A cyclic group of order $p$ contains an element of order two if and only if $2$ divides $p$. Necessity follows from Lagrange's theorem; for even $p$, the unique such element is $c=g_1^{p/2}$. HenceBecause $g_1$ has order $p>2$ and the representation is faithful, $D(g_1)$ cannot be a reflection, which has order two. It is therefore a rotation through a primitive angle $\vartheta =2\pi q/p$, with $\gcd (q,p)=1$. The condition on the greatest common divisor is exactly what makes its order $p$ rather than a proper divisor. Since $p$ is even, $q$ is odd, so $D(c)$ is the half-turn:The character is the trace:Choosing the primitive generator with $q=1$ gives the customary rotation representation. The general coprime value is included because faithfulness alone does not select that generator. These are the two-dimensional cases in the irreducible real representations of a finite cyclic group.

Since $a^2=I$, a representation satisfies $D(a{)}^2=I_2$. Orthogonality gives $D(a{)}^{-1}=D(a{)}^T$, while the involution relation gives $D(a{)}^{-1}=D(a)$. Hence $D(a)$ is a symmetric matrix, so $\gamma =\delta$.

Its entries obeyIf $\gamma \ne 0$, then $\beta =-\alpha$; putting $\alpha =\cos \theta$ and $\gamma =\sin \theta$ gives a reflection. If $\gamma =0$, the diagonal entries are independently $\pm 1$: the opposite-sign cases are already included among these reflections, while equal signs give $I_2$ or $-I_2$. Faithfulness excludes $I_2$, since $a\ne I$. Therefore the orthogonal involutions of the plane giveThe first is rotation by $\pi$. The second is reflection in the line making angle $\theta /2$ with the first axis: the unit vector along that line is fixed, and its perpendicular is negated. These are all the nonidentity orthogonal transformations of order two in the plane.

For fixed $k$ and $\ell$, the line $-z/\kappa$ becomes increasingly negative, so each intersection approaches its branch’s left-hand vertical asymptote. Write $z_n=a_n+{\epsilon }_n$, where $a_n=(n-1/2)\pi$ and ${\epsilon }_n>0$. The eigenvalue equation givesSince $\cot \epsilon ={\epsilon }^{-1}-\epsilon /3+O({\epsilon }^3)$, it follows that ${\epsilon }_n=\kappa /a_n+O(a_n^{-3})$. ThusThe half-integer phase identifies the sharper high-frequency behaviour of these Dirichlet-Robin eigenvalues on a finite interval. The expansion holds with the Robin coefficient fixed; it is not a uniform expansion in $k\to \infty$.

By part (c), write $\lambda ={\mu }^2$ with $\mu >0$. The equation is $y^{\prime \prime }+{\mu }^{2}y=0$, and the left Dirichlet boundary condition eliminates the cosine term. Thus $y=A\sin (\mu x)$, with $A\ne 0$. The right Robin boundary condition becomesPut $z=\mu \ell$ and $\kappa =k\ell >0$. A root cannot have $\cos z=0$, so this is equivalent to $\tan z=-z/\kappa$.

On each interval $((n-1/2)\pi ,n\pi )$, $n\ge 1$, the function $\tan z+z/\kappa$ rises from $-\infty$ to a positive value and has derivative ${\sec }^{2}z+1/\kappa >0$. The intermediate value theorem and strict monotonicity therefore give exactly one root $z_n$ there. On the intervening positive-tangent intervals both terms are positive, so there are no further positive roots. ConsequentlyThese intervals establish ${\lambda }_1<{\lambda }_2<\cdots$ and an infinite sequence of eigenvalues. The apparent algebraic root $z=0$ of the sine equation is excluded: it gives a zero sine function, and the direct zero-eigenvalue solution $y=Ax$ would require $A(1+k\ell )=0$.

For vector normalization to unit norm, the Dirichlet-Robin eigenvalues on a finite interval giveThe eigenfunctions can otherwise be scaled by any nonzero real constant.

The left boundary condition is $y(0)=0$, while the right one is $y^{\prime }(\ell )=-ky(\ell )$. Hence $[y{y}^{\prime }{]}_0^{\ell }=-ky(\ell {)}^2$. The Rayleigh quotient becomesEvery term on the right is nonnegative. If the whole expression were zero, in particular ${\int }_0^{\ell }(y^{\prime }{)}^{2}dx=0$, so $y$ would be constant. Its zero value at the left endpoint would force $y\equiv 0$, contrary to its being an eigenfunction. ThereforeThis is the positivity mechanism for Dirichlet-Robin eigenvalues on a finite interval. The sign of the boundary contribution is positive for the convention $y^{\prime }+ky=0$ with $k>0$.

Multiply the eigenvalue equation of the Sturm-Liouville problem by $y$ and integrate. Integration by parts givesSince $y$ is a nonzero eigenfunction and $\omega >0$, the denominator is strictly positive. Thus the Rayleigh quotient isThe endpoint term must be retained for a general Robin boundary condition; it vanishes for homogeneous Dirichlet boundary conditions but need not vanish in the present general problem.

