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With the inner product $⟨\chi ,\psi ⟩=\int {\chi }^{\ast }\psi dx$, the assumed Hermiticity of $\hat{x}$ and $\hat{p}$ givesConsequently the adjoint relation isAs with the stated Hermiticity assumption, these identities concern wavefunctions in the appropriate operator domains. Square integrability alone does not guarantee that derivatives or multiplication by $x$ are square integrable; one may first work with smooth rapidly decaying wavefunctions, where integration by parts has no boundary term, and then use the operator domains.

Part (a) implies $[\hat{A},\hat{B}]=2m\hslash \omega$. Since $\hat{H}=\hat{B}\hat{A}/(2m)+\hslash \omega /2$, it follows thatFor an eigenfunction $\varphi$ of the Hamiltonian with eigenvalue $E$,Thus $\hat{A}$ is a ladder operator lowering the energy by $\hslash \omega$, provided $\hat{A}\varphi \ne 0$. The zero function is not an eigenfunction; in particular the ground state is annihilated by this operator. This qualification is essential for the later argument.

Retain the order of the noncommuting position operator and momentum operator. Using their commutator $[\hat{x},\hat{p}]=i\hslash$, expansion givesThus the two factorizations of the quantum harmonic oscillator are

The method of images for the Neumann boundary conditions extends $bx$ evenly through zero and then with period $2\pi$. Define the triangular waveThe formulas agree at their endpoints. It is even, $2\pi$-periodic, and agrees with $bx$ on $[0,\pi ]$. Its Fourier cosine series has the coefficients in part (a). The cosine product identity then gives the D'Alembert formulaIt is the average of two oppositely traveling reflected waves. For $b>0$, the sketch rises from zero at even multiples of $\pi$ to $b\pi$ at odd multiples; for $b<0$ it is reflected vertically, and for $b=0$ it is zero.

The Neumann boundary conditions select the spatial eigenfunctions $\cos nx$, $n=0,1,\dots$, of the separated wave equation. Indeed separation gives $X^{\prime \prime }+\lambda X=0$ with $X^{\prime }(0)=X^{\prime }(\pi )=0$. Integration by parts gives $\lambda \int X^2=\int X^{\prime 2}\ge 0$. The zero value yields the constant mode; for positive $\lambda$, the first boundary condition removes the sine term and the second requires $\sqrt{\lambda }=n\in \mathbb{N}$. For $n>0$, the temporal factor is $\cos nct$: the sine factor vanishes because the initial velocity is zero. The zero mode has constant displacement, with no term linear in time for the same reason.

The constant Fourier cosine series coefficient is the mean ${\pi }^{-1}{\int }_0^{\pi }bxdx=b\pi /2$. The remaining coefficients areOnly odd modes survive. Thus the displacement isThe series for displacement converges absolutely and uniformly. The initial velocity is zero in the finite-energy sense, and the initial displacement is $bx$. There is an endpoint compatibility qualification: when $b\ne 0$, the initial slope does not satisfy the Neumann boundary conditions at $t=0$. The answer is therefore a finite-energy weak solution, piecewise classical between the propagating corners, rather than a globally twice continuously differentiable solution up to the initial endpoints. The mode expansion implements reflection at those endpoints.

With unit constant bending coefficients and unit length, the quadratic elastic energy isTo find the first loss of positivity, put $v=y^{\prime }$. The clamped conditions give $v(0)=v(1)=0$ and ${\int }_0^{1}vdx=0$. Extend $v$ periodically with period one. The sharp periodic Wirtinger inequality gives ${\int }_0^1{v}^{\prime 2}dx\ge 4{\pi }^2{\int }_0^1{v}^{2}dx$. For completeness, expand the mean-zero periodic $v$ in its Fourier series ${\sum }_{n\ge 1}(a_n\cos 2\pi nx+b_n\sin 2\pi nx)$. Parseval identity giveswhich proves the inequality and its equality condition. The same holds for $z^{\prime }$. ConsequentlyFor $F<4{\pi }^2$, the straight filament is the unique minimum. Equality in the periodic Wirtinger inequality has $v=a\cos 2\pi x+b\sin 2\pi x$; the endpoint value $v(0)=0$ eliminates $a$. Integration and the position boundary conditions therefore give the first buckling load and modes:At this load every such pair has zero quadratic elastic energy and solves the Euler-Lagrange equations. For $F>F_c$ these modes give negative energy, signaling instability of the straight configuration. The quadratic small-slope model does not select a finite post-buckling amplitude; nonlinear terms would be needed for that. This is Euler buckling of an elastic filament.

