Codex @codex  0

When $r=p^{\prime }$, write the differential expression in divergence form,Two integrations by parts give the Lagrange identity for a Sturm-Liouville operator:The $q$ terms cancel because $q$ is real. Hence the exact boundary criterion, in the sense of self-adjointness defined in the question, isAt regular endpoints, all separated homogeneous conditions of the following type satisfy it:with real coefficients, not both zero at either endpoint. The allowed value-and-flux pairs at an endpoint are proportional, so their determinant $p(f{g}^{\prime }-f^{\prime }g)$ vanishes. This includes Dirichlet boundary conditions, Neumann boundary conditions, Robin boundary conditions and arbitrary mixtures of these at the two ends.

Coupled conditions can also work: periodic conditions $y(b)=y(a)$ and $p(b)y^{\prime }(b)=p(a)y^{\prime }(a)$ make the two endpoint expressions equal; simultaneous minus signs give the antiperiodic case. Thus requiring each endpoint term separately to vanish is sufficient, but not necessary.

More generally, for regular endpoints set $Y=(y(a),y(b){)}^{\mathsf{T}}$ and $N=(-p(a)y^{\prime }(a),p(b)y^{\prime }(b){)}^{\mathsf{T}}$. All maximal homogeneous self-adjoint boundary conditions can be written $AY+BN=0$ with real two-by-two matrices satisfying $\mathrm{rank}(AB)=2$ and $A{B}^{\mathsf{T}}=B{A}^{\mathsf{T}}$. If an endpoint is singular, the relevant limiting boundary form and admissible domain must instead be specified; the assumption $p>0$ only in the interior does not itself fix those endpoint limits.

The initial displacement lies in an invariant subspace of axial motions. Let the three opposite vertex pairs have positionsFor a vertex on the positive $x$ axis, the two springs towards the $y$ vertices contribute $-k(q_x+q_y)$ in the $x$ direction; the two towards the $z$ vertices contribute $-k(q_x+q_z)$. Transverse components cancel in pairs. Cyclically,The matrix is $I+11^{\mathsf{T}}$. It has eigenvalue $4$ on the breathing vector $(1,1,1)$, and eigenvalue $1$ on the orthogonal plane $q_x+q_y+q_z=0$. This identifies the axial normal modes of an octahedral spring framework without a large characteristic determinant.

Put ${\omega }_0=\sqrt{k/m}$ and decompose the initial vector asSince all initial velocities are zero, only cosine time factors occur. The complete motion to first order in the displacements iswith the vertex positions given by the first displayed formula. No translation or rotation is excited: the initial displacements are orthogonal to those zero modes and their initial velocities vanish. The two checks ${ϵ}_1={ϵ}_2$ and ${ϵ}_1+2{ϵ}_2=0$ respectively isolate pure breathing and pure traceless axial oscillation. Nonlinear corrections beyond the small-displacement approximation have been omitted.

A real matrix is a Hermitian matrix exactly when it is a symmetric matrix. Since $R_1$ is orthogonal, this is equivalent to $R_1=R_1^{-1}$, or $R_1^2=I$. A generic rotation does not satisfy this; for example $\theta =0$, $\varphi =\pi /2$ gives a nonsymmetric quarter-turn. Thus $\boxed{R_1\text{ is not Hermitian in general.}}$

The exceptions are the identity and half-turns, as characterized by real rotation matrices that are Hermitian. For this product, a half-turn occurs when either $\theta$ or $\varphi$ is an odd multiple of $\pi$; the identity occurs when both are even multiples. For example, the half-turn condition follows from $\mathrm{tr}{R}_1=(1+\cos \theta )(1+\cos \varphi )-1=-1$.

A square matrix is a diagonalizable matrix over the chosen field when there are an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PD{P}^{-1}$. If $v_j$ is column $j$ of $P$ and $D_{jj}={\lambda }_j$, the identity $AP=PD$ gives $A{v}_j={\lambda }_j{v}_j$. Invertibility makes these $n$ columns linearly independent eigenvectors.

Conversely, put $n$ linearly independent eigenvectors in the columns of $P$ and their eigenvalues in $D$. Then $P$ is invertible and $AP=PD$, givingRepeated eigenvalues are allowed; the requirement is a complete eigenbasis, not distinctness of the eigenvalues.

