the centred initial density has $L^2(\pi )$ norm at most $\pi (x{)}^{-1/2}$; laziness puts all nonconstant eigenvalues between zero and $1-{\rho }^{-1}$. Their powers contract that norm, and the Cauchy-Schwarz inequality bounds total variation distance by half the density's $L^2(\pi )$ norm.

Here is a precise sufficient density condition for the permanent application: every row and every column has more than $n/2$ ones. Interpret the permanent of a matrix as the number $m_n$ of perfect matchings in its bipartite graph, with $n$ vertices per side. Write $m_k$ for the number of $k$-edge matchings in a graph.

First prove the short augmenting paths in a dense bipartite graph bound. Given a nonperfect matching in a graph, choose unmatched $u,v$ on opposite sides. If either has an unmatched neighbour, augment with one edge. Otherwise the matched partners of $N(u)$ and the neighbours $N(v)$ are subsets of the same $n$-element side whose total size exceeds $n$. Their intersection gives an alternating path in a graph $u,b,a,v$ of length three, with $ab$ a matching edge. Repeating proves $m_n>0$. Choose augmentations by a fixed deterministic rule. A reverse augmentation is specified by at most four vertices, so a next-size matching in a graph has at most $C_0=2n^4$ preimages. Therefore

Use the matching generating function $Z(\lambda )={\sum }_{k=0}^n{m}_k{\lambda }^k$ and the weighted matching Markov chain on all matchings in a graph, with stationary weights ${\pi }_{\lambda }(M)={\lambda }^{|M|}/Z(\lambda )$. It stays put with probability $1/2$; otherwise it chooses a graph edge uniformly and proposes its deletion, its addition if both endpoints are unmatched, or its exchange for the unique incident matching edge if precisely one endpoint is matched. Other proposals stay put. Accept with probability $\min (1,{\lambda }^{|M^{\prime }|-|M|})$. Reverse proposals use the opposite exchanged edge, so detailed balance holds. Deletions connect every state to the empty matching in a graph.

For completeness, the canonical paths for the weighted matching chain give a polynomial bound without assuming a mixing theorem. The symmetric difference $I△F$ consists of alternating graph paths and graph cycles. Order its components deterministically, orient open components from their smaller endpoint, and convert them from $I$ to $F$ by moving an unmatched endpoint with exchanges. If the first edge belongs to $I$, delete it first. For a graph cycle, delete one $I$-edge, move the unmatched endpoint around the opened cycle, then insert the last $F$-edge. Each intermediate state is a matching in a graph, and each entire transition path in a graph has length at most $4n$.

Fix a directed transition with initial state $M$. The complementary edge set $K=I△F△M$ agrees with one of the two matchings in a graph outside the active component and has at most two degree-two vertices on that component. Remove at most two offending edges to obtain a matching in a graph $K^{\prime }$. Store those edges and one bit specifying the initial alternating colour of the active component. This encoding is injective for the fixed transition: it restores $K$ and $I△F=M△K$, identifies the active component from the changed edges, and recovers the two colours using the fixed component order. Completed components have $F$'s colour in $M$, pending components have $I$'s colour, and the stored bit resolves the active component. Common edges are $M\cap K$.

There are at most $2(m+1{)}^2$ auxiliary records per $K^{\prime }$, where $m$ is the number of graph edges. If $d=|K|-|K^{\prime }|\le 2$, thenAlso $Q(M,M^{\prime })\ge {\pi }_{\lambda }(M)/(2m\max (\lambda ,{\lambda }^{-1}))$. Sum over the encodings and use the path length bound to obtainThis is polynomial whenever $\lambda$ and ${\lambda }^{-1}$ are polynomially bounded. Starting at the empty matching in a graph, $\log (1/{\pi }_{\lambda }(\varnothing ))=\log Z(\lambda )\le m\log (1+\lambda )$, so the Discrete Poincare inequality gives polynomial-time sampling to any prescribed total variation distance. The addition/deletion/exchange construction is also described in Section 12.4 of Jerrum and Sinclair's survey.

It remains to turn sampling into a relative count; merely sampling perfect matchings would not do that. For $0<\epsilon \le 1/2$, choose ${\lambda }_0=\epsilon /(8m)$ and ${\lambda }_{\ast }=8C_0/\epsilon$. The elementary bounds above giveChoose a schedule ${\lambda }_0<{\lambda }_1<\cdots <{\lambda }_J={\lambda }_{\ast }$ whose successive ratios are at most $1+1/n$. It has $J=O(n\log (n/\epsilon ))$ stages. The annealing ratios for a matching generating function satisfyEach observable lies in $[1,e]$, so the Hoeffding inequality gives relative accuracy $\eta =\epsilon /(10J)$ with $O({\eta }^{-2}\log (J/\delta ))$ independent samples per stage. Run independent copies of the chain from the empty matching in a graph; make the total variation distance per sample at most $\delta /(3JK)$, where $K$ is that sample count. The coupling characterization of total variation distance and the union bound make the entire sample collection coincide with exact stationary samples except with probability at most $\delta /3$. Choose the sample-count constant so all stationary sample means are accurate except with probability at most $\delta /3$.

