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Put $\beta =\Vert b\Vert >0$ and choose an oriented orthonormal basis with third vector $b/\beta$. The Levi-Civita symbol term has action ${ϵ}_{ijk}{b}_k{x}_j=x\times b$, the negative of $b\times x$. Thus this axisymmetric linear map has block formThe transverse eigenvalues $1\pm i\beta$ are not real. The only real eigenvalue and its eigenvectors are thereforeThe determinants are $\det B^{+}=(1+{\beta }^2{)}^2>0$ and $\det B^{-}=(1-{\beta }^2)(1+{\beta }^2)$, proving that $B^{+}$ is always invertible, and $B^{-}$ is invertible exactly when $\Vert b\Vert \ne 1$. The complex transverse pair is important for this determinant even though the question asks for real eigenvalues.

For small oscillations with quadratic kinetic and potential energies, a normal mode has a fixed spatial amplitude pattern and one angular normal frequency. Its amplitude is a normal coordinate; choosing all modes diagonalizes the two quadratic forms and decouples their equations. A zero frequency describes a free symmetry motion.

Let $q_j=R[{\theta }_j-j\pi /2]$ be the small signed arc displacements, with cyclic indices $j=0,1,2,3$. The springs follow arcs, so their extensions are $q_{j+1}-q_j$. The Lagrangian of the normal modes of a four-bead spring ring isThe stiffness matrix divided by $k$ is the cycle graph Laplacian, with diagonal entries two, adjacent entries minus one, and zero opposite entries. An orthonormal basis of mode vectors and their frequencies isThe normal mode coordinates are $Q_{\alpha }=u_{\alpha }\cdot q$, with equations ${\ddot{Q}}_{\alpha }+{\omega }_{\alpha }^2{Q}_{\alpha }=0$. The zero mode rotates all beads together without stretching a spring and has $Q_0=A+Bt$. In each of the two degenerate middle modes, opposite beads have opposite signed arc displacements while the other pair remains stationary. Any fixed linear combination is another mode of that frequency. The last mode alternates the signed displacement of adjacent beads. The radius cancels from the frequencies because arc length supplies the same factor $R^2$ to the angular kinetic and spring energies.

A two-index Cartesian tensor has components transforming as $T_{ij}^{\prime }=R_{ik}{R}_{jl}{T}_{kl}$ under an orthogonal change of orthonormal basis. The tensor order here counts indices, rather than the matrix rank. Its scalar, symmetric-traceless and axial tensor decomposition follows by settingThen $T_{ij}=Y{\delta }_{ij}+{\Omega }_{ij}+S_{ij}$, with $\Omega$ antisymmetric and $S$ symmetric and traceless. There are three independent antisymmetric components and five independent symmetric traceless components: three off-diagonal entries and two independent diagonal ones in the latter case.

For the symmetric tensor relative to a nonzero vector $A$, put $q=A_i{A}_i>0$, $s=A_i{T}_{ij}{A}_j$, and $t=T_{ii}$. The symmetric tensor decomposition relative to an axis givesIn particular $A_i{B}_i=s/q-s/q=0$. Define the orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/q$ onto the plane perpendicular to $A$, and setThis tensor is symmetric and kills $A$. Its trace is $t-s/q-2\alpha =0$. To verify the full decomposition, set $u=A/\sqrt{q}$ and split $T$ into its longitudinal component $(u^{T}Tu)u{u}^T$, two mixed components, and $PTP$. The mixed terms are exactly $B{A}^T+A{B}^T$, and $\alpha +\beta q=s/q$ gives the longitudinal coefficient. Thus all requested constraints and the reconstruction hold.

On the sphere, constant $\theta$ gives a latitude circle of radius $a\sin \theta$ and height $a\cos \theta$. Constant $\varphi$ gives a meridian semicircle in a vertical plane through the $z$ axis. The orthonormal basis of spherical coordinates isHere ${\mathbf{e}}_r$ points outward, ${\mathbf{e}}_{\theta }$ along increasing polar angle, and ${\mathbf{e}}_{\varphi }$ along increasing azimuth. At either pole the radial unit vector is unambiguous, but the two tangential coordinate vectors depend on the undefined longitude. The figure shows their directions away from the poles.