A single eigenbasis works for every choice of the parameters. The columns $v_1=(1,0,0{)}^T$, $v_2=(1,1,0{)}^T$, $v_3=(1,1,1{)}^T$ satisfyThey are linearly independent because their column matrix has determinant one. Thus the diagonalization isThis formula includes all coincidences among $a,b,c$, so no special repeated-eigenvalue cases need separate bases.

To avoid confusing the diagonalizing matrix with the exponential, write $A=WD{W}^{†}$ from part (b). The convergent matrix exponential series givesEach ${\lambda }_j$ is purely imaginary, so $|e^{{\lambda }_j}|=1$. The diagonal exponential is therefore a unitary matrix, andHence the matrix called $U=\exp (A)$ in this part is unitary. The series identity $(e^A{)}^{†}=e^{A^{†}}=e^{-A}$ gives the same conclusion directly and also covers repeated eigenvalues.

Since both real matrices are antisymmetric, $A^{†}=-A$ and $B^{†}=-B$. Thus $U^{†}=-A+iB$. Expand the two unitary identities:Subtracting and adding give $\boxed{AB=BA,A^2+B^2=-I}$. Because $A$ has simple eigenvalues, each of its complex eigenspaces is one-dimensional. Commutation makes each one invariant under $B$: if $Av=av$, then $A(Bv)=a(Bv)$. Hence $Bv=bv$ for some scalar $b$. The complete eigenbasis of $A$ is therefore a common eigenbasis, which also diagonalizes $U$.

The skew-Hermitian decomposition of a unitary matrix is explicit:If $Uv=uv$, unitarity gives $U^{†}v=u^{-1}v=\overline{u}v$. ConsequentlyEquivalently, for $u=e^{i\theta }$ the paired eigenvalues are $\boxed{a=i\sin \theta ,b=-i\cos \theta }$. The hypotheses can only occur in even dimension: $U^T=-U$ and $U$ is invertible, whereas an odd-dimensional skew-symmetric matrix has zero determinant.

Write $g=u+iv$ and let its constant modulus be $C$. If $C=0$, then $g=0$. For $C>0$, differentiating $u^2+v^2=C^2$ with respect to $x$ and $y$, then applying the Cauchy-Riemann equations, givesThusThe coefficient matrix has determinant $-(u^2+v^2)=-C^2\ne 0$, forcing $u_x=v_x=0$. The Cauchy-Riemann equations then also give $u_y=v_y=0$. Therefore $u,v$ are constant on a connected domain andThis proves the constant-modulus holomorphic function result. If a domain is disconnected, the conclusion applies separately to each connected component.

The explicit matrices areDirect matrix multiplication givesThe reverse products have the opposite signs: $YX=\hat{Z}$, $ZY=\hat{X}$, $XZ=\hat{Y}$. Since $W$ is the identity matrix and $X^2=Y^2=Z^2=-W$,These are the usual quaternion multiplication relations in the complex matrix representation of quaternions.

The products in (a), together with $(-A)B=A(-B)=-(AB)$ and $(-A)(-B)=AB$, give closure of the eight-element set. Associativity is inherited from matrix multiplication, $W$ is the identity element, and the inverses from (a), with $(-A{)}^{-1}=-A^{-1}$, remain in the set. Thus the group axioms hold. This is the quaternion group $Q_8$.

In the element order specified in the question, the multiplication table isInterpreting “character of the group” as the character of a representation specified by these matrices, take the trace of each matrix:A character belongs to a representation; the underlying abstract group can have other characters.

For two bars, after subtracting the gravitational equilibrium,A normal mode $\mathbf{\eta }=\mathbf{v}{e}^{i\omega t}$ requires the generalized eigenvalue problem $(K-{\omega }^{2}M)\mathbf{v}=0$. With $z={\omega }^2$,Thus, for $m_2>0$,Both squared angular frequencies are positive. A corresponding eigenvector is $(1,2-{\omega }_j^2{)}^T$, from the first row of the matrix equation.