Let $\hat{R}$ be the product of the sample means, and output $\hat{p}=\hat{R}/{\lambda }_{\ast }^n$. On the successful event its multiplicative error relative to $Z({\lambda }_{\ast })/Z({\lambda }_0)$ lies between $e^{-\epsilon /5}$ and $e^{\epsilon /5}$. Combining the two endpoint bounds provesActivities, sample counts, simulation lengths and rational bit lengths are polynomial in $n,{\epsilon }^{-1},\log ({\delta }^{-1})$. Thus this is a fully polynomial randomized approximation scheme. Small dimensions can be evaluated directly.

Over the Grassmannian ${\mathrm{Gr}}_k({\mathbb{C}}^N)$, the complex tautological bundle on a Grassmannian isIts fibre at $W$ is the k-plane $W$ itself. A neighbourhood of a fixed plane consists of graphs of linear maps $A:W\to W^{\perp }$; $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The inclusions of coordinate spaces give the infinite Grassmannian and its tautological rank-k bundle ${\gamma }_k^{\infty }$.

Use the customary compact Hausdorff base convention for the classification of complex vector bundles by a Grassmannian; equivalently the argument uses finite subordinate partitions of unity. Give a rank-k complex vector bundle $E\to X$ a continuous fiber metric given by Hermitian inner products. Choose a finite trivializing cover with local orthonormal coordinates $t_i:E{|}_{U_i}\to U_i\times {\mathbb{C}}^k$, and a subordinate partition of unity ${\rho }_i$ with support contained in $U_i$. Defineextending each component by zero off its support. This is continuous and satisfies $\Vert J_x(v){\Vert }^2={\sum }_i{\rho }_i(x)\Vert v{\Vert }^2=\Vert v{\Vert }^2$, so it is a fibrewise embedding. The image planes vary continuously: in local frames, their orthogonal projection matrix is $J(J^{\ast }J{)}^{-1}{J}^{\ast }$. Thus they define a map $c_E:X\to {\mathrm{Gr}}_k({\mathbb{C}}^{kr})$, and $v\mapsto (x,J_{x}v)$ identifies $E$ with $c_E^{\ast }{\gamma }_k^{\infty }$. This proves that every bundle is a tautological pullback vector bundle, by the finite-dimensional embedding of a complex vector bundle.

We next prove the two independence assertions that turn this construction into a bijection. First, homotopic maps have isomorphic pullback vector bundles. Pulling back along a homotopy gives a bundle over compact Hausdorff $X\times I$; embed that bundle in a finite trivial vector bundle by the construction just given. Its image has a continuous family of rank-k orthogonal projection matrices $P(x,t)$. Compactness gives a subdivision $0=t_0<\cdots <t_s=1$ withRestriction of $P(x,t_{j+1})$ to the image at $t_j$ is injective: if it killed $v$, then $v=(P(x,t_j)-P(x,t_{j+1}))v$, impossible by the strict norm bound unless $v=0$. Equal ranks make it an isomorphism, with continuous local inverses. Composing these bundle maps identifies the endpoint vector bundles. The fact that close orthogonal projections identify their image bundles therefore makes the pullback assignment well-defined on homotopy classes.

Conversely, suppose two classifying maps give isomorphic vector bundles. The infinite Grassmannian has its standard weak CW topology; the compact-subset property of CW complexes puts the compact images of the two maps in finite stages. Their tautological pullback vector bundles therefore give two finite-dimensional embeddings $J_0:E\to {\mathbb{C}}^{N_0}$ and $J_1:E\to {\mathbb{C}}^{N_1}$ of the same bundle, after using the vector bundle isomorphism. In orthogonal coordinate blocks interpolate byAt least one coefficient is nonzero, so every $J_t$ is injective and its image planes give a continuous Grassmannian homotopy. One endpoint uses the first block and the other the second block. Moving the latter coordinate copy to the standard one is a fixed unitary change of coordinates, connected to the identity by a path in the unitary group. Such a path exists by diagonalizing the unitary matrix and continuously varying its eigenvalue phases. Hence the original two maps are homotopic in the infinite Grassmannian. This also proves independence of all embedding and metric choices. Together the two directions establishFor arbitrary non-Hausdorff compact spaces without the required numerability, that usual bundle-classification statement needs additional hypotheses; the construction above specifies precisely the conventional setting being used.

