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Suppose that $Q$ were not surjective. Its image would be a proper subspace of $F^n$, so there would be a nonzero functionalvanishing on $\mathrm{im}Q$. Thenfor every $x\in V$, contradicting the linear independence of the $q_j$. Hence the surjectivity of independent linear functionals gives

Let $K={\bigcap }_j\ker q_j=\ker Q$ and suppose $K\subseteq \ker f$. DefineThis is well-defined: if $Qx=Qy$, then $x-y\in K$, so $f(x)=f(y)$. It is linear, and surjectivity of $Q$ means it is defined on all of $F^n$. Thus there are scalars $a_1,\dots ,a_n$ such that $g(y)={\sum }_j{a}_j{y}_j$. Thereforeand hence

Vanishing net charge density does not require vanishing current. Positive and negative charge carriers can cancel in charge density while their oppositely directed motions add to a nonzero current. Charge conservation only requiresfor magnetostatics this becomes $\nabla \cdot J=0$.

Because $\nabla \cdot B=0$, one may introduce a magnetic vector potential withIt is not unique: $A+\nabla \chi$ gives the same field for any scalar $\chi$.

For the stated current,so it is consistent with stationary charge conservation. Direct calculation givesThus $J$ is a Beltrami field. For $\lambda \ne 0$, chooseThen $\nabla \cdot B=0$ and $\nabla \times B={\mu }_{0}J$. A convenient Coulomb-gauge potential issince $\nabla \times A=B$ and $\nabla \cdot A=0$. Gradient gauge terms may of course be added.

If $\lambda =0$, the current is the constant field $J=(0,J_0,0)$. The formulas involving $1/\lambda$ do not apply; one valid choice is

Writing the velocity as $(u,v)=(y,ax)$ givesso the flow is incompressible. With the conventiona stream function isThe streamlines are its level sets. For $a>0$, they are the hyperbolaswith separatrices $y=\pm \sqrt{a}x$ and a saddle at the origin. For $a<0$, they are concentric ellipsestraversed clockwise. This is the streamline classification of a planar linear saddle or centre.

The scalar vorticity isHence the flow is irrotational exactly whenThen $(u,v)=(y,x)=\nabla \varphi$, and a velocity potential is

The coefficient condition says precisely that every nonconstant monomial has positive powers of both variables. ThusIf $c+XYf$ and $d+XYg$ lie in $R$, so do their sum and their productso $R$ is a subring of $K[X,Y]$.

For $n\ge 1$, letThen $I_n\subseteq I_{n+1}$. The containment is strict because $X{Y}^{n+1}\notin I_n$: multiplying a generator $X{Y}^j$ by an element of $R$ produces either a scalar multiple of $X{Y}^j$, or terms divisible by $X^2$. It cannot produce the monomial $X{Y}^{n+1}$ when $j\le n$. Henceis a strictly increasing ideal chain. The constant-plus-ideal non-Noetherian subring therefore satisfies

The steady Euler equations for an inviscid fluid areTaking the scalar product with $u$ and usinggivesThusThe Bernoulli function $H$ is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.

Let $y(t)$ be the downward displacement of the free surface from its initial level, and let $U$ be the speed in the tube. Conservation of volume givesThe free surface and the outlet are both at atmospheric pressure, and their vertical separation is $H+h_0-y$. Applying the Bernoulli equation between them, while retaining the small free-surface speed, givesThereforeThe surface reaches the upper tube end when $y=h_0$. Hence the draining time of a uniform tank through a siphon is

Fix $y\in V$. The map $x\mapsto \theta (x,y)$ is a linear functional, so the finite-dimensional Riesz representation theorem gives a unique vector $\beta (y)$ such thatFor scalars $\lambda ,\mu$,Uniqueness givesso $\beta$ is linear.

For $\alpha \in \mathrm{End}(V)$, apply this result toIt produces a unique linear map ${\alpha }^{\ast }$ satisfyingwhich proves existence and uniqueness of the adjoint.

Differentiate the spatial equation with respect to $t$ and the temporal equation with respect to $x$. Substitution ofinto ${\psi }_{xt}-ik{\psi }_t$ cancels every term involving $\psi$, $u$, and $u_x$, leavingThe left side is $u_t$ by differentiating ${\psi }_x-ik\psi =u$, so consistency is exactly $u_t=i{u}_{xx}$. This is a scalar Lax pair formulation.

Taking the half-line Fourier transform and integrating twice by parts givesMultiplication by $e^{i{k}^{2}t}$ and integration in time yields

Replace $k$ by $-k$ and subtract. The unknown Neumann boundary value cancels:For $x>0$, Fourier inversion of the zero extension givesInverting the preceding identity therefore givesThusThis elimination is the global relation for the half-line free Schrodinger equation.

Lorentz invariance requires $A^{\mu }$ and $J^{\mu }$ to transform as four-vectors, $\chi$ as a scalar, $m$ and ${\mu }_0$ as scalars, and the integration domain and boundary conditions to respect the transformation. Then every contracted term and $d^{4}x$ are Lorentz invariant.

Under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$, the field strength is unchanged. The source term changes by a boundary term pluswhich vanishes for a conserved current and suitable boundary behaviour of $\chi$. The mass term is not gauge invariant. Thus the action has the usual local gauge invariance whenwith the gauge function or its boundary contribution suitably controlled. For $m>0$, the mass term breaks this gauge symmetry.