This coordinate ambiguity does not make a vector field singular if its tangential coefficient vanishes suitably. In Cartesian coordinates,The first is smooth at the north pole and zero there; the second is smooth and zero at the south pole. Their actual singular axes are the opposite ones. These are the regular patches of a two-patch vector potential for a radial monopole.

The spectral theorem for real symmetric matrices supplies an orthonormal basis of real eigenvectors. Put these in the columns of $Q$, so $Q^{T}Q=I$ and $Q^{T}AQ=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. In the question's coordinate convention choose $O=Q^T$ and $y=Ox$. Since $x=O^{T}y$,Reorder the eigenvalues with $m$ positive, $\ell -m$ negative and $n-\ell$ zero. For nonzero eigenvalues set $z_i=\sqrt{|{\lambda }_i|}{y}_i$, and retain the remaining coordinates as $z_i=y_i$. This is an invertible real transformation $z=DOx$, where $D$ is the diagonal rescaling; the combined matrix $L=DO$ gives $z=Lx$. It giveswith the zero coordinates absent. The general allowed range is $0\le m\le \ell \le n$. The printed restriction $m\ge 1$ is not valid for every real symmetric matrix: a negative-definite or zero form has no positive square. The invariance of these counts is Sylvester's law of inertia.

For the specified form, a particularly simple change isSubtracting the first two squared coordinates verifies the identity directly, and $\det L=1$ proves invertibility. Thus $m=1$, $\ell =2$ and there is one zero direction. The final rescaling need not remain orthogonal.

An isotropic tensor has unchanged components under every proper rotation of the Cartesian axes. Every rank-zero scalar is isotropic, and $c{\delta }_{ij}$ is a rank-two example. The zero rank-one tensor is isotropic, but no nonzero vector is: a half-turn about an axis perpendicular to it reverses its sign.

For the inertia tensor, write $u=x-G$ and $X=P-G$, so $x-P=u-X$. Expand its defining integrand. Every cross term has an integral containing ${\int }_{B}u\rho d^{3}x=0$ by the definition of the centre of mass. The remaining terms give the tensor parallel-axis theorem:For a uniform unit ball, rotational symmetry gives $\int \rho u_i{u}_j{d}^{3}u=c{\delta }_{ij}$. Its trace is $\int \rho |u{|}^2{d}^{3}u=3M/5$, so $c=M/5$ and $I(G)=2MI/5$. ThereforeAssuming positive mass, it is isotropic exactly when $P=G$. For $X\ne 0$, its eigenvalues and directions areTo give an explicit orthonormal eigenbasis when $X_1^2+X_2^2>0$, take $e_1=(-X_2,X_1,0)/\sqrt{X_1^2+X_2^2}$ and $e_2=e_{\parallel }\times e_1$. If $X$ is a nonzero vector along the third axis, take the first two coordinate unit vectors instead. At $X=0$, all three eigenvalues are $2M/5$ and every orthonormal basis works. The inertia tensor of a displaced uniform ball describes the same result for an arbitrary radius.

Let $Q$ be the orthogonal matrix taking old components to new ones, so $x_i^{\prime }=Q_{ip}{x}_p$. With repeated indices summed, the tensor component transformation law isAn isotropic tensor has the same components after every proper rotation. Since $Q_{ip}{Q}_{jp}={\delta }_{ij}$, the Kronecker delta is invariant. Applying this identity to each factor shows that $a{\delta }_{ij}$ and every linear combination $b{\delta }_{ij}{\delta }_{kl}+c{\delta }_{ik}{\delta }_{jl}+d{\delta }_{il}{\delta }_{jk}$ are invariant as well.

To show that an isotropic second-rank tensor must be scalar, start with any matrix $t$, without assuming symmetry. Invariance under quarter-turns about axes 1 and 2 implies invariance under their squares, the half-turnsThe first forces $t_{12},t_{21},t_{13},t_{31}$ to vanish. The second also forces $t_{23},t_{32}$ to vanish. The matrix is therefore diagonal. A quarter-turn about axis 1 interchanges its 22 and 33 entries; one about axis 2 interchanges its 11 and 33 entries. All three must agree, giving $t=aI$.

