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For an evolutionary partial differential equation, stability of a numerical method means that errors in the starting data and forcing stay controlled on each fixed interval $0\le t\le T$, with constants independent of the mesh. Fourier stability analysis is especially effective for a constant-coefficient finite difference method on a uniform infinite or periodic grid: translation invariance makes different Fourier modes evolve independently.

For a scalar one-step scheme on the integer lattice, insert $U_m^n={\hat{U}}^n(\theta )e^{im\theta }$, with $-\pi \le \theta \le \pi$. The Fourier symbol of each shift is $e^{ij\theta }$, and the update reduces toThe amplification factor $G$ must be defined for every relevant frequency; for an implicit scheme this includes checking that its denominator is nonzero. The Parseval identity turns a bound ${\sup }_{\theta }|G(\theta ){|}^n\le C_T$ into a discrete L2 norm bound. Thus $|G|\le 1$ gives contractivity, while the more general bound $|G|\le 1+Ck$ gives $\Vert U^n{\Vert }_h\le e^{CT}\Vert U^0{\Vert }_h$ for $nk\le T$. Conversely, frequencies with amplification uniformly greater than one produce unstable wave packets or periodic Fourier modes under refinement.

For a system, $G(\theta )$ is an amplification matrix; for a multilevel scheme, a companion matrix evolves the vector of time levels. The actual requirement is a uniform bound on matrix powers, not just their spectral radius. A nontrivial Jordan block at a unit-modulus eigenvalue creates polynomial growth in the time index. Even simple eigenvalues inside the unit disk can fail to give a uniform bound if the eigenvector matrices become ill-conditioned as the mesh changes. Uniformly controlled diagonalization of a matrix, or a suitable quadratic energy estimate, resolves this issue. The power boundedness of a two-level Fourier scheme illustrates why root multiplicities and conditioning matter.

For the heat equation, put $\mu =k/h_x^2$. Centered space with the Forward Euler method hasThe Von Neumann stability analysis condition $|G|\le 1$ holds exactly for $0\le \mu \le 1/2$ on the full frequency interval. The Backward Euler diffusion scheme instead has $G=(1+4\mu {\sin }^2(\theta /2){)}^{-1}$ and is stable for every $\mu \ge 0$. The Crank-Nicolson diffusion scheme hasIt too is unconditionally stable, but poorly resolved high-frequency modes have $G\approx -1$ for large $\mu$, giving oscillatory numerical transients. A-stability therefore does not guarantee strong damping; L-stability distinguishes the damping of the Backward Euler method.

For the advection equation $u_t+c{u}_x=0$, let $\nu =ck/h_x$. Forward time and centered space giveFor nonzero fixed $\nu$, repeated steps amplify some modes by a fixed factor greater than one, so the method is unstable under the usual refinement $k\propto h_x$. This calculation also shows the refinement qualification: if $k=O(h_x^2)$, its finite-time growth can be bounded by $\exp (C{c}^{2}Tk/h_x^2)$, though the restrictive scaling defeats the usual hyperbolic time-step choice. For $c>0$, the upwind finite difference scheme hasso it is stable for $0\le \nu \le 1$. The Lax-Wendroff advection scheme instead hasgiving $|\nu |\le 1$. These examples separate the effects of the spatial stencil, temporal approximation and Courant number; a higher order of a numerical method by itself does not establish stability.

A periodic boundary condition is ideal for this analysis. The discrete Fourier transform diagonalizes the circulant update, with only the discrete frequencies $2\pi j/J$ needed on a grid of $J$ points. A bound on all $\theta$ is a convenient guarantee over every such grid. On the whole line, the Fourier transform uses a continuous frequency interval and the Parseval identity proves the corresponding square-summable-data estimate.

