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For $f=u+iv$ holomorphic in $z=x+iy$, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. Therefore the normal vectors to the two level curves obeyAt a point where $f^{\prime }\ne 0$, both gradients are nonzero, so their perpendicular tangent lines are also perpendicular. At critical points this argument does not assert that regular level curves exist.

TakeIts real part is $u=x^2-y^2$, so the specified first family is $u=-\alpha$. Its imaginary part is $v=2xy$, so the second family is $v=2\beta$. Since $\beta \ne 0$, their intersections exclude the origin, where $f^{\prime }(z)=2z$ would vanish. Thus these hyperbolas meet orthogonally as required.

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.

The quadratic Lagrangian gives $\partial L/\partial \dot{q}=T\dot{q}$ and $\partial L/\partial q=-Vq$. Thus the Euler-Lagrange equations areA nonzero mode vector $v$ has a common harmonic time factor and satisfies the generalized eigenvalue problem for small oscillations $Vv={\omega }^{2}Tv$; $\omega$ is its normal frequency. A zero-frequency mode instead admits a constant or linear time factor.

Measure the three coordinates as displacements from a configuration with both springs unstretched. The kinetic energy is $\frac{1}{2}{\dot{x}}_1^2+\frac{1}{2}M{\dot{x}}_2^2+\frac{1}{2}{\dot{x}}_3^2$, and the potential energy is $\frac{1}{2}(x_2-x_1{)}^2+\frac{1}{2}(x_3-x_2{)}^2$. Expansion givesFor $M>0$, direct substitution yields the three normal modes:The three vectors are independent and orthogonal in the mass inner product $v^{T}Tw$, so they exhaust the motion. Their squared mass norms are $M+2,2,2+4/M$; division by the corresponding square roots gives normalized modes. This is the unit-spring, unit-outer-mass case of the normal modes of three masses connected by two equal springs.

As $M\to \infty$, the central displacement in the symmetric oscillatory mode tends to zero and both nonzero frequencies tend to one. The two limiting shapes $(1,0,1)$ and $(1,0,-1)$ span independent motions of the outer masses against an effectively fixed centre; each unit mass has its own unit spring frequency. The translational mode remains at zero frequency. Under a bounded kinetic-energy condition its common velocity tends to zero as $M^{-1/2}$, while a static common displacement is still an equilibrium shift. One must specify that energy condition before concluding that an arbitrary freely translating heavy mass remains stationary.

Since $M$ is real, $(M^2{)}^T=(M^T{)}^2=(-M{)}^2=M^2$. For every real vector $x$,Thus $M^2$ is a real symmetric matrix and is negative semidefinite. When $n$ is odd, part (i) supplies a nonzero $x$ in the kernel of $M$, and then $M^{2}x=0$. Hence zero is an eigenvalue of $M^2$. In fact $\ker M^2=\ker M$, as the displayed norm identity shows.

Define the outer product $T_{ij}=u_i{v}_j$. Transforming each vector givesThus $u_i{v}_j$ is a rank-two tensor, since it obeys the two-index tensor component transformation law. This is the component form of the tensor product of two vectors; its two indices are independent, not contracted.

Start from the three-index transformation law and contract the first two Cartesian indices:Orthogonality supplies ${\sum }_i{Q}_{ia}{Q}_{ib}={\delta }_{ab}$, where $\delta$ is the Kronecker delta. The remaining factor is exactly the vector transformation rule. Therefore $C_k$ is a vector.

This tensor contraction removes two indices by pairing them through the Euclidean metric. In general nonorthonormal coordinates, contraction must use the appropriate metric or pair a covariant and a contravariant index; the bare repeated-lower-index sum relies on Cartesian orthonormal coordinates.