A rotation preserves the ball $V_R$, $|x{|}^2$ and the volume element. Changing integration variables by that rotation shows that $A$ and $B$ are isotropic tensors. Write $A_{ij}=a{\delta }_{ij}$. Taking its tensor contraction givesso $a=4\pi R^7/21$. Every permutation of the indices of $B$ leaves its integrand unchanged. Its complete symmetry therefore makes all three isotropic coefficients equal, say $d$:Double contraction gives $15d=B_{iikk}={\int }_{V_R}|x{|}^{4}dV=4\pi R^7/7$. Thus the isotropic fourth moment of a ball and its radial contraction are

Put $d=4\pi R^7/105$, so that $a=5d$. Contract $B$ with the prescribed unit vector twice:This has the spectrum of a scalar matrix plus a rank-one projector. In the direction $n$, the outer product $n{n}^T$ acts as the identity; on the plane perpendicular to $n$, it acts as zero. ThereforeAny nonzero multiple of $n$ is a parallel eigenvector, and any nonzero vector in the perpendicular plane is a transverse eigenvector. The transverse eigenvalue has multiplicity two; an orthonormal basis of that plane completes a set of principal axes.

For $B{x}_i={\lambda }_i{x}_i$, Hermitian symmetry givesDistinct eigenvalues therefore imply $x_i^{†}{x}_j=0$ for $i\ne j$. Together with the unit normalizations,The $n$ mutually orthogonal nonzero vectors are linearly independent and span the $n$-dimensional space, so they form an orthonormal basis. Here ${\delta }_{ij}$ is the Kronecker delta.

Normalize $e_1$ and put $f_1=M{e}_1/{\mu }_1$. The two vectors are an orthonormal basis of a plane $E$ invariant under $M$, since $M{e}_1={\mu }_1{f}_1$ and $M{f}_1=-{\mu }_1{e}_1$. Its orthogonal complement is also invariant: for $w\perp E$ and $z\in E$,Apply the spectral theorem for real symmetric matrices to the restriction of $M^2$ to $E^{\perp }$. Any remaining negative eigenvalue supplies another orthonormal pair by the same construction. Induction leaves only the kernel. Thus every nonzero eigenvalue of $M^2$ has even multiplicity. The pairs can share an eigenvalue, so this does not assert distinctness of the different $-{\mu }_j^2$.

By the spectral theorem, a Hermitian matrix has real eigenvalues and an orthonormal eigenbasis. Choose column vectors $u_j$ with $H{u}_j={\lambda }_j{u}_j$ and $u_i^{†}{u}_j={\delta }_{ij}$, and form $U=(u_1\cdots u_n)$. Then $U^{†}U=I$, so $U$ is unitary, and $HU=U\Lambda$ givesThe entries of $\Lambda$ are the real eigenvalues, repeated according to their multiplicities. Within a degenerate eigenspace, any orthonormal basis may be used. Equivalently $H=U\Lambda U^{†}$.

The second coordinate direction is already an eigenvector with eigenvalue three. The first and third coordinates form a symmetric block, whose sum and difference directions giveFor example, the sum of the block's diagonal and off-diagonal entries is $(1+c)+(5-c)=6$, while their difference is $(1+c)-(5-c)=2c-4$. Equivalently the characteristic polynomial is $(\lambda -6)(\lambda -3)(\lambda -2c+4)$.

The displayed basis works for every real $c$. At $c=7/2$, the eigenvalue three has eigenspace spanned by the second and third listed vectors; at $c=5$, the eigenvalue six has eigenspace spanned by the first and third. Every nonzero vector in a repeated eigenspace is an eigenvector, and any orthonormal basis of it may replace the displayed choices.

Differentiate the position vector, using the angular factor from the scale factors of orthogonal coordinates:The basis is an orthonormal basis, so all cross terms in its squared length vanish. The Euclidean line element is thereforeIn particular, an angular change $d\varphi$ at fixed radius traverses a physical distance $\rho |d\varphi |$.

Dividing the coordinate tangents by their scale factors of orthogonal coordinates gives an orthonormal basisEvaluate the Cartesian cross products componentwise:ConsequentlyThe ordered cylindrical basis is right-handed: its scalar triple product ${\mathbf{e}}_{\rho }\cdot ({\mathbf{e}}_{\varphi }\times {\mathbf{e}}_z)$ equals $+1$.