For the static point source, take $A=0$ and $\alpha =0$. The scalar equation becomesIts spatial Fourier transform satisfiesThe stated inverse transform therefore giveswhere $\lambda$ absorbs the transform normalization and source strength. This is the Yukawa potential: a nonzero vector-field mass screens the interaction beyond the range $m^{-1}$. As $m\to 0$, it tends to the long-range Coulomb potential $\lambda /r$. The same limit removes the mass term that broke gauge invariance in part (b).

For scalar $\theta$, Taylor's theorem about the MLE, where
${\ell }_n^{\prime }({\hat{\theta }}_n)=0$, gives some ${\tilde{\theta }}_n$ between ${\theta }_0$ and ${\hat{\theta }}_n$ such thatThe Wald statistic isUnder the null, consistency gives
${\hat{\theta }}_n,{\tilde{\theta }}_n\to {\theta }_0$ in probability. A uniform law of large numbers for the observed information on a neighborhood of ${\theta }_0$ givesCancellation of the common squared displacement and Slutsky's theorem prove the Wald and likelihood-ratio asymptotic equivalence

Suppose the spectrum were empty. The resolventwould then be an entire operator-valued function. For $|\lambda |>\Vert T\Vert$,because the series converges in operator norm and multiplication by $\lambda I-T$ telescopes to $I$. Hence $\Vert R(\lambda )\Vert \le (|\lambda |-\Vert T\Vert {)}^{-1}$.

For fixed $x,y\in {\ell }^2$, the scalar function $⟨R(\lambda )x,y⟩$ is entire, bounded outside a disc by the estimate and bounded inside by compactness. Liouville's theorem makes it constant, and its limit at infinity makes that constant zero. This for all $x,y$ would imply $R(\lambda )=0$, impossible for an inverse. Thus the spectrum is nonempty.

An eigenvalue $\lambda$ is a scalar for which $\ker (A-\lambda I)$ contains a nonzero vector; that kernel is its eigenspace. The characteristic polynomial has degree $n$, so the fundamental theorem of algebra supplies at least one complex root and hence an eigenvalue. Its eigenspace dimension can be any integer from $1$ to $n$.

The fixed points are $0$ and, when $a\ge -1$, $1\pm \sqrt{1+a}$. For a scalar autonomous equation, a simple fixed point is stable when $f^{\prime }<0$. Since $f^{\prime }(0)=a$ and $f^{\prime }(r)=2r(1-r)$ at either nonzero root, excluding the bifurcation values as requested,and

The disc model isand the upper half-plane model isThe Cayley mapmaps $D$ bijectively to $\mathfrak{h}$. Since $C^{\prime }(z)=2i/(1-z{)}^2$ and $\mathrm{Im}C(z)=(1-|z{|}^2)/|1-z{|}^2$, direct substitution gives $C^{\ast }{g}_{\mathfrak{h}}=g_D$.

Representing $C$ by $K=\left(\begin{array}{cc}i& i\\ -1& 1\end{array}\right)$, the disc isometry corresponding to $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is represented, up to a nonzero scalar, by

The dual $X^{\ast }$ is the vector space of bounded linear maps $f:X\to \mathbb{F}$, with normIf $(f_n)$ is Cauchy in this norm, then $(f_n(x))$ is Cauchy in the complete scalar field for every $x$. Define $f(x)={\lim }_n{f}_n(x)$. Pointwise passage to the limit preserves linearity. Given $\epsilon >0$, choose $N$ such that $\Vert f_n-f_m\Vert <\epsilon$ for $m,n\ge N$; sending $m\to \infty$ givesThus $f$ is bounded and $\Vert f_n-f\Vert \le \epsilon$. Hence $X^{\ast }$ is Banach, even if $X$ was only normed.

Let $E_2=(e_1,e_2)$ andBecause $Q_k{R}_k$ is the QR factorization of $H^{k+1}$, $S_k$ is the span of the first two columns $q_1^{(k)},q_2^{(k)}$ of $Q_k$.

The first row of the tridiagonal eigenvalue equation isIn the generic case ${\lambda }_{n-1}\ne {\lambda }_n$, the two dominant coefficient vectors are independent, sinceHere $h_{12}\ne 0$, since otherwise the displayed eigenvalue equation and the nonzero $b_j$ would force all eigenvalues to equal $h_{11}$. The strict spectral gap below the dominant pair now impliesThis is two-dimensional subspace iteration. If the dominant eigenvalue is repeated, $H$ acts as a scalar on $W$; the same QR argument either captures all of $W$ or separates its missing direction from the strictly smaller eigenspaces, and the conclusion below is unchanged.

Since $q_2^{(k)}\in S_k$ and $q_3^{(k)}\perp S_k$,The space $W$ is invariant under $H$, soShifting the index does not affect the limit, and thereforeThus the unshifted QR algorithm asymptotically deflates a $2\times 2$ block associated with the equal-modulus dominant pair.

Isotropic means invariant under every proper rotation. The general ranks $0,1,2,3$ are respectively a scalar $a$, zero, $b{\delta }_{ij}$, and $c{\epsilon }_{ijk}$ (for full orthogonal invariance the last must also vanish).

The Möbius group consists of maps $z\mapsto (az+b)/(cz+d)$ with $ad-bc\ne 0$, including the usual action on $\infty$. Scalar matrices act trivially and are the whole kernel, so $M\cong G{L}_2(\mathbb{C})/({\mathbb{C}}^{\ast }I)$.

By Schur lemma, $z\in Z$ acts on irreducible $V$ by a scalar. Induced representation shows that $V$ occurs in ${\mathrm{Ind}}_Z^G\Theta$ iff that scalar character is $\Theta$, and then with multiplicity $\dim V$.