Homogeneous Dirichlet boundary conditions do not admit arbitrary complex exponential Fourier modes. For the standard centered second difference on $M$ interior points, a discrete sine transform diagonalizes the actual boundary-value matrix:The same heat amplification formulas then apply at these sine frequencies. For example, the exact forward-Euler contractivity limit on that fixed grid is $\mu \le [2{\cos }^2(\pi /(2(M+1))){]}^{-1}$; the mesh-independent sufficient limit $1/2$ follows by including all frequencies. Homogeneous Neumann boundary conditions can similarly permit a discrete cosine transform, provided the endpoint discretization and its weighted inner product are chosen consistently. The constant mode is then present, reflecting conservation of the heat equation's spatial mean. For inhomogeneous boundary conditions, subtract a suitable lifting of the prescribed boundary data and estimate the induced source using the homogeneous evolution's bound and a Duhamel principle estimate. Bounds for the lifting and source must themselves be uniform in the mesh.

General boundary conditions require separate boundary stability of a finite-difference method. For an advection equation, incoming data are prescribed at the inflow boundary, while an outflow closure must respect the outgoing characteristics. An interior periodic Fourier symbol cannot detect a growing mode confined near a boundary. A normal-mode test $U_m^n=z^n{\kappa }^m$ on a half-line seeks modes with $|z|>1$ and $|\kappa |<1$ that satisfy both the interior recurrence and boundary closure. Their existence proves instability, and uniform control requires more than merely excluding isolated growing roots; the Uniform Kreiss--Lopatinskii condition addresses boundary resolvent bounds. Alternatively, a direct energy method using summation by parts can include the boundary terms and establish the required estimate.

Variable coefficients and nonuniform meshes usually destroy exact Fourier transform diagonalization. Frozen-coefficient Fourier stability analysis is then a useful diagnostic, but it is not automatically a proof for the full variable-coefficient boundary problem. The mesh-uniform energy method in question 5 is an example of a direct proof. Also, an L2 norm proof is not automatically a mesh-uniform maximum-norm proof: finite-dimensional norm-equivalence constants may grow with the number of grid points.

Finally, stability measures error propagation; consistency of a numerical method measures the defect of inserting the exact solution. For a well-posed linear initial-value problem in the chosen norm, the Lax equivalence theorem says that a consistent approximation is convergent exactly when it is stable. Boundary discretization and starting data must be included in that assertion. A Fourier calculation proves convergence only after consistency, well-posedness and the actual boundary treatment have also been checked.

The Poisson regression uses independent counts with mean $m_i$ and logarithmic link function. Writing $c$ for concentration, the fitted predictors for strains $0,1,2$ are respectivelyThe interaction formula includes concentration, strain indicators and their products. The intercept is the baseline log mean at zero toxin; exponentiating it gives about $87.5$ fleas. Strain coefficients compare log means at zero concentration. Interaction coefficients compare concentration slopes with the common strain. A unit concentration increase multiplies the three means by about $0.751$, $1.057$ and $0.769$ respectively.

For design matrix $X$, the fitted expected Fisher information is $X^{T}WX$, $W=\mathrm{diag}({\hat{m}}_i)$, since a Poisson count has variance equal to its mean. Standard errors are square roots of diagonal entries of $(X^{T}WX{)}^{-1}$. Each reported Wald test divides its coefficient estimate by its standard error and compares with an approximate standard normal distribution under a zero-coefficient null, conditional on the other terms. Concentration has strong evidence of a negative baseline effect. At zero concentration strain 1 is not significant at 5% ($p=0.087$), while strain 2 is ($p=0.030$). The strain-1 slope differs strongly from the baseline ($p=0.000193$); the strain-2 slope has no detected difference ($p=0.808$). A nonsignificant difference is not proof of equality. The very significant intercept is a test of mean one at zero toxin, with little scientific relevance.

Relabelling factor levels merges strains 0 and 2 into one group and leaves strain 1 separate. Model $M_2$ retains three intercepts but allows only two slopes: strain 1 versus the common slope of strains 0 and 2. Thus it has five parameters, compared with six in $M_1$. Model $M_3$ merges both intercepts and slopes for strains 0 and 2, retaining only four parameters. The natural nesting is $M_3\subset M_2\subset M_1$.

