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Write the symmetry group of a square as $D_4=⟨r,s\mid r^4=s^2=e,srs=r^{-1}⟩$, with $r$ a counterclockwise quarter-turn and $s$ reflection across the horizontal axis. The conjugacy classes of the symmetry group of a square areHere $r^2$ is the half-turn and $r^3$ the clockwise quarter-turn. The two members of the fourth class reflect across the horizontal and vertical axes. Those of the fifth reflect across $y=x$ and $y=-x$. The identity and half-turn are central elements, giving singleton classes. The relations $srs=r^3$, $rs{r}^{-1}=r^{2}s$, and $r(rs)r^{-1}=r^{3}s$ show that each displayed pair consists of conjugate elements. More generally, conjugating $r^{k}s$ by $r$ changes $k$ to $k+2$, and conjugating by $s$ changes $k$ to $-k$; parity is preserved, so the two reflection pairs cannot merge.

In a one-dimensional representation, the relation $srs=r^{-1}$ forces the image of $r$ to equal its inverse. Thus $r$ and $s$ can be assigned independent signs $\alpha ,\beta \in \{1,-1\}$. These four choices give four irreducible characters. The ordinary two-dimensional action on the plane gives one more character: a rotation through $\theta$ has trace $2\cos \theta$, and a reflection has trace zero. Therefore the character table of the symmetry group of a square isFor instance the last row has squared character norm $(4+4)/8=1$, so the planar representation is irreducible; its inner product with each one-dimensional row is zero. All five rows obey character orthogonality. Their squared dimensions sum to $1+1+1+1+4=8$, so they exhaust the irreducible representations of this order-eight group.

If $Hv=\lambda v$ for a Hermitian matrix and a nonzero eigenvector, then $v^{†}Hv=\lambda v^{†}v$. The scalar on the left equals its complex conjugate because $H^{†}=H$, while $v^{†}v>0$. Thus $\lambda$ is real. If also $Hw=\mu w$, thenConsequently distinct eigenvalues force $v^{†}w=0$, proving orthogonality of their eigenvectors.

For a unitary matrix, $U^{†}U=I$ preserves the norm induced by the standard Hermitian inner product. If $Uv=\zeta v$ with $v\ne 0$, then $\Vert v{\Vert }^2=\Vert Uv{\Vert }^2=|\zeta {|}^2\Vert v{\Vert }^2$, so $\boxed{|\zeta |=1}$.

A normal mode is a fixed displacement pattern whose components share a common harmonic time dependence in the equations for small oscillations. Its normal frequency is the associated oscillation frequency; below, $\omega$ denotes the angular frequency, so the frequency in cycles per unit time is $\omega /(2\pi )$. A normal coordinate is the amplitude of one such pattern, chosen so that the linear equations decouple. For mass matrix $M$ and stiffness matrix $K$, the patterns solve $Kv={\omega }^{2}Mv$. Zero eigenvalues describe free rigid motions rather than periodic restoring oscillations.

For a piecewise continuous function of exponential order on $t\ge 0$, its unilateral Laplace transform isThe Bromwich inversion formula, at a point of continuity, isThe Bromwich contour is a straight vertical line, traversed upwards, with $\gamma$ to the right of the abscissa of convergence and hence of all singularities of the transform. Under the usual piecewise smoothness assumptions, at a jump the symmetric inversion limit gives the mean of the two one-sided values.

In the transform in the question, put $\lambda =e^{-t}$. Then $d\lambda =-e^{-t}dt$ andThus the restricted Mellin transform on the unit interval is the Laplace transform of $f(t)=h(e^{-t})$. Apply the Bromwich inversion formula with $t=-\log \lambda$, where $0<\lambda <1$, to obtainHere $\gamma$ lies in the convergence half-plane of $H$. At a discontinuity this formula has the same mean-value interpretation. The real natural logarithm defines ${\lambda }^{-s}$ unambiguously.

Commuting inverse. Assume $I-M$ is invertible, as required by the displayed inverse. Multiplication givesMultiply on both sides by $(I-M{)}^{-1}$ to obtainMore generally, a matrix commuting with an invertible matrix also commutes with its inverse.

Orthogonality. For a real skew-symmetric matrix $A$, $x^{T}Ax=0$. If $(I-A)x=0$ for a real vector $x$, multiply by $x^T$ to obtain $\Vert x{\Vert }^2=0$. Thus $I-A$ is invertible; the same argument applies to $I+A$. For the Cayley transform of a real skew-symmetric matrix,The factors commute by the first result, soTherefore $\boxed{N\text{ is an orthogonal matrix}.}$

Eigenvector mapping. If $Av=av$, then $(I-A{)}^{-1}v=(1-a{)}^{-1}v$. ConsequentlyThe denominator cannot vanish. In fact the eigenvalues of a real skew-symmetric matrix are purely imaginary: $v^{†}Av$ is the negative of its complex conjugate, so $a=-\overline{a}$. For $a=i\beta$ the corresponding eigenvalue of $N$ is $(1+i\beta )/(1-i\beta )=e^{2i\arctan \beta }$, consistent with orthogonality.