The columns of $R_z$ are an orthonormal basis, since ${\cos }^2\varphi +{\sin }^2\varphi =1$ and the first two columns have zero dot product. Thus $R_z^T{R}_z=I$: it is an orthogonal matrix. Consequentlyand hence $\boxed{|{\mathbf{v}}^{\prime }|=|\mathbf{v}|}$. This is the preservation of the Euclidean norm by a rotation matrix.

Let $L\varphi =\lambda \varphi$ with $\varphi \ne 0$ in the self-adjoint domain. By the boundary identity from part (i),Since its squared weighted norm is positive, $\boxed{\lambda \in \mathbb{R}}$.

For $L{\varphi }_n={\lambda }_n{\varphi }_n$ and $L{\varphi }_m={\lambda }_m{\varphi }_m$, the same identity givesThus ${\lambda }_n\ne {\lambda }_m$ impliesThis is orthogonality of Sturm-Liouville eigenfunctions. Degenerate eigenspaces can be given an orthonormal basis, although vectors in a degenerate eigenspace are not automatically orthogonal.

Under the specified passive change of orthonormal basis, each Cartesian tensor index acquires one copy of $L$. Thus a tensor of order $n$ transforms asRepeated old-basis indices are summed from one to three. Orthogonality of the new basis gives $L{L}^T=I$. Taking determinants yields $(\det L{)}^2=1$, hence $\boxed{\det L=+1\text{ or }-1}$. The positive case preserves orientation and the negative case reverses it.

Let $A{v}_i={\lambda }_i{v}_i$ and $A{v}_j={\lambda }_j{v}_j$. The skew-Hermitian matrix property giveswhere the last equality uses part (a). Distinct eigenvalues therefore imply $\boxed{v_i^{†}{v}_j=0\text{ for }i\ne j}$. Normalize each eigenvector to length one. Since there are $n$ distinct eigenvalues, these vectors form an orthonormal basis.

Put them in the columns of $U$. Then $U$ is a unitary matrix and $AU=UD$, so

The required actions have already been identified:In three dimensions choose a unit vector $w$ perpendicular to both. Its one-dimensional orthogonal complement to the pair is invariant under $M$. On a one-dimensional real space, a skew-symmetric matrix must act as zero: $Mw=\eta w$ would imply $\eta =w^{T}Mw=0$. Thus $Mw=0$.

The columns of $W=(u_1{v}_{1}w)$ form an orthonormal basis, so $W$ is an orthogonal matrix. The displayed actions giveand thereforeIf $M=0$, take ${\mu }_1=0$ and any orthogonal matrix $W$. Otherwise its nonzero two-dimensional block supplies the pair used above.

For a real symmetric matrix $A$, find its real eigenvalues from $\det (A-\lambda I)=0$, then solve $(A-\lambda I)v=0$ for the corresponding eigenspaces. Distinct eigenspaces are orthogonal. Within each repeated eigenspace, apply the Gram-Schmidt process and normalize to obtain an orthonormal basis of eigenvectors.

Put these vectors in the columns of $O$. The spectral theorem for real symmetric matrices guarantees a full basis, so $O^{T}O=I$ andThis is the orthogonal diagonalization of $A$ and the associated orthogonal change of coordinates.

Use the standard regular Sturm-Liouville problem hypotheses: positive weight $w$, real coefficients, and real self-adjoint boundary conditions. The positive weight and boundary-domain assumptions are implicit in the requested real orthonormal basis; real differential coefficients alone would not guarantee them for arbitrary complex boundary conditions.

Multiply the eigenfunction equation by $\overline{y_n}$ and integrate. By self-adjointness, the number ${\int }_a^b\overline{y_n}\mathcal{L}{y}_{n}dx$ is real, whereasThe second integral is strictly positive, so $\boxed{{\lambda }_n\in \mathbb{R}}$. Because the coefficients and boundary conditions are real, the real and imaginary parts of any complex eigenfunction separately obey the same equation and conditions. At least one is nonzero, so a real eigenfunction can be chosen. More generally, each eigenspace has a real basis, which can be orthonormalized using the weighted inner product.

The orthogonality of Sturm-Liouville eigenfunctions, with normalization chosen to one, isFor the real choice, omit the complex conjugate.