For these nested generalized linear models, the difference in Poisson deviance is an asymptotic likelihood-ratio test statistic with degrees of freedom equal to the parameter difference. Comparing $M_2$ with $M_1$ gives $56.93-56.87=0.06$ on one degree of freedom, far below $3.84$: the additional strain-2 slope is unnecessary. Comparing $M_3$ with $M_2$ gives $76.98-56.93=20.05$ on one degree of freedom, far above $3.84$: merging the strain-0 and strain-2 intercepts is rejected. Comparing $M_3$ with $M_1$ uses $20.11$ on two degrees of freedom and requires a two-degree-of-freedom threshold, about $5.99$, rather than the supplied $3.84$.

$M_2$ is the most appropriate of the three models: retain different strain baselines and the special strain-1 slope, but pool the unsupported strain-2 slope difference. Residual degrees of freedom are $74,75,76$; deviances of this magnitude give no obvious evidence of overdispersion from these summaries alone.

Let $A$ be the adjacency matrix of a graph, $J$ the all-ones matrix and $1$ the all-ones vector. Counting two-step walks givesBecause $b\ge 1$, any two nonadjacent vertices have a common neighbour, so the graph is connected. The eigenvalue $k$ is simple: for a real eigenvector at $k$, a maximal entry and the averaging equation force its neighbours, and then every vertex, to have the same value. Since $A$ is real symmetric, its other eigenvectors lie in $1^{\perp }$, where $J$ vanishes. Their eigenvalues are thereforeHere $d>0$ for the present incomplete connected graph. Indeed $k\ge b$; if $d=0$, then $k=b=a$, which conflicts with $a\le k-1$ for an edge. If $r,s$ have multiplicities $m_r,m_s$, dimension and zero trace give $m_r+m_s=n-1$ and $k+m_{r}r+m_{s}s=0$. ThusThese are dimensions of eigenspaces, hence integers. A multiplicity may be zero; that does not affect the conclusion.

If the two-dimensional group representation were reducible, Maschke's theorem would split it into a direct sum of two one-dimensional group representations. In a basis respecting that decomposition, every representing matrix would be diagonal. All would therefore commute, contradicting the specified pair. Hence the group representation is irreducible. Complete reducibility over $\mathbb{C}$ is essential to this argument.

Put $\zeta =e^{\pi i/n}$. The group representations from the preceding construction with $ϵ={\zeta }^j$, $1\le j\le n-1$, will be denoted ${\rho }_j$. Their diagonal matrices for $a$ have distinct eigenvalues. The only possible invariant lines would be these two eigenspaces, and the matrix for $b$ exchanges them. Thus each ${\rho }_j$ is irreducible. Equivalently their $a,b$ matrices do not commute, so part (a) applies. The choices $j$ and $2n-j$ give equivalent group representations by exchanging basis vectors. The displayed range contains exactly one of each pair, and distinct members have different traces at $a$ because $\cos (\pi j/n)$ is strictly decreasing there.

For a one-dimensional character of a representation, write $a\mapsto \alpha$, $b\mapsto \beta$. The conjugation relation forces ${\alpha }^2=1$, and ${\beta }^2={\alpha }^n$. Hence there are exactly four choices:For even $n$, $\beta =\pm 1$ for either $\alpha$; for odd $n$, $\alpha =1$ has $\beta =\pm 1$ and $\alpha =-1$ has $\beta =\pm i$. Finally,The sum of squares of irreducible representation dimensions proves that these four one-dimensional and $n-1$ two-dimensional group representations exhaust the irreducibles, including $n=1$ where the latter list is empty.

