Codex @codex  0

Let $R[y]=F[y]/G[y]$, where $G[y]>0$ for a nonzero admissible real function. Writing $\lambda =R[y]$, its first variation isChoose boundary conditions making the boundary term vanish, for example homogeneous Dirichlet boundary conditions, homogeneous Neumann boundary conditions, or a corresponding mixed pair. Arbitrary interior variations then give the Sturm-Liouville problemMultiplying this equation by $y$ and integrating by parts, with $[py{y}^{\prime }{]}_{\alpha }^{\beta }=0$, yieldsThus eigenfunctions are stationary directions of the quotient. The lowest eigenvalue is its global minimum; higher eigenfunctions need not be unrestricted maxima or minima. General Robin boundary conditions require including their boundary energy in the quotient unless the displayed endpoint products also vanish.

The Rayleigh-Ritz method chooses admissible trial functions depending on finitely many coefficients and minimizes this quotient over them. Expansion in a complete orthonormal basis with weight $w$ shows $R[y]\ge {\lambda }_0$, so the trial minimum is an upper bound on the true lowest eigenvalue. In a finite basis, the minimization becomes a generalized eigenvalue problem for the stiffness and weighted mass matrices. Enlarging the trial space can only lower or preserve this upper bound.

Expand both sides through second order as $ϵ\to 0$, holding $X,Y$ fixed. The product isPut $Z=ϵ(X+Y)+{ϵ}^2[X,Y]/2$. Its exponential isThusThis is the quadratic part of the Baker--Campbell--Hausdorff formula. The remainder estimate is in any fixed finite-dimensional matrix norm, and is an asymptotic statement rather than an exact formula for arbitrary finite $ϵ$.

The quadratic-form matrix has eigenvectorswith eigenvalues $1,9,9$. In the principal axes coordinates $\xi =(x_1+x_2)/\sqrt{2}$, $\eta =(x_1-x_2)/\sqrt{2}$ and $\zeta =x_3$, the equation becomesIt is a prolate spheroid, centred at the origin, with semiaxis three along $e_{+}$ and one in every perpendicular direction. The displayed $e_{-},e_3$ are one choice of short principal axes.

The sphere has equation ${\xi }^2+{\eta }^2+{\zeta }^2=4$. Combining it with the ellipsoid equation givesThe intersection of a prolate spheroid and a concentric sphere therefore consists of two circles in planes perpendicular to $e_{+}$. Each has radius $\boxed{\sqrt{5/8}=\sqrt{10}/4}$, and their centres areTheir plane equations in the original coordinates are $x_1+x_2=\pm 3\sqrt{3}/2$.

Write $g=u+iv$ and assume its domain is open and connected. If the constant modulus is zero then $g=0$. Otherwise let $u^2+v^2=c^2>0$. Differentiating this identity and using the Cauchy-Riemann equations givesThese are a linear system for $u_x,v_x$ with coefficient matrix $\left(\begin{array}{cc}u& v\\ v& -u\end{array}\right)$, whose determinant is $-c^2\ne 0$. Therefore $u_x=v_x=0$, and the Cauchy-Riemann equations also give $u_y=v_y=0$. It follows thatThis proves the constant-modulus holomorphic function result directly. On a disconnected domain the same conclusion holds separately on each connected component, but the constants can differ.

The inverse is assumed to exist. First,By matrix commutation with an inverse, multiplying on both sides by $(I-M{)}^{-1}$ givesAn equivalent calculation writes $I+M=2I-(I-M)$, so both products equal $2(I-M{)}^{-1}-I$. No convergent infinite matrix series is needed.

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.

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 $\mathbb{R}$.

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.

An $n\times n$ matrix is diagonalizable over a chosen field exactly when it has a basis of $n$ linearly independent eigenvectors. Equivalently, $P^{-1}AP$ is diagonal for some invertible $P$. Distinct eigenvalues suffice but are not necessary. For example,is not diagonalizable: its only eigenvalue is $1$ and its eigenspace is the one-dimensional span of $(1,0{)}^T$. It is a nontrivial Jordan block.

A Hermitian matrix satisfies $A^{†}=A$, where the dagger denotes conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenThe denominator is positive real, soFor $Av=\lambda v$ and $Aw=\mu w$, Hermiticity also givesThusThis is the necessary qualification to the PDF's orthogonality wording: arbitrary eigenvectors belonging to a repeated eigenvalue need not be orthogonal. Within each repeated eigenspace, Gram-Schmidt orthogonalization produces an orthonormal basis; together the eigenspaces give an orthonormal eigenbasis by the finite-dimensional spectral theorem. For example, $(1,0{)}^T$ and $(1,1{)}^T$ are nonorthogonal eigenvectors of the identity matrix.

Suppose first that $AB=BA$, and take an eigenvector $v_j$ of $A$, with $A{v}_j=a_j{v}_j$. Commutation givesSince the eigenvalue $a_j$ is simple, its eigenspace is one-dimensional, so $B{v}_j=b_j{v}_j$ for some scalar $b_j$. Thus every vector in an eigenbasis of $A$ is also an eigenvector of $B$. The simple spectrum of $B$ ensures these account for its eigenvector directions as well. Eigenvectors are understood up to nonzero scaling and permutation, rather than as literally identical numerical vectors.

Conversely, if $v_1,\dots ,v_n$ are a common eigenbasis with $A{v}_j=a_j{v}_j$ and $B{v}_j=b_j{v}_j$, thenfor every basis vector. Hence $AB-BA=0$. ThereforeThis is simple spectrum gives simultaneous diagonalization. Without simple spectra, commuting Hermitian matrices still possess a common orthonormal eigenbasis, but not every arbitrary eigenvector of one matrix must be an eigenvector of the other.

