Yes. A surjective $T_A$ has $\mathrm{im}{T}_A={\mathbb{R}}^n$, so every $\mathbf{b}$ is attained. Its rank of a linear map is $n$, and the rank-nullity theorem givesThe solution set of a linear system is therefore a translate of a nonzero vector subspace, containing infinitely many solutions as in iii.

No. The rank-nullity theorem givesbut this guarantees infinitely many solutions only when the linear system is consistent. For a counterexample takeThere is no solution, since the left side always vanishes.

No. Take $A=I_m$ and $\mathbf{b}=0$. The homogeneous linear system then has only the solution $\mathbf{x}=0$. Homogeneity ensures existence, but infinitely many solutions would additionally require a nonzero kernel.

No. Here $A$ has more rows than columns, which does not force a nonzero kernel. For example, with $m=1$, $n=2$,the linear system has the unique solution $x=0$. With the same $A$ and $\mathbf{b}=(0,1{)}^T$ it has no solution at all.

If $\mathbf{b}$ is outside the image of a linear map $T_A$, there is no solution. Otherwise choose ${\mathbf{x}}_0$ with $A{\mathbf{x}}_0=\mathbf{b}$. By ii, a zero kernel gives a unique solution. If the kernel contains a nonzero vector $\mathbf{z}$, allare solutions. They are distinct, since $(t-s)\mathbf{z}=0$ forces $t=s$. Hence there are infinitely many. These alternatives exhaust all possibilities.

Suppose ${\mathbf{x}}_0$ is one solution. By linearity,Thus the solution set of a linear system is exactly ${\mathbf{x}}_0+\ker T_A$. It has one member if the kernel is $\{0\}$. Conversely, if there is a unique solution and $\mathbf{z}\in \ker T_A$, then ${\mathbf{x}}_0+\mathbf{z}$ is also a solution, forcing $\mathbf{z}=0$. Combining this with i gives

By definition, the image of a linear map $T_A$ isConsequently $\mathbf{b}\in \mathrm{im}{T}_A$ means exactly that some $\mathbf{x}\in {\mathbb{R}}^m$ satisfies $A\mathbf{x}=\mathbf{b}$. This is precisely the existence of a solution of the linear system.

Put $w_1=v_1$ and $w_j=v_j-v_1$ for $j\ge 2$. Since $v_1=w_1$ and $v_j=w_j+w_1$, the two families have the same linear span. Moreover,If the $v_j$ are linearly independent, a vanishing expression forces $a_j=0$ for $j\ge 2$ and then $a_1=0$. Conversely,so linear independence of the $w_j$ forces all $b_j=0$. Thus both the spanning set condition and linear independence are equivalent for the two families, proving the desired equivalence of bases.

The Steinitz exchange lemma states that if $u_1,\dots ,u_r$ are linearly independent and $w_1,\dots ,w_s$ span a vector space $V$, then $r\le s$ and some $r$ of the $w_j$ can be replaced by $u_1,\dots ,u_r$ while preserving the spanning set property.

Here is the exchange argument. Suppose that after $k$ replacements, a spanning set iswhere the remaining $w_j$ have been relabelled if necessary. Express $u_{k+1}$ as a linear combination of this family. At least one remaining $w_j$ has a nonzero coefficient; otherwise $u_{k+1}$ belongs to the span of $u_1,\dots ,u_k$, contradicting linear independence. Relabel this entry $w_{k+1}$. Solving for $w_{k+1}$ shows that it lies in the span ofThe new family therefore still spans $V$. Starting at $k=0$ repeats this step until all $r$ independent vectors have been inserted. If $r>s$, after $s$ steps the next $u_{s+1}$ would lie in the span of $u_1,\dots ,u_s$, a contradiction. Thus $r\le s$, and the construction proves the replacement assertion as well.

The vectors are linearly independent ifThey span $V$ if every $v\in V$ is a linear combination $v={\sum }_{j=1}^n{a}_j{v}_j$ with real coefficients. They form a basis of $V$ if they are linearly independent and span $V$. Equivalently, every $v\in V$ has exactly one such representation: existence is the spanning set condition, and subtracting two representations shows that uniqueness is precisely linear independence.

For the thin spherical charge layer limit, smoothness of $h$ givesThus the shell's total electric charge tends to $4\pi {\rho }_0{a}^3{h}^{\prime }(1)$ while its thickness tends to zero. At each fixed $r>a$, the observation point eventually lies outside the shell, whereas at each fixed $r<a$ the interior electric field is unchanged. The limiting electric field away from the sphere is thereforeIts outward-minus-inward jump isPhysically, the limit is a uniformly charged ball of charge density ${\rho }_0$, together with an infinitesimally thin spherical shell of uniform surface charge densityIndeed its electric charge is $4\pi a^2\sigma$, and the standard normal electric field jump is $\sigma /{ϵ}_0$. The surface charge density can have either sign, or vanish, according to $h^{\prime }(1)$.

The order of limits matters. For every finite $\delta >0$, taking the two one-sided limits at $r=a$ first gives zero jump. The displayed nonzero jump is obtained by first taking $\delta \to 0$ at fixed points on either side, then approaching $a$. The value at the surface itself need not coincide with either trace of the limiting electric field.

Set $r=R$ in the enclosed electric charge formula:Outside the entire charge distribution, Gauss's law gives $E_r(r)=Q/(4\pi {ϵ}_0{r}^2)$. Integrating $E_r=-{\phi }^{\prime }$ with $\phi (\infty )=0$ givesThus the exterior electric potential and electric field are exactly those of a point charge $Q$ placed at the center. This is the electrostatic shell theorem: outside a spherically symmetric charge distribution, only its total electric charge affects the electric field, not how that electric charge is distributed radially. If positive and negative electric charges cancel to give $Q=0$, both exterior quantities vanish.

By the assumed spherical symmetry of