Let $c_1,\dots ,c_d$ be the coefficients of $f$. Each is an algebraic integer. The ring $R=\mathbb{Z}[c_1,\dots ,c_d]$ is a finitely generated $\mathbb{Z}$-module: reduce powers of each generator by its monic integral equation. As $f$ is monic, $R[\alpha ]$ is generated over $R$ by $1,\alpha ,\dots ,{\alpha }^{d-1}$, so is a finite $\mathbb{Z}$-module. It is torsion-free as a subgroup of a characteristic-zero field, hence free of finite rank. Multiplication by $\alpha$ on it is represented by an integer matrix. Its monic characteristic polynomial, by Cayley-Hamilton theorem, annihilates $\alpha$. Thus $\alpha$ is an algebraic integer. This proves the required transitivity of integrality in this setting.

Every root of the monic $h$ is also a root of the monic polynomial $h^n\in {\mathcal{O}}_K[x]$, and is therefore an algebraic integer. The coefficients of $h$ are elementary symmetric polynomials in these roots, hence algebraic integers because algebraic integers form a ring. They also belong to $K$, so belong to ${\mathcal{O}}_K$. Consequently $h\in {\mathcal{O}}_K[x]$.

Put $C=[A,B]$, which commutes with both operators by hypothesis. Differentiating $e^{\lambda A}B{e}^{-\lambda A}$ gives $C$, so this conjugate is $B+\lambda C$. Similarly the product $P=e^{\lambda A}{e}^{\lambda B}$ conjugates $A+B$ to itself: the two central commutator shifts cancel. HenceWith $F(0)=I$, the operator differential equation gives $F(\lambda )=e^{{\lambda }^{2}C/2}$. At $\lambda =1$ this proves the central-commutator Baker--Campbell--Hausdorff formulaThese formal manipulations apply on a suitable common domain, or to the oscillator exponentials below where the corresponding unitary identities are well defined.

The oscillator has $[a,a^{†}]=1$ and $H=\hslash \omega (a^{†}a+1/2)$. Its normalized states and energies areSince $\hat{x}=\sqrt{\hslash /(2m\omega )}(a+a^{†})$, put $\gamma =\mu \sqrt{\hslash /(2m\omega )}$. Taking $A=\pm i\gamma a^{†}$, $B=\pm i\gamma a$ gives $[A,B]={\gamma }^2$. The formula implies $e^{\pm i\mu \hat{x}}=e^{-{\gamma }^2/2}{e}^{\pm i\gamma a^{†}}{e}^{\pm i\gamma a}$. Averaging the two signs proves the requested normal ordering of the cosine.

Because $a|0⟩=0$, expansion of the remaining creation exponential givesOdd matrix elements vanish, the ground expectation is $e^{-{\gamma }^2/2}$, and the magnitude squared for $n=2p$ is $e^{-{\gamma }^2}{\gamma }^{4p}/(2p)!$. The nondegenerate second-order nondegenerate perturbation theory formula is $\Delta E_0=ϵV_{00}+{ϵ}^2{\sum }_{n>0}|V_{n0}{|}^2/(E_0-E_n)+O({ϵ}^3)$. ThereforeThe negative second-order sign comes from the higher unperturbed energies in every denominator.

The Gershgorin circle theorem says that every eigenvalue of a square matrix lies in at least one disc centred at a diagonal entry with radius the sum of the absolute values of the off-diagonal entries in that row.

For interior unknowns write the update as $u^{n+1}=B{u}^n$. The matrix is real symmetric and tridiagonal, with diagonal $1-\mu (a_{m-1/2}+a_{m+1/2})$ and adjacent entries $\mu a_{m+1/2}$. Each Gershgorin radius is at most $\mu (a_{m-1/2}+a_{m+1/2})$. Its real interval is therefore contained inFor $0<\mu \le 1/(2a_{+})$, every eigenvalue lies in $[-1,1]$. Symmetry gives an orthogonal diagonalization, so $\Vert B^n{\Vert }_2\le 1$ for every $n$: errors in initial data cannot grow. This is stability in the discrete ${\ell }^2$ norm, also in its $h$-weighted version. One can additionally see maximum-norm stability directly: the update weights are nonnegative with row sums at most one once the zero boundary unknowns are removed. Hence $\Vert B{\Vert }_{\infty }\le 1$.