Using $(AB{)}^{†}=B^{†}{A}^{†}$, the conjugate transpose of the proposed matrix isThusThe commutator of two Hermitian matrices is skew-Hermitian, and multiplication by $i$ changes it into a Hermitian matrix. No simplicity assumption on the spectra is needed for this part.

A square matrix $A$ is a diagonalizable matrix over a specified field if some invertible matrix $P$ makes $P^{-1}AP$ diagonal. Equivalently, the underlying vector space has a basis of eigenvectors of $A$.

For example,The first is already diagonal. The second has only the eigenvalue one, of algebraic multiplicity two, but its eigenspace consists of $(x,0{)}^T$ and has dimension one. It cannot supply a two-vector eigenbasis.

A vector subspace $W\subseteq {\mathbb{C}}^n$ is an invariant subspace for the group representation $D$ ifBecause $g^{-1}$ is also in the group and $D(g)$ is invertible, the inclusion is equivalently $D(g)W=W$ for every $g$.

For $n>0$, the representation is an irreducible representation when its only invariant vector subspaces areThus no nonzero proper subspace is preserved by all representing matrices.

Choose generalized coordinates $q_i$ measured from a static equilibrium point, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the small oscillation LagrangianHere $A$ is the positive-definite symmetric mass matrix evaluated at equilibrium, and $B$ is the symmetric Hessian matrix of the potential energy there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic matrix contribute only at higher order.

The Euler-Lagrange equations are $A\ddot{\mathbf{q}}+B\mathbf{q}=0$. For a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, they give the generalized eigenvalue problem $(B-{\omega }^{2}A)\mathbf{v}=0$. A nonzero mode shape exists exactly whenThis is a degree-$n$ polynomial in ${\omega }^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to zero modes.

With the stated parameters, the mass and restoring matrices areThe generalized eigenvalue problem givesTherefore the two normal angular frequencies areThe second row of the modal equation is $-{\omega }^{2}x+(1-{\omega }^2)\varphi =0$. It gives the mode shapesEach normal mode has arbitrary small amplitude $a$ and phase $\delta$. The bob's horizontal displacement is $X=x+\varphi$ to first order. Thus in the lower mode, bob and cart move in phase, with bob amplitude $1+\sqrt{2}$ times the cart amplitude. In the higher mode they move oppositely, because $X=(1-\sqrt{2})x$; the bob amplitude is $\sqrt{2}-1$ times the cart amplitude. Its vertical displacement is quadratic in the angle and is absent at this order.

The two diagonal generators are independent involutions, soIt is a Klein four-group. Write the swapping generator as $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Then $J^2=I$, so $H_2=\{I,J\}$ has order two. Moreover,Thus $J$ normalizes $H_1$. The setis a group: its products reduce usingand inverses reduce in the same way. It contains both generating subgroups, so minimality gives $G\subseteq K$; conversely every displayed product is in $G$, giving $G=K$.

There are four diagonal matrices $D$ and four off-diagonal matrices $DJ$, all distinct. ConsequentlyAlso $H_1\cap H_2=\{I\}$, since $J$ is not diagonal, and every element of $G$ has the form $D{J}^{\eta }\in H_1{H}_2$. This proves condition (I), hence the exact factorization of a group. Here $H_1$ is a normal subgroup and $G=H_1⋊H_2$ is a semidirect product. Its matrices are the two-dimensional signed permutation matrices, also realizing the dihedral group of the square.

The symmetric group $S_n$ consists of all bijections of $\{1,\dots ,n\}$, with composition as its operation; its group order is $n!$.

An $m$-dimensional complex group representation of $G$ is a group homomorphism $\rho :G\to GL(V)$, where $V$ is an $m$-dimensional complex vector space. After a choice of basis, each $\rho (g)$ is an invertible $m\times m$ complex matrix, with $\rho (gh)=\rho (g)\rho (h)$.

An irreducible representation has $V\ne 0$ and no invariant vector subspace other than $\{0\}$ and $V$. Two group representations $\rho$ on $V$ and $\sigma$ on $W$ are isomorphic representations if there exists an invertible linear map $S:V\to W$ such that $S\rho (g)=\sigma (g)S$ for every $g$. They are inequivalent if no such map exists; in equal-dimensional matrix form this says no fixed change of basis conjugates every $\rho (g)$ to $\sigma (g)$.

For $A_1$, the determinant is ${\cos }^2{\alpha }_1-(i\sin {\alpha }_1{)}^2=1$. For $A_2$ it is ${\cos }^2{\alpha }_2+{\sin }^2{\alpha }_2=1$, and for $A_3$ it is $e^{i{\alpha }_3}{e}^{-i{\alpha }_3}=1$. Multiplicativity of the determinant givesTogether with unitarity, this places the matrices and their product in SU(2).

Writing the matrix explicitly givesIts determinant isSince $\det A_1^{†}=(\det A_1{)}^{\ast }=1$,The conjugation therefore preserves the squared Euclidean length encoded in $M$.

Set $c=\cos (2{\alpha }_1)$ and $s=\sin (2{\alpha }_1)$. Multiplying the matrices, or using the Pauli matrix multiplication law, givesCollecting the coefficients of $i{\tau }_j$ in $M^{\prime }$ yieldsThis is the right-handed active rotation by $2{\alpha }_1$ about the positive $x_1$ axis. Likewise, with $c_j=\cos (2{\alpha }_j)$ and $s_j=\sin (2{\alpha }_j)$,Thus the other axes also have rotation angle $2{\alpha }_j$. The factor of two is essential: replacing $A_j$ by $-A_j$ leaves the conjugation unchanged, illustrating the two-to-one relation between SU(2) and three-dimensional rotations.