Write $B={\mathrm{Gr}}_k({\mathbb{C}}^{\infty })$. For the assumed fiber bundle $U_k\to \nu \to B$ with contractible total space, the long exact sequence of homotopy groups of a fibration has segmentFor $q\ge 2$ this gives group isomorphisms. For $q=1$, the relevant pointed-set part and connectedness of $U_k$ give ${\pi }_1(B)=0$, corresponding to the single component ${\pi }_0(U_k)$. Thus the claimed formula includes its endpoint:Finally the determinant gives a fiber bundle $S{U}_2\to U_2\to S^1$. Since ${\pi }_3(S^1)={\pi }_4(S^1)=0$, its exact sequence yields ${\pi }_3(U_2)\cong {\pi }_3(S{U}_2)$. The given identification $S{U}_2\cong S^3$ and the degree calculation ${\pi }_3(S^3)=\mathbb{Z}$ therefore giveThe equality of free and based homotopy classes uses the simple connectedness just proved. HenceThe classification of rank-two complex vector bundles over the four-sphere is a bijection of isomorphism classes, labelled by the integer clutching function homotopy class after choosing a generator. It is not an assertion that Whitney sum gives a group operation on bundles of fixed rank two.

Let $M={\mathbb{R}}^d/\Lambda$ be a flat torus, where $\Lambda$ is a full-rank Euclidean lattice, and use the nonnegative Laplace-Beltrami operator $\Delta =-{\sum }_{k=1}^d{\partial }_k^2$. The dual lattice isFor $w\in {\Lambda }^{\ast }$, the function ${\varphi }_w(x)=\mathrm{vol}(M{)}^{-1/2}{e}^{2\pi i⟨w,x⟩}$ is periodic under $\Lambda$, so it defines a function on $M$, and direct differentiation gives $\Delta {\varphi }_w=4{\pi }^2|w{|}^2{\varphi }_w$. Integrating characters on a fundamental parallelepiped proves that these functions are orthonormal. After choosing a lattice basis, the ordinary Fourier series theorem on ${\mathbb{R}}^d/{\mathbb{Z}}^d$ proves completeness, so there are no additional eigenvalues. Thus the spectrum of a flat torus iswith multiplicity equal to the number of dual vectors of the given length; the zero eigenvalue occurs once.

It remains to show that a two-dimensional Euclidean lattice is determined up to an orthogonal transformation by its vector-length multiset. This is the two-dimensional lattice reconstruction from vector lengths. Let $v$ be a shortest nonzero vector, of length $a$. It is primitive: $v=ku$ with integer $|k|>1$ would give a shorter nonzero vector. The lattice vectors on its line are therefore precisely the integer multiples of $v$. Subtract two occurrences of each length $ka$, $k=1,2,\dots$, from the length multiset. This removes exactly one primitive lattice line, even when other vectors happen to have the same lengths. The smallest remaining length $b$ is the length of a shortest vector $w$ independent of $v$.

These two vectors form a basis of the lattice. Otherwise a nonzero coset of $\mathbb{Z}v+\mathbb{Z}w$ has a lattice representative $u=sv+tw$ with $|s|,|t|\le 1/2$. It cannot have $t=0$, since primitivity of $v$ would then force $u=0$. It is therefore independent of $v$, whileThe inequality is strict: if both coefficients are nonzero, the vectors are not parallel and the triangle inequality is strict; if $s=0$, then $|u|\le b/2$. This contradicts the choice of $w$.

The covolume $A$ is also determined by the length multiset. If $N(R)$ counts lattice vectors in a radius-$R$ disk, thenTo justify the estimate, tile the plane by a bounded fundamental parallelogram. Only cells meeting a fixed-width neighbourhood of the disk boundary contribute an error; their total number is $O(R)$. The lattice basis $v,w$ has Gram matrixThus $|c|=\sqrt{a^2{b}^2-A^2}$ is determined. Replacing $w$ by $-w$ makes $c\ge 0$, so the Gram matrix is determined entirely. Equal Gram matrices give an orthogonal map between the lattice bases and hence between the lattices.

Two isospectral two-dimensional flat tori consequently have orthogonally equivalent dual lattices. Taking duals shows that their original lattices are orthogonally equivalent too, and the orthogonal map descends to an isometry of the quotient tori. Two isospectral flat tori of dimension two are isometric.