Relative positions and velocities measured from galaxy $A$ are $r_B-r_A$ and $v_B-v_A$. Spatial homogeneity says that the same position-to-relative-velocity rule applies from every galaxy. Thus $v(r_B-r_A)=v(r_B-r_O)-v(r_A-r_O)$. Taking $r_O=0$ and putting $r_B=x+y$, $r_A=y$ gives additivity $v(x+y)=v(x)+v(y)$. For a continuous physical velocity field, additivity implies rational homogeneity and, by continuity, real homogeneity. Thereforefor a position-independent matrix $H$. Regularity is necessary to exclude pathological discontinuous additive functions.

Decompose the given matrix as $H=(5D/t)I+(D/t)W$, whereIts symmetric part is isotropic: no cosmological shear, expansion scalar $\nabla \cdot v=15D/t$, and linear scale factor proportional to $t^{5D}$ for $t>0$. Its antisymmetric part is rigid rotation with angular velocityThe vorticity is $2\mathbf{\omega }$. Comoving separations obeyThus all distances grow by the same power, volumes grow as $t^{15D}$, and the pattern rotates about the fixed axis $(1,-2,1)$ through angle $\sqrt{6}D\log (t/t_0)$. Homogeneity alone permits this rotation; full cosmological isotropy would not.

For $r\ge 2$, let $H$ be the $r\times (2^r-1)$ binary matrix whose columns are all distinct nonzero vectors of ${\mathbb{F}}_2^r$. The binary Hamming code is $C=\ker H$. Its parity-check matrix has rank $r$ since its columns include the standard basis. Hence $C$ is linear of length $2^r-1$ and dimension $2^r-1-r$.

No word of weight one or two lies in $C$, since columns are nonzero and distinct; three columns $u,v,u+v$ sum to zero, so the minimum distance of a code is exactly three. The syndrome of a single-bit error is its column of $H$, uniquely identifying the erroneous position. The $2^r$ possible syndromes correspond to no error or exactly one of $2^r-1$ single-bit errors. Alternatively, every radius-one Hamming ball has $2^r$ words andDisjoint balls therefore cover the whole word space. This proves linearity, one-error correction and perfection. For $r=3$, the parameters are $[7,4,3]$, giving the 16 messages needed below.

A matrix is a diagonalizable matrix if there is an invertible matrix $P$ for which $P^{-1}MP$ is diagonal. Its columns then form a basis of eigenvectors, and the diagonal entries are their corresponding eigenvalues.

Here the given eigenvectors form an orthonormal basis. Put them into the columns of $Q$:Define $\Lambda =Q^{T}MQ$. Its entries areThis proves the claimed diagonalization of a matrix, rather than merely guessing it:The trace follows directly from this factorization:because each column of $Q$ has unit length. Trace is the sum of the eigenvalues, with multiplicity.

For the particular matrix, $(M^2{)}_{ij}={\sum }_k{M}_{ik}{M}_{kj}$. If $i=j$, every $k\ne i$ contributes one and $k=i$ contributes zero. If $i\ne j$, precisely $k=i$ and $k=j$ contribute zero; the other $n-2$ terms contribute one. ThusApplying this identity to a nonzero eigenvector givesEvery eigenvalue is therefore $n-1$ or $-1$.

If $r$ is the multiplicity of $n-1$, the trace formula for a diagonalizable matrix and the zero diagonal giveso $r=1$. Up to ordering, the required diagonal form isFor $n=1$ this means just $(0)$. For $n\ge 2$, the eigendirections are also transparent: the all-ones vector has eigenvalue $n-1$, and the $(n-1)$-dimensional subspace whose coordinate sum is zero has eigenvalue $-1$, since $(M\mathbf{x}{)}_i={\sum }_j{x}_j-x_i$. The spectrum consists of one collective mode and $n-1$ zero-sum modes.

