A real two-dimensional group representation is a group homomorphism $D:G\to GL(2,\mathbb{R})$, so $D(AB)=D(A)D(B)$ and $D(I)=I$.

Substitute $\tilde{x}=\alpha x+\beta y$ and $\tilde{y}=\gamma x+\delta y$ into the quadratic expression. Its coefficients of $x^2$ and $y^2$ are ${\alpha }^2\hat{a}+{\gamma }^2\hat{b}$ and ${\beta }^2\hat{a}+{\delta }^2\hat{b}$. The cross term is zero for every matrix in this particular cyclic group, because each has one nonzero entry per row. ThusLet $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ generate the group. For $A=\pm I$, $M=I$; for $A=\pm J$, $M=S:=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Since $S^2=I$, each $M$ is its own inverse and $D(J^k)=S^k$. The group law is respected: $S^{k+l}=S^k{S}^l$, and $S^4=I$ makes it independent of the representative modulo four. This proves that $D=M^{-1}$ is a group representation.

This quarter-turn action on diagonal quadratic forms hasso it is not faithful. Quadratic forms cannot distinguish a simultaneous sign reversal of both coordinates. The invariant coefficient directions $(1,1)$ and $(1,-1)$ correspond to the fixed form $x^2+y^2$ and the form $x^2-y^2$ whose sign changes under a quarter-turn.

Put $\Phi (r,\theta )=R(r)Y(\theta )$ in Laplace's equation, multiply by $r^2/(RY)$, and separate the two variables:With $s=\cos \theta$, the angular equation becomes $(1-s^2)Y_{ss}-2s{Y}_s+\lambda Y=0$. Requiring a smooth axisymmetric harmonic function at both poles selects $\lambda =n(n+1)$ and the Legendre polynomial $Y=P_n(s)$, for nonnegative integers $n$. The radial Cauchy-Euler equation has solutions $r^n$ and $r^{-n-1}$. Superposition over the angular eigenfunctions therefore givesThis expansion assumes regularity on each whole angular sphere; without it, singular angular solutions would also be possible. The boundary data in the shell involve only the $n=0,1$ modes. Define the spherical-shell harmonic interpolation weightsEach $p_n$ equals one at the inner boundary and zero at the outer one, and $q_n$ does the reverse. Hence the steady temperature isEach term is a harmonic function, and substitution verifies both prescribed boundary values. Uniqueness follows from the maximum principle for harmonic functions.

Take a variation $y_{\epsilon }=y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. Differentiating the functional at zero and using integration by parts givesThe boundary term vanishes. Since the remaining integral must vanish for every smooth endpoint-zero variation, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation $d(f_{y^{\prime }})/dx=f_y$.

For the mechanical problem put $A(r)=1-2m/r>0$. The Lagrangian is independent of $\theta$, a cyclic coordinate, so its conjugate momentum $\partial L/\partial \dot{\theta }=2r^2\dot{\theta }=2J$ is conserved. Therefore $J$ is constant, including the factor of two between $J$ and the canonical momentum for this normalization of $L$.

For the radial coordinate, $\partial L/\partial \dot{r}=2\dot{r}/A$ and $\partial L/\partial r=-{\dot{r}}^2{A}^{\prime }/A^2+2r{\dot{\theta }}^2-\mu A^{\prime }/A^2$. Its Euler-Lagrange equation isFor a circular orbit $r=R>2m$, the radial velocity and acceleration vanish. Using $A^{\prime }(R)=2m/R^2$ and $\dot{\theta }=J/R^2$ leaves $2J^2/R^3=2m\mu /[R^{2}A(R{)}^2]$, henceThis is the circular orbit condition for a radial kinetic Lagrangian applied to the coefficients here.

Choose the left homogeneous solution $v(x)=\sinh x$ and the right one $w(x)=\sinh (1-x)$. Their Wronskian is $v{w}^{\prime }-v^{\prime }w=-\sinh 1$. The Dirichlet modified Helmholtz Green function on an interval, with operator sign $D_x^2-1$, is thereforeIt vanishes at both endpoints, is continuous at $x=\xi$, and its first derivative jumps by $+1$, giving $(D_x^2-1)G=\delta (x-\xi )$. The Green-function representation then givesThe two integrals are $(e^{2x}-1)/4-x/2$ and $e(1-x)/2-(e-e^{2x-1})/4$, respectively. Substituting them and simplifying givesDirect differentiation verifies $u^{\prime \prime }-u=e^x$ and both zero boundary values. The negative kernel also explains why the solution is negative inside the interval despite the positive right-hand side.

Use the complex inner product $⟨a,b⟩={\sum }_j\overline{a_j}{b}_j=a^{†}b$, conjugate-linear in its first argument. For $a\ne 0$, put $c=⟨a,b⟩/⟨a,a⟩$. Positivity givesHence the Cauchy-Schwarz inequality is $|⟨a,b⟩|\le \Vert a\Vert \Vert b\Vert$. If $a=0$ it is immediate; equality holds precisely when the two vectors are linearly dependent.

