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In the characteristic coordinates, the chain rule gives ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ and ${\partial }_t=c({\partial }_{\xi }-{\partial }_{\eta })$. HenceIntegrating this mixed derivative gives $\zeta =F(\xi )+G(\eta )$, a sum of oppositely traveling profiles. The initial data require $F+G=u_0$ and $c(F^{\prime }-G^{\prime })=v_0$, soIntegrating the profiles and matching their total additive constant gives the D'Alembert formula with initial velocity:Here $u_0$ names the initial surface displacement, rather than the water velocity $u$ appearing in the original shallow water equations. The formula verifies both initial conditions directly.

Linearize about rest and a flat surface. An appropriate small-amplitude scaling has $|\zeta |/h\ll 1$ and $|u|/\sqrt{gh}\ll 1$, with gradients on fixed common length/time scales. Products such as $(\zeta u{)}_x$ and $u{u}_x$ are then higher order. The linearized shallow water equations becomeDifferentiate the first with respect to time and substitute the second:The shallow-water dispersion relation therefore has a constant wave speed. In these equations $g$ is gravitational acceleration, despite the source's label “gravitational constant”; $gh$ has units of speed squared.

Let $r=e^{i\theta }\ne 1$. The geometric series from any starting index is bounded:Using summation by parts,Since the coefficients are nonincreasing and positive, the modulus is at most $K(a_N+a_m-a_N)=K{a}_m\to 0$. The Cauchy criterion proves convergence for every $\theta \notin 2\pi \mathbb{Z}$. This is a direct proof of the Dirichlet test in this setting.

For $a_n=1/n$, the imaginary parts of the convergent partial sums give convergence of the sine series at those angles. If $\theta \in 2\pi \mathbb{Z}$, every sine term is zero. Hence the sine series converges for all real angles.

Use $\sin 0=0$, the derivative identity ${\sin }^{\prime }=\cos$, continuity of cosine at zero, and $\cos 0=1$. For $u\ne 0$, the mean value theorem gives $\sin u/u=\cos c_u$ for a point between zero and $u$. Hence $\sin u/u\to 1$ as $u\to 0$. For $x\ne 0$, apply this with $u=x/3^k$ to obtain the limit $\boxed{3^k\sin (x/3^k)\to x}$. For $x=0$ every term is zero.

The additional bound $|\cos t|\le 1$ and the same mean value theorem give $|\sin u|\le |u|$. ThereforeThe right side is summable. By the comparison test for series the stated series has absolute convergence for every real $x$.

Every traceless skew-Hermitian matrix of size two has the formso it equals $\mathcal{M}(c)$ for $c=(2p,2\mathrm{Re}w,2\mathrm{Im}w)$. If $c=0$, take $A=B=0$. Otherwise choose a unit vector $u$ perpendicular to $c$ and set $a=u$, $b=c\times u$. The vector triple product gives $a\times b=c$.

By part (iii), choosing $A=\mathcal{M}(a)$ and $B=\mathcal{M}(b)$ gives the required representationIn fact both factors can themselves be chosen as traceless skew-Hermitian matrices. The cross-product model of su(2) explains this geometric commutator construction.

For $a=(a_1,a_2,a_3)$ the map has the explicit formIts diagonal entries sum to zero and its conjugate transpose is its negative, because the three coordinates are real. Hence it is traceless and a skew-Hermitian matrix.

Direct multiplication of the given basis matrices yieldsThe reverse products give the negative commutators, and equal-index commutators vanish. Bilinearity then givesThus the cross product is represented by the commutator:This realizes the cross-product model of su(2) with exactly the normalization and cyclic orientation used here.

The conjugate transpose reverses multiplication. If $A^{†}=-A$ and $B^{†}=-B$, thenThus the commutator is again a skew-Hermitian matrix. The order reversal is what produces the required minus sign.

Expanding the commutator immediately gives $[A,A]=0$, $[A,B]=-[B,A]$, and $[A,\lambda B]=\lambda [A,B]$. For its matrix trace, writeConsequently every matrix commutator is traceless: $\boxed{\mathrm{tr}[A,B]=0}$. This is the trace of a matrix commutator identity, valid for arbitrary square matrices over the complex numbers.

If $Ax=b$ and $h\in \ker A$, symmetry gives $b\cdot h=(Ax)\cdot h=x\cdot Ah=0$, proving necessity. Conversely, expand $b$ in the orthonormal basis from part (i), with $A{y}_i={\lambda }_i{y}_i$. The zero-eigenvalue vectors span the kernel of a linear map. If $b$ is orthogonal to that kernel, then $b\cdot y_i=0$ whenever ${\lambda }_i=0$, so the remaining coordinates can be solved independently.

For a nonzero eigenvalue and any associated nonzero eigenvector $v$, symmetry gives $b\cdot v=\lambda x\cdot v$. Thus the component vector along $v$ isFor the unit vectors $y_i$, this coefficient is $(b\cdot y_i)/{\lambda }_i$. Consequently the general solution isThis proves sufficiency as well as the claimed kernel criterion. If the compatibility condition fails, there is no solution; if it holds, the freedom is exactly an arbitrary kernel vector.