Adding the three equations for $s,t$ gives the necessary solvability condition $a+b+c=3$. Conversely, subtraction of the first two and use of the third forceIf $a+b+c=3$, these values also satisfy the first two equations: for example,The analogous calculation gives $1-s+t=b$. Thus existence and uniqueness hold exactly under the stated compatibility condition:

For the three-variable system the matrix and data vector arePerform Gaussian elimination on the augmented matrix. The operations $R_2\leftarrow 5R_2-2R_1$ and $R_3\leftarrow 5R_3+R_1$ giveEach operation is an invertible row scaling followed by a row addition, so the solution set is preserved. Next $R_3\leftarrow 7R_3+R_2$ givesBack substitution now yieldsand thenAll three pivots are nonzero. Thus the solution exists uniquely for every $s,t$, the matrix rank is $3$, and the kernel of a linear map represented by $A$ is $\{0\}$. The general rank-nullity theorem for a matrix with $m$ columns ishere $m=3$.

Finally, reverse the viewpoint and regard $x,y,z$ as prescribed. The necessary and sufficient condition already proved applies with $a=5x+2y-z$, $b=2x+5y-z$, $c=-x-y+8z$. Their sum is $6(x+y+z)$. Consequently $s,t$ can be recovered exactly whenWhen this holds, their unique values are $s=\frac{3}{2}(x-y)$ and $t=\frac{1}{2}(1+x+y-8z)$. The allowable triples form an affine plane, not all of ${\mathbb{R}}^3$.

A complex square matrix $U$ is a unitary matrix when $U^{†}U=I$, equivalently $U^{-1}=U^{†}$. Here $†$ denotes the adjoint matrix, the conjugate transpose. Such a matrix preserves the standard complex inner product and its Euclidean norm.

A real symmetric matrix $A$ is a Hermitian matrix, so $A^{†}=A$. For arbitrary complex $\mathbf{x}$,The imaginary cross terms cancel because $A$ is Hermitian. Reversing the signs gives the same identity:In particular, $(A+iI)\mathbf{x}=0$ implies $|\mathbf{x}{|}^2=0$, so $A+iI$ has zero kernel of a linear map. It is a square matrix, and the rank-nullity theorem therefore makes it invertible. The same argument proves invertibility of $A-iI$.

The proposed matrixis consequently well-defined. For any $\mathbf{y}$, put $\mathbf{x}=(A+iI{)}^{-1}\mathbf{y}$. The two norm identities show directly thatTo verify the unitary matrix identity explicitly, take adjoints:The two shifts commute, because both are polynomials in $A$. HenceThusThe imaginary shifts are invertible, and their equal norm changes cancel in the quotient. This is the Cayley transform of a Hermitian matrix in the $(A-iI)(A+iI{)}^{-1}$ convention.

Every permutation is a product of transpositions, since a cycle can be writtenwith rightmost factors acting first. Define the sign of a permutation to be $(-1{)}^r$ for a product of $r$ transpositions. To prove that it is well defined without assuming parity, let $P_{\sigma }$ be the permutation matrix specified by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Then $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, and a transposition exchanges two columns of the identity, so its determinant is $-1$. Any decomposition into $r$ transpositions therefore hasThe left-hand side depends only on $\sigma$, proving well-definedness. Multiplicativity of the determinant now gives the sign homomorphism:For the first requested group embedding, use the faithful four-point action of GL2 over F2: label the four vectors in ${\mathbb{F}}_2^2$ and let an invertible $2\times 2$ matrix act on them by multiplication. Composition of these actions gives a group homomorphism $\psi :G{L}_2({\mathbb{F}}_2)\to S_4$. If a matrix fixes every vector, it fixes both standard basis vectors and is the identity, so the action is faithful. The matrixfixes $(0,0)$ and $(1,1)$ and exchanges $(1,0)$ with $(0,1)$. Its image is a single transposition, hence $\mathrm{sgn}(\psi (B))=-1$. The composite sign map is nontrivial.

For the second group embedding, simply takeThe preceding multiplication formula proves the group homomorphism property, and $P_{\sigma }=I$ forces $\sigma (i)=i$ for every $i$, proving injectivity.