Taking the adjoint of $U=M+iN$ gives $U^{†}=M-iN$. In particular, the Hermitian parts of a unitary matrix are uniquely determined:Expanding both unitary identities givesSubtracting shows $MN=NM$, and adding gives $M^2+N^2=I$. If $Mv=\mu v$, commutation implies $M(Nv)=\mu Nv$. The assumed simple spectrum makes the eigenspace one-dimensional, so $Nv=\nu v$. Thus $M,N$ have a common orthonormal eigenbasis.

On such a vector, $Uv=(\mu +i\nu )v$. If the corresponding unitary eigenvalue is $u=e^{i\theta }$, thenThe formulas for $M,N$ also show the same eigenvalue correspondence directly from $Uv=uv$ and $U^{†}v=u^{\ast }v$. Degeneracies do not prevent existence of a common basis, although the simple-spectrum assumption makes the elementary proof immediate.

The matrix is $M=2I-e{e}^T$, where $e=(1,1,1{)}^T$. Along $\hat{e}=e/\sqrt{3}$, its eigenvalue is $2-3=-1$. Every vector perpendicular to $e$ has eigenvalue two. For example, use the orthonormal basisThe principal-coordinate equation isIt depends only on distance from the $u_3$ axis and position along it, so every rotation about that axis preserves the surface. ThereforeThis is a one-sheet hyperboloid with waist radius $\sqrt{3/2}$. The repeated positive eigenvalue makes the two transverse principal axes nonunique.

The spectral theorem for real symmetric matrices supplies real eigenvalues and an orthonormal eigenbasis. Compute eigenvectors, normalize them, and within each repeated-eigenvalue eigenspace choose an orthonormal basis. Distinct eigenspaces are already orthogonal because symmetry gives $(\lambda -\mu )v^{T}w=0$ for eigenvectors with eigenvalues $\lambda ,\mu$.

Use the resulting basis vectors as columns of $O$. ThenThe order and signs of the columns can be chosen freely; rotations within a repeated-eigenvalue eigenspace are also allowed.

The coordinate tangent vectors are the partial derivatives of the position vector:Their pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, are $1,r,1+r\sin \alpha$. Since $R<1$, the last is positive throughout the body. Thus these circular torus coordinates are orthogonal wherever $r>0$; at $r=0$ the cross-section angle degenerates, as in ordinary polar coordinates.

The ordered Jacobian determinant isIts sign is positive for $r>0$, so the volume element is $dV=r(1+r\sin \alpha )drd\alpha d\beta$. The coordinate degeneracy on the central circle has zero volume and does not obstruct volume integration.

Let $A{v}_j=a_j{v}_j$. Commutation gives $A(B{v}_j)=B(A{v}_j)=a_{j}B{v}_j$, so $B$ preserves each eigenspace of $A$. Since the eigenvalues of $A$ are nondegenerate, each eigenspace is one-dimensional and $B{v}_j=b_j{v}_j$. The spectral theorem for a Hermitian matrix lets us choose these vectors as an orthonormal eigenbasis. With $U=(v_1,\dots ,v_n)$,This is simple spectrum gives simultaneous diagonalization. The common eigenvectors can be chosen up to their phases and ordering. In fact degeneracy does not prevent a common eigenbasis for commuting Hermitian matrices, but the one-dimensional argument already suffices here.

The convergent matrix exponential series respects similarity, so $e^A=U{e}^{{\Lambda }_A}{U}^{†}$ and likewise for $B$. Scalar exponentials multiply on each diagonal entry, givingThe essential hypothesis is commutation; Hermiticity supplies this convenient diagonalization proof.

The spectral theorem for a real symmetric matrix gives real eigenvalues and an orthonormal eigenbasis. Compute the roots of $\det (A-\lambda I)=0$, find a basis for each eigenspace, and orthonormalize within any repeated-eigenvalue eigenspace. Eigenvectors belonging to distinct eigenvalues are already orthogonal: if $Av=\lambda v$ and $Aw=\mu w$, symmetry gives $(\lambda -\mu )v^{T}w=0$.

Putting the normalized basis vectors in the columns of an orthogonal matrix $Q$ yieldsFor repeated eigenvalues, the choice of axes inside their eigenspace is not unique.

Taking determinants and using invariance under transpose,This determinant identity actually holds in every dimension. In dimension three, skew symmetry additionally givesso $\det A=0$. Hence there is a real nonzero vector $x$ in the kernel:The eigenvalue relation then gives $Nx=x$. Thus the Cayley transform fixes an axis and is a proper three-dimensional rotation, including the identity as a degenerate case.

First, $I-A$ is invertible. If $(I-A)x=0$ for a real vector, then $Ax=x$, whereas the skew-symmetric matrix identity gives $x^{T}Ax=0$; hence $x^{T}x=0$ and $x=0$. Thus its real kernel is trivial, which suffices for invertibility. The same argument applies to $I+A$.

Using $A^T=-A$ and the rule for the transpose of a product,The factors $I+A$ and $I-A$ commute, as do the required inverses. ConsequentlyThus $\boxed{N\text{ is orthogonal}}$. This is the Cayley transform of a real skew-symmetric matrix.

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.

A second-order Cartesian tensor assigns components in each orthonormal coordinate system according toHere $R$ is the matrix taking old vector components to new ones, and repeated indices are summed. This is the tensor component transformation law; an arbitrary array of nine numbers is not a tensor unless its components in different frames obey this rule. No symmetry $T_{ij}=T_{ji}$ is implied by being a second-order tensor.

In orthonormal Cartesian coordinates, a rank-$n$ tensor has components transforming under a change of axes $R$ asRepeated indices are summed. This is the tensor component transformation law; for rank one it is precisely the transformation law of a vector.

Use the vector triple product to write the double cross product asBoth dot products are invariant under an orthogonal change of axes, while $u^{\prime }=Ru$ and ${\Omega }^{\prime }=R\Omega$. ThereforeHence $W$ is a vector. This argument also avoids the sign issue for an individual cross product under an orientation-reversing orthogonal transformation: the double product has two such factors, whose signs cancel.

Take the centre of the ball as the origin. An orthogonal change of variables leaves both this domain and its volume measure invariant. Consequently its Cartesian moment integrals are isotropic tensors. Equivalently, inversion $x\mapsto -x$ immediately makes every odd moment zero.

There is no nonzero invariant rank-one vector, soAn isotropic tensor of rank two is a scalar multiple of the Kronecker delta, giving ${\int }_V{x}_i{x}_{j}dV=\alpha {\delta }_{ij}$. Contract the indices and use spherical coordinates:Thus the Cartesian moments of a three-dimensional ball giveFor the last identity, one may also use the general rotation-invariant rank-three tensor $c{ϵ}_{ijk}$: the moment is fully symmetric in its indices, whereas the Levi-Civita symbol is antisymmetric, forcing $c=0$.

The origin convention is necessary. For a ball centred at a nonzero vector $c$, its first moment would instead be $(4\pi a^3/3)c_i$, so the displayed claims implicitly refer to coordinates measured from its centre.

Use the vector triple product expression from part (a). Put $s=\Omega \cdot x$ and $r^2=x\cdot x$. Since $u=\Omega +\beta x$,Expanding $W=(u\cdot u)\Omega -(u\cdot \Omega )u$ and canceling the zeroth-order terms givesThe first bracket is odd in $x$ and integrates to zero over the centred ball. With $\alpha =4\pi a^5/15$, the Cartesian moments of a three-dimensional ball giveThereforeThe resulting vector is parallel to $\Omega$, as the rotational symmetry of the centred domain requires.

Take any constant vector $\mathbf{a}$. Its curl is zero, so part (a) and the divergence theorem givewhere $\mathbf{n}$ is the outward unit normal vector. The scalar triple product identity rewrites the last integrand as $\mathbf{a}\cdot (\mathbf{F}\times \mathbf{n})$. Since $\mathbf{a}$ is arbitrary, equality of these scalar products implies the vector identity