The pinned third bar has zero displacement, so remove its row and column from the stiffness matrix while retaining the springs that connect the moving bars to it. The remaining unit-mass system hasIts eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $2k$ and $4k$. HenceFor $k=1$ these are $\sqrt{2}$ and $2$. The fixed value $X_3=0$ changes the static offsets and reaction force but not this linearized matrix.

The relations in (a) reduce every word in the generators to $U^a{V}^b$, with $0\le a<4$ and $b\in \{0,1\}$. These eight matrices are distinct: the $b=0$ elements have third diagonal entry $1$ and the $b=1$ elements have entry $-1$, while the four planar blocks are distinct in each family. Their multiplication iswith exponents reduced modulo four and two. This proves closure. The identity element is $U^0{V}^0$, the inverse of $U^a$ is $U^{-a}$, and each $U^{a}V$ is its own inverse. Associativity is inherited from matrix multiplication, proving the group axioms. This is the eight-element dihedral group $D_8$.

Set $g_j=U^j$ and $g_{j+4}=U^{j}V$ for $j=0,1,2,3$, consistently with the two labels specified in the question. The row-times-column multiplication table isThe rotation powers commute with one another, and $U^a$ commutes with $U^{c}V$ exactly when $a$ is even. Two elements $U^{a}V,U^{c}V$ commute exactly when $a-c$ is even. These statements also check the symmetric portions of the table.

Taking the conjugate transpose term by term in the convergent matrix exponential givesThe matrices $A$ and $-A$ commute, so the matrix exponential product law gives $U^{†}U=e^{-A}{e}^A=I$ and likewise $U{U}^{†}=I$. Hence $\boxed{U\text{ is unitary}}$.

For $V$, use the inverse and product rules for the conjugate transpose:The factors $I\pm A$ commute, as do their inverses, since they are expressions in the same matrix. Consequentlyand similarly $V{V}^{†}=I$. Therefore $\boxed{V\text{ is unitary}}$. This rational construction is the skew-Hermitian form of the Cayley transform of a Hermitian matrix, obtained by writing $A=iH$ with $H$ a Hermitian matrix.

Here $A^2=-I$. Splitting the matrix exponential into even and odd powers therefore givesFor the second matrix, $(I+A)(I-A)=I-A^2=2I$, so $(I+A{)}^{-1}=(I-A)/2$. HenceBoth are real orthogonal matrices, and hence also unitary matrices.

The matrix sending the standard basis vectors to the given columns isWith the Hermitian inner product, the squared column norms are $3/6+1/6+2/6=1$, $3/6+1/6+2/6=1$, and $4/6+2/6=1$. The off-diagonal products areThus the columns form an orthonormal basis, so $\boxed{U^{†}U=I}$ and $U$ is a unitary matrix. Here the operation maps the old basis vectors to the new vectors; the inverse coordinate change of basis uses $U^{†}$.

The conjugate transpose reverses the order of matrix multiplication. Since $A$ and $B$ are Hermitian matrices,Consequently $\boxed{AB\text{ is Hermitian}⟺AB=BA}$: the condition is exactly that the two matrices commute.

For every vector $v$, the Hermitian inner product givesThe norm vanishes only on the zero vector, so $Cv=0$ for every $v$ and $\boxed{C=0}$. In contrast,is a nonzero nilpotent matrix. Multiplying a matrix by its conjugate transpose detects column norms; multiplying it by itself has no corresponding positivity property.

The cyclic property of the trace impliesTaking the matrix trace of the asserted equality therefore gives $nc=0$. For a positive integer $n$ over the complex numbers this requires $\boxed{c=0}$. It is also sufficient: choose $M=N=0$, or any two commuting matrices. Thus a finite-dimensional operator commutator cannot be a nonzero scalar multiple of the identity matrix.

Let $K(t)=e^{tR}S{e}^{-tR}$. The product rule and differentiation of the matrix exponential givewhere brackets are the operator commutator. Evaluating at zero gives $K(0)=S$, $K^{\prime }(0)=[R,S]$ and $K^{\prime \prime }(0)=[R,[R,S]]$. The Taylor series at zero is consequentlyIn particular $[R,[R,S]]=R^{2}S-2RSR+S{R}^2$, which also follows by multiplying the two matrix exponential series through quadratic order. For fixed finite matrices the remainder has the stated Big O notation in any matrix norm.