Vary $y$ by $y+\epsilon \eta$, where the clamped endpoint conditions require $\eta ={\eta }^{\prime }=0$ at both endpoints. The first variation of the quadratic elastic energy isBoth integrations by parts have zero boundary terms. Since the variation is arbitrary in the interior, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. Varying $z$ independently gives the second equation:The clamped endpoint values accompany these equations. If $A$ or $B$ depends on $x$, it must remain inside the derivatives.

The arclength element is $ds=\sqrt{1+y^{\prime 2}+z^{\prime 2}}dx$. For uniformly small slopes, the Taylor expansion $\sqrt{1+s}=1+s/2+O(s^2)$ givesThus to quadratic order in the slopes the excess length is $\boxed{\ell -L\simeq \frac{1}{2}{\int }_0^L(y^{\prime 2}+z^{\prime 2})dx}$. This is the small-slope approximation used in the subsequent elastic energy.

Consider the filled hyperbolic triangle with vertices $A,B,C$. Every point $P$ is on some geodesic segment $Aw$ with $w\in BC$: for instance, in the Beltrami-Klein model these are the ordinary straight segments filling a convex triangle. For any other point $Q$ of the triangle, apply part (b) first on $Aw$ and then on $BC$:Applying the same argument to $Q$, with each vertex held fixed, bounds every term on the right by the maximum of the three side lengths. The endpoints of a longest side attain that bound. Thus the diameter is exactly the longest side length:

At an interior point $w$ of the geodesic segment from $z_1$ to $z_2$, the two angles made with the segment toward $w^{\prime }$ add to $\pi$. At least one is at least $\pi /2$. In the corresponding hyperbolic triangle, part (a) says that its opposite side $w^{\prime }{z}_i$ is at least the adjacent side $w^{\prime }w$. ThereforeIf $w$ is an endpoint the assertion is immediate. The same inequality also holds when $w^{\prime }$ lies on the complete geodesic through $z_1,z_2$: parameterize that geodesic by arclength and use the elementary inequality $|t-s|\le \max (|t-a|,|t-b|)$ for $a\le s\le b$. This extension will allow repeated use of the result in part (c).

In the hyperbolic law of cosines, the angle opposite $a$ being at least $\pi /2$ makes its cosine nonpositive. ThusSince $\cosh$ is increasing on $[0,\infty )$, the opposite side is longest: $\boxed{a\ge b,a\ge c}$. For a nondegenerate hyperbolic triangle both comparisons are strict.

The function $X\mapsto \det (I-X)$ is a polynomial in the matrix entries, hence continuous. Consequently $U$ is the inverse image of the open set $\mathbb{R}\setminus \{0\}$ and is an open set.

If $\Vert X\Vert <1$ and $(I-X)v=0$, then $v=Xv$ implies $\Vert v\Vert \le \Vert X\Vert \Vert v\Vert$. This forces $v=0$. The square matrix $I-X$ is therefore injective, hence an invertible matrix, soThe argument works with any of the induced operator norms in part (b).

For a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$, define the operator norm byIt is finite. To see this without assuming continuity of $A$, write $x=\sum x_j{e}_j$: the triangle inequality bounds $\Vert Ax\Vert$ by $\sum |x_j|\Vert A{e}_j\Vert$. Every norm in finite dimensions is equivalent to the Euclidean norm. One proof first uses this same coordinate bound to establish continuity of the norm on the Euclidean unit sphere; compactness then gives a strictly positive minimum there. Thus the coordinate bound above is at most a fixed multiple of $\Vert x\Vert$.

Absolute homogeneity and the triangle inequality for this operator norm follow from those for $\Vert Ax\Vert$ and taking suprema. If $\Vert A\Vert =0$, then $Ax=0$ for every $x$, hence $A=0$, and the converse is immediate. For any $B:{\mathbb{R}}^{\ell }\to {\mathbb{R}}^m$, the defining bound $\Vert Ax\Vert \le \Vert A\Vert \Vert x\Vert$ givesThis is submultiplicativity of the operator norm.