Use the ordinary real inner product $⟨u,v⟩={\int }_a^{b}uvdx$ on the domain of functions obeying the two separated boundary conditions. Integration by parts gives the Lagrange identity for a Sturm-Liouville operatorAt $a$, the two boundary-value vectors $(u(a),u^{\prime }(a))$ and $(v(a),v^{\prime }(a))$ satisfy the same homogeneous linear relation. If $A_2=0$, both function values vanish; if $A_2\ne 0$, their derivatives are the same constant multiple $-A_1/A_2$ of their values. Either way $v{u}^{\prime }-u{v}^{\prime }=0$. The same argument applies at $b$. HenceThis proves symmetry, and the regular separated Sturm-Liouville problem has the same adjoint domain, so $L$ is self-adjoint. To see the domain point, require the boundary form to vanish against every admissible $u$: the independently variable endpoint data force an adjoint-domain function to obey those same endpoint relations.

As usual, each endpoint pair $(A_1,A_2)$ and $(B_1,B_2)$ must be nonzero, and the coefficients must have the regularity needed for these operations. An identically zero pair would impose no condition at that endpoint and would not give the claimed self-adjoint operator. The related operator ${\omega }^{-1}L$ is self-adjoint in the weighted inner product ${\int }_a^b\omega uvdx$.

Assume $c(y)>0$ is differentiable along the ray, with sufficient regularity for the following variation, and fix the endpoints. Fermat principle makes the travel time stationary:For the variational integrand $f=\sqrt{1+y^{\prime 2}}/c$, the Euler-Lagrange equation isExpanding the derivative and multiplying by $c^2(1+y^{\prime 2}{)}^{3/2}$ givesand thereforeEquivalently, the Beltrami identity gives $1/[c(y)\sqrt{1+y^{\prime 2}}]=\text{constant}$. Mere continuity of $c$, as stated in the wording, does not by itself justify the appearance of $c^{\prime }$; the displayed classical ray equation requires this additional differentiability.

The origin is an ordinary point. Substitute $y={\sum }_{n\ge 0}{a}_n{x}^n$ into the equation and collect the coefficient of $x^n$:Hence the recurrence relation isThere are two free initial coefficients. The independent even and odd power series beginThe general solution is $\boxed{y=a_0{y}_{\mathrm{e}}+a_1{y}_{\mathrm{o}}}$. As a useful independent check, the substitution $x=\sin t$ reduces the equation to $d^{2}y/d{t}^2+k^{2}y=0$. ThusFor a nonterminating parity series, $|a_{n+2}/a_n|\to 1$, so the ratio test in $x^2$ gives $\boxed{R=1}$. The regular singular points at $\pm 1$ explain the natural obstruction.

There is an important exception to an unconditional radius-one answer. If $k^2=m^2$ for an integer $m\ge 0$, the series of parity $m$ terminates at degree $m$ and is a multiple of a Chebyshev polynomial. Its radius of convergence is $\boxed{R=\infty }$; the other independent parity solution still has radius one. For $m=0$ the even solution is constant. Any nonzero combination containing the nonterminating solution has radius one. These are the full power series of the Chebyshev differential equation, including the terminating cases.

The Wronskian isSuppose constants $A,B$ satisfy $Af+Bg=0$ throughout the interval. Differentiating gives $A{f}^{\prime }+B{g}^{\prime }=0$. At $x_0$, these form the homogeneous linear systemIts determinant is nonzero by hypothesis, so $A=B=0$. This is exactly linear independence of the two functions on the interval. No differential equation is needed for this implication; the converse for arbitrary differentiable functions would require additional hypotheses.

For the normalized second-order equation, $x_0$ is an ordinary point if both coefficient functions $p$ and $q$ are holomorphic near $x_0$. If it is singular but $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ are holomorphic there, it is a regular singular point. Thus $p$ may have at most a simple pole and $q$ at most a double pole.

The distinction tells us which local series method is justified. At an ordinary point, two independent solutions are represented by ordinary Taylor series. At a regular singular point, the Frobenius method uses $(x-x_0{)}^r{\sum }_{n\ge 0}{a}_n(x-x_0{)}^n$; the indicial equation determines the exponents, and logarithmic terms may appear in a second solution. Singularities worse than these are irregular and need not have this Frobenius form.

A complex function is holomorphic at a point if it is complex differentiable throughout some open neighbourhood of that point. Writing $\zeta =\xi +i\eta$ and $\psi =P+iQ$, a convenient sufficient and necessary criterion under continuously differentiable real components is the Cauchy-Riemann equationsThey must hold throughout an open neighbourhood, rather than at an isolated point. Merely having partial derivatives that satisfy these equalities at one point is not enough without an appropriate differentiability hypothesis. For the composition $\psi =G\circ f$, holomorphy of $f$ and $G$ on the relevant open sets suffices; the additional condition $f^{\prime }\ne 0$ is needed when using a local inverse conformal map, not for holomorphy of the composition itself.

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.

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 $\mathbb{R}$.

A square matrix $A$ is diagonalizable over a specified scalar field if an invertible matrix $P$ exists such thatEquivalently, the underlying vector space has a basis of eigenvectors of $A$: those basis vectors are the columns of $P$, and their eigenvalues are the diagonal entries of $D$. Here the field is $\mathbb{C}$ unless specified otherwise.