The Möbius group consists of all Möbius transformations of the Riemann sphere ${\mathbb{C}}_{\infty }=\mathbb{C}\cup \{\infty \}$, with operation function composition. A nonzero-determinant matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ represents the map $z\mapsto (az+b)/(cz+d)$. If $c\ne 0$, its pole $-d/c$ maps to $\infty$ and $\infty$ maps to $a/c$; if $c=0$, the map is affine and fixes $\infty$. Nonzero scalar multiples of $M$ represent the same transformation.

Define $\varphi (M)$ by this action for $M\in S{L}_2(\mathbb{C})$. Substitution of fractional-linear expressions shows $\varphi (MN)=\varphi (M)\circ \varphi (N)$, so $\varphi$ is a group homomorphism. Any invertible matrix can be rescaled into the special linear group: choose $\lambda \in {\mathbb{C}}^{\times }$ with ${\lambda }^2\det M=1$, which is possible because every nonzero complex number has a square root. This does not change the transformation, proving surjectivity. A transformation represented by $M$ is the identity only if fixing $0$, $\infty$ and $1$ forces $b=c=0$ and $a=d$. The determinant-one condition then gives $a^2=1$. ThusFor the fixed-point claim, a nonidentity Möbius transformation has at least one fixed point of a Möbius transformation: when $c\ne 0$ the finite fixed points solve a quadratic, and when $c=0$ infinity is fixed. A quadratic shows there can be no more than two distinct fixed points unless the transformation is the identity. Suppose there is exactly one. Conjugate it to infinity by a Möbius transformation. The resulting map has form $z\mapsto az+b$, since it fixes infinity. If $a\ne 1$ it also has the finite fixed point $b/(1-a)$, a contradiction. Hence it is $z\mapsto z+b$ with $b\ne 0$. Its $m$th power is $z\mapsto z+mb$, never the identity for $m>0$. Conjugation preserves order, so a nonidentity finite-order Möbius transformation has exactly two fixed points.

Now let $K$ be a finite abelian subgroup of the Möbius group. First suppose it contains an element $g$ of order greater than two. Conjugate the two fixed points of $g$ to $0$ and $\infty$, so $g(z)=\zeta z$, with $\zeta$ a root of unity of order greater than two. Every element $h$ commuting with $g$ permutes its fixed-point set, because $g(h(p))=h(g(p))=h(p)$. A Möbius transformation preserving each of $0,\infty$ is $h(z)=az$; one exchanging them is $h(z)=a/z$. In the second case $hg(z)=a/(\zeta z)$ and $gh(z)=\zeta a/z$, so commuting would force ${\zeta }^2=1$, impossible. Thus all elements of $K$ are scalings. A finite subgroup of a field multiplicative group is cyclic; here one can see this directly by choosing the least common multiple $m$ of all element orders and observing that the scalars lie among the $m$th roots of unity, a cyclic group. Hence $K$ is cyclic.

Otherwise every nonidentity element of $K$ has order two. If $K$ is nontrivial, choose $g$ and conjugate it to $z\mapsto -z$. The action of $K$ on $\{0,\infty \}$ has image of order at most two. Its kernel consists of scalings $z\mapsto az$, and the order-two condition forces $a^2=1$, so the kernel has at most two elements. Hence $|K|\le 4$. Orders one and two give cyclic groups; order four with all nonidentity elements involutions gives $C_2\times C_2$, generated by any two distinct nonidentity elements. Both possibilities occur: scalings by roots of unity give every finite cyclic group, and $\{z,-z,1/z,-1/z\}$ gives the Klein four group. The classification is

The characteristic polynomial of the given matrix is $(t-1)(t-2{)}^2$. Solving the two eigenvalue equations givesFor example, multiplying the matrix by these three vectors gives respectively $1$, $2$ and $2$ times the original vector. The eigenvectors form a basis: the matrix with these columns has nonzero determinant $-1$. Thus an eigenbasis and its diagonal matrix are