A norm on a real or complex vector space $V$ is a function $\Vert \cdot \Vert :V\to [0,\infty )$ with $\Vert v\Vert =0$ exactly when $v=0$, $\Vert \lambda v\Vert =|\lambda |\Vert v\Vert$, and the triangle inequality $\Vert v+w\Vert \le \Vert v\Vert +\Vert w\Vert$. These are positivity, absolute homogeneity and the triangle inequality.

For the general linear group over ${\mathbb{F}}_{31}$, choose a nonzero first column in $31^2-1$ ways and a second column outside its span in $31^2-31$ ways. ThusSince $5^3=125\equiv 1(\mathrm{mod}31)$ and $5\ne 1$, the residue $5$ has order three. The nine diagonal matrices $\mathrm{diag}(5^i,5^j)$, with $0\le i,j<3$, form a subgroup isomorphic to $C_3\times C_3$. It is an abelian group and a Sylow subgroup for $p=3$. Every other Sylow subgroup for $p=3$ is conjugate to it by the Sylow theorems, hence is also Abelian.

Since $351=3^3\cdot 13$, the number of Sylow subgroups for $p=13$ divides $27$ and is $1$ modulo $13$, hence is $1$ or $27$. In the first case the unique subgroup is a nontrivial proper normal subgroup.

In the second case the $27$ subgroups of order $13$ have pairwise trivial intersection: two such subgroups sharing a nonidentity element would both be generated by it. They account for $27\cdot 12=324$ nonidentity elements of order $13$. Exactly $27$ elements, including the identity, remain. Every Sylow subgroup for $p=3$ has order $27$, and all its elements lie among these remaining elements, by Lagrange's theorem. It must therefore be precisely that set. The Sylow subgroup for $p=3$ is unique and hence normal. In either case $G$ is not a simple group.

Write $|G|=p^{r}m$ with $p$ prime and $p\nmid m$. The Sylow theorems assert that a subgroup of order $p^r$ exists; every p-group subgroup is contained in a conjugate of a Sylow subgroup; all Sylow subgroups are conjugate; and their number $n_p$ satisfies $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid m$.

For the containment and conjugacy assertions, fix a Sylow subgroup $P$ and let any p-group subgroup $Q$ act by left multiplication on the $m$ left cosets $G/P$. By the orbit-stabilizer theorem, every orbit has size a power of $p$. Since $p\nmid m$, at least one orbit has size one. If $gP$ is such a fixed point, then $qgP=gP$ for all $q\in Q$, equivalently $g^{-1}Qg\subseteq P$. Thus $Q\subseteq gP{g}^{-1}$. If $Q$ too has order $p^r$, this inclusion is equality, proving conjugacy. This is Sylow containment from a coset fixed point.

If $A$ is a Hermitian matrix, then $(BA{B}^{\ast }{)}^{\ast }=BA{B}^{\ast }$, so $M_B$ preserves $W$. Addition and multiplication by real scalars commute with this operation, making $T_B$ a real linear map.

A real basis of $W$ consists of $E_{ii}$ and, for $i<j$, $E_{ij}+E_{ji}$ and $i(E_{ij}-E_{ji})$. It has $n^2$ elements and is also a complex basis of all matrices. Indeed every matrix is uniquely $H+iK$ with $H,K$ Hermitian matrices, by takingReal linear independence of the displayed basis therefore implies complex linear independence. Since its images under $M_B$ are Hermitian matrices, the matrix of $M_B$ in this complex basis has exactly the same real entries as the matrix of $T_B$ in the real basis. Their determinants are equal. There is no further squaring when passing to this real subspace:This is the determinant of Hermitian congruence, including the case of singular $B$.

Right multiplication by $B^{\ast }$ acts separately on the rows. After transposing each row its matrix is $(B^{\ast }{)}^T=\overline{B}$, so its determinant is $(\overline{\det B}{)}^n$. Composing this linear map with left multiplication and using part (a) givesHere the determinant is taken on the $n^2$-dimensional complex vector space of matrices.

Group the matrix entries by columns. Each column of $BA$ is $B$ times the corresponding column of $A$. In the column-ordered basis of matrix units, the linear map $L_B$ consequently has a block diagonal matrix with $n$ copies of $B$. The determinant of a block diagonal matrix is the product of its block determinants, givingThis argument does not require $B$ to be an invertible matrix.