Spectrum in three dimensions. Let $Q$ be any real three-by-three orthogonal matrix. For a complex eigenvector $v$, $\Vert Qv\Vert =\Vert v\Vert$ implies $|\lambda |=1$. A real cubic characteristic polynomial has a real root, necessarily $+1$ or $-1$. Any nonreal roots form a complex-conjugate pair of unit modulus. If all three roots are real, two of the three signs agree and can be grouped as a pair with angle zero or $\pi$. Thus in every case the eigenvalues can be writtenTheir product is $\epsilon =\det Q$.

Determinant and fixed axis for the Cayley transform. Since $A^T=-A$,so $\boxed{\det N=1}$. This determinant argument works in every dimension. In dimension three the preceding spectrum result then guarantees an eigenvector with $Nx=x$. Set $z=(I-A{)}^{-1}x$. The fixed-vector equation says $(I+A)z=(I-A)z$, so $Az=0$ and $x=(I-A)z=z$. HenceThis is the axis fixed by the three-dimensional rotation. Equivalently, a real odd-dimensional skew-symmetric matrix has zero determinant, because $\det A=\det A^T=\det (-A)=-\det A$.

Local errors of the Euler schemes. Let $h=\Delta t>0$, and evaluate $f$ and its partial derivatives at an exact starting point $(y(t),t)$. DefineThe exact Taylor expansion is $y(t+h)=y+hf+\frac{1}{2}{h}^{2}A+O(h^3)$. The forward Euler method retains only $y+hf$, so its one-step defect, exact minus computed value, isFor the Backward Euler method, substitution of the exact solution into the scheme gives the residualbecause $f(y(t+h),t+h)=f+hA+O(h^2)$. Under local smoothness, solving the implicit step changes this residual by a factor $1+O(h)$, so the one-step defect has the same leading term:These are unnormalized local truncation errors, as in the question. A convention dividing by $h$ instead gives $O(h)$. Both schemes have first-order global error under the usual smoothness and stability hypotheses.

Euler stability. For the Dahlquist test equation $y^{\prime }=-\Lambda y$, put $z=h\Lambda$, with $\mathrm{Re}z>0$. The numerical amplification factors areStrict decay requires $|R|<1$; the boundary $|R|=1$ gives bounded but generally nondecaying values. For Forward Euler method,For positive real $\Lambda$, this reduces to $0<h\Lambda <2$. The absolute stability region in the $z$ plane is the disk $|z-1|\le 1$. For Backward Euler method, $|1+z|>1$ whenever $\mathrm{Re}z>0$, so it decays for every step size. It is A-stable and also L-stable, since $R_{BE}(z)\to 0$ for large $z$ in this half-plane. Here $z$ has the opposite sign from the alternative convention $z=h\lambda$ for $y^{\prime }=\lambda y$.

Predictor-corrector accuracy. The printed two-stage construction is Heun's method, equivalentlyAlthough the second stage resembles a Backward Euler method evaluation, it uses the predicted value, so the combined method is explicit. Expand that stage around the exact starting point, puttingThenand the numerical step is $y+hf+\frac{1}{2}{h}^{2}A+\frac{1}{4}{h}^{3}B+O(h^4)$. Along the exact solution, $y^{\prime \prime \prime }=B+f_{y}A$, so comparison with the exact Taylor expansion givesThus the local defect improves by one power of $h$, giving second-order global error under the usual hypotheses. For the linear test equation the defect reduces to $-h^3{\Lambda }^{3}y/6+O(h^4)$.

Predictor-corrector stability. Applied to the test equation, the predictor is $y^{(1)}=(1-z)y_n$ and the second value is $y^{(2)}=(1-z+z^2)y_n$. Their average givesFor real positive $z$, $R_H(z)=\frac{1}{2}[(z-1{)}^2+1]$, so strict decay holds exactly for $0<z<2$, the same real-axis step bound as Forward Euler method. For complex $z=x+iy$ the precise condition isThe region is bounded and does not contain the full right half-plane: for example $z=3$ gives $R_H=5/2$. It is therefore not A-stable. Its imaginary-axis values satisfy $|R_H(iy){|}^2=1+y^4/4>1$ for $y\ne 0$. Better accuracy has not inherited the unconditional stability of an actual implicit Backward Euler method step.