For a unitary matrix, $U^{†}U=I$, so $\Vert Ua{\Vert }^2=a^{†}{U}^{†}Ua=\Vert a{\Vert }^2$. If $Uv=\lambda v$ with a nonzero eigenvector, this implies $|\lambda |\Vert v\Vert =\Vert v\Vert$. Thus every eigenvalue satisfiesFor the one-parameter family, expand unitarity to first order: $(I+t{A}^{†}+O(t^2))(I+tA+O(t^2))=I$, so $A^{†}+A=0$. Thus $A$ is a skew-Hermitian matrix, an element of the unitary Lie algebra. If $Av=\mu v$, then $v^{†}Av=\mu \Vert v{\Vert }^2$, whereas complex conjugation gives $(v^{†}Av{)}^{\ast }=v^{†}{A}^{†}v=-v^{†}Av$. Consequently $\overline{\mu }=-\mu$, proving that every eigenvalue of $A$ is purely imaginary, with zero permitted.

The matrix trace is $\mathrm{Tr}A={\sum }_i{A}_{ii}$. Expanding the matrix products gives $\mathrm{Tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}={\sum }_{j,i}{B}_{ji}{A}_{ij}=\mathrm{Tr}(BA)$. Thus the cyclic property of the trace gives zero trace to every commutator. If $AB-BA=I$, taking the trace would give $0=n$, impossible for positive finite $n$ over the real or complex field.

For real $A,B$, let $p(\lambda )=\det (A+\lambda B)$, a real polynomial of degree at most $n$. Since $p(i)=\det C\ne 0$, it is not the zero polynomial. It has at most finitely many real roots of a polynomial, so $A+\lambda B$ is an invertible matrix for some real $\lambda$, in fact for all but finitely many.

Finally write the invertible intertwiner as $R=A+iB$. The relation $PR=RQ$ and the reality of $P,Q$ give $PA=AQ$ and $PB=BQ$. Choose a real $\lambda$ for which $S=A+\lambda B$ is invertible. Then $PS=SQ$, so $P=SQ{S}^{-1}$. This proves that complex similarity of real matrices implies real similarity.

For a nonzero real function, write $\Lambda =F/G$ with $G={\int }_a^{b}w{y}^{2}dx>0$. Multiplying the Sturm-Liouville equation by $y$ and integrating givesThus the stated vanishing boundary product proves $\lambda =\Lambda [y]$.

For the variational statement, the first variation isWith admissible boundary conditions that also make $[p{y}^{\prime }\eta {]}_a^b=0$ for every allowed variation, the fundamental lemma of the calculus of variations identifies stationary functions with the eigenfunctions and stationary values with their eigenvalues. Zero Dirichlet boundary conditions, zero Neumann boundary conditions, and matched periodic data give familiar examples. The single printed condition $[py{y}^{\prime }{]}_a^b=0$ proves the quotient identity but, by itself, does not specify the admissible variations needed for stationarity; the boundary conditions for variational Rayleigh quotients make this distinction explicit.

For the half-line equation, multiplication by $x^{n-1}$ gives $p=w=x^{n-1}$ and $q=x^{n+1}$. Use the regular radial boundary condition at zero and square-integrable decay at infinity, under which the proposed Gaussian is admissible. With $y=e^{a{x}^2}$, $a<0$, $y^{\prime }=2axy$, soThe ratio of the two Gaussian integrals is $n/(-4a)$, from a change of variables and the Gamma function recurrence. Differentiating gives ${\Lambda }^{\prime }(a)=-n[1-1/(4a^2)]$, so the minimum on $a<0$ occurs at $a=-1/2$. The Gaussian variational bound for a radial harmonic oscillator therefore givesHere the trial actually satisfies the differential equation with $\lambda =n$. Moreover, on the same regular decaying domain, integration by parts gives $F[y]-nG[y]={\int }_0^{\infty }{x}^{n-1}(y^{\prime }+xy{)}^{2}dx\ge 0$. Hence the estimate is exact, with $\boxed{{\lambda }_0=n}$. For $n=1$, regular radial data mean the Neumann boundary condition $y^{\prime }(0)=0$; imposing a zero Dirichlet boundary condition there would exclude this trial and give a different problem.

A normal subgroup $H$ satisfies $gH{g}^{-1}=H$ for every $g\in G$. A group homomorphism preserves multiplication, $\Phi (ab)=\Phi (a)\Phi (b)$. It follows that $\Phi (e_1)=e_2$ and $\Phi (g^{-1})=\Phi (g{)}^{-1}$.

The kernel of a group homomorphism $K=\{g:\Phi (g)=e_2\}$ contains the identity and satisfies $a{b}^{-1}\in K$ for $a,b\in K$, so is a subgroup. For $k\in K$, $\Phi (gk{g}^{-1})=\Phi (g)e_2\Phi (g{)}^{-1}=e_2$. Thus $gK{g}^{-1}\subseteq K$, and applying the same inclusion to $g^{-1}$ gives equality. Hence the kernel is a normal subgroup.

The product $f={\prod }_{i<j}(x_i-x_j)$ is a Vandermonde determinant up to a fixed overall sign. A transposition exchanging $x_p,x_q$ reverses their mutual factor. Every other factor involving one of these variables is paired with the corresponding factor involving the other; those paired products are unchanged. ThereforeIf a permutation has decompositions into $r$ and $s$ transpositions, they act as $(-1{)}^{r}f$ and $(-1{)}^{s}f$. This polynomial is nonzero, for example at pairwise distinct real inputs, so the signs must agree. Thus all decompositions have the same parity, and an even decomposition forces every decomposition to be even. This defines the sign of a permutation; its kernel is the alternating group, linking the parity argument to the normal-kernel result above.