View a possible complex eigenvector $v$ using the Hermitian inner product. For a real symmetric matrix, $A^{†}=A$, so $v^{†}Av$ is real. Since $v^{†}Av=\lambda v^{†}v$ and $v^{†}v>0$, every eigenvalue is real.

For eigenvectors $v,w$ with distinct eigenvalues $\lambda ,\mu$, symmetry givesso $v^{†}w=0$. In particular, real eigenvectors are orthogonal in the real dot product.

Use the permitted diagonalizability assumption to take bases of the real eigenspaces whose union spans ${\mathbb{R}}^n$. Apply the Gram-Schmidt process within each eigenspace. Its linear combinations remain eigenvectors of that same eigenvalue; vectors in different eigenspaces are already orthogonal. The resulting union is an orthonormal basis of eigenvectors, giving an orthogonal diagonalization of a real symmetric matrix.

Take points $p=(3,0,4)$ and $q=(-2,3,3)$ with directions $u=(1,3,-1)$ and $v=(0,1,-1)$. A common normal is $u\times v=(-2,1,1)$, of length $\sqrt{6}$. The planes through the two lines with this common normal have separationThis is a lower bound for the distance between skew lines. It is attained at $s=-1$ and $t=-4$: the two points are $(2,-3,5)$ and $(-2,-1,7)$, whose difference $(4,-2,-2)$ is perpendicular to both directions and has length $2\sqrt{6}$. Thus the shortest distance is $\boxed{2\sqrt{6}}$.

The plane through $a_i$ has the constant normal coordinate $r\cdot \hat{n}=a_i\cdot \hat{n}$. Since $\hat{n}$ is a unit vector, the separation of these normal coordinates is a length. Every displacement joining the planes has this fixed normal component, so its length is at least its absolute value; a displacement parallel to $\hat{n}$ attains that bound. Hence the distance between parallel planes isThe absolute value, present in the PDF, is necessary because distance is nonnegative.

The first coordinate of the vector triple product isThe other two coordinates follow by cyclically permuting the indices, proving the identity.

A point on the line has the form $r=b+sm$. Substituting in the plane equation gives $s(m\cdot n)=(a-b)\cdot n$. When $m\cdot n\ne 0$ the parameter, and therefore the intersection point, is unique. The vector triple product gives $n\times (b\times m)=(n\cdot m)b-(n\cdot b)m$, so the unique intersection isFor the parallel case $m\cdot n=0$, the line is entirely in the plane if $(b-a)\cdot n=0$, and otherwise has no intersection. As usual, a genuine line and plane require nonzero direction and normal vectors. This is the line-plane intersection criterion.

Let $a_n=n!/n^n$. Its consecutive coefficient ratio isThe ratio test for the full terms therefore has limit $|z|/e$: the series has absolute convergence when $|z|<e$ and its terms eventually grow in magnitude when $|z|>e$. Thus the radius of convergence is $\boxed{R=e}$. No boundary decision is needed to determine the radius of convergence.

Expanding the characteristic polynomial givesSolving the three nullspace systems gives the eigenvalues and eigenspacesFor example, at $\lambda =1$ the first two equations force $z=0$ and $x=y$; at $\lambda =2$ they force $x=0$ and $y=-z$; at $\lambda =-1$ they force $x=y=z$.

Putting these eigenvectors in columns proves their linear independence:Hence $A$ is a diagonalizable matrix over the reals.

An eigenvector of a matrix $A$ is a nonzero vector $v$ for which $Av=\lambda v$. The scalar $\lambda$ is its eigenvalue. For a fixed eigenvalue, the zero vector together with all corresponding eigenvectors forms the eigenspace $\ker (A-\lambda I)$. The requirement $v\ne 0$ prevents every scalar from being an eigenvalue merely because $A0=\lambda 0$.

For a real line $ux+vy+w=0$, with $(u,v)\ne (0,0)$, choose $B=(u-iv)/2$ and $C=w$. Direct multiplication gives $Bz+\overline{B}\overline{z}=ux+vy$, so every line has the required form with $B\ne 0$.

For a circle of centre $c$ and radius $\rho >0$, expand its modulus equation:Thus the required circle parameters are $\boxed{B=-c,C=|c{|}^2-{\rho }^2}$. Conversely, completing the square in the circle form gives $|z+B{|}^2=|B{|}^2-C$, which describes a genuine circle when $|B{|}^2-C>0$. The complex conjugate terms make both equations real-valued.

The exponent in the PDF is $2z^2$. A complex exponential equals one precisely when its exponent is an integer multiple of $2\pi i$: its real part must be zero, and its imaginary part must be a full rotation. Therefore $z^2=\pi ik$ for $k\in \mathbb{Z}$. Taking both square roots yields all solutionsThe first nonzero family corresponds to positive exponent index, and the second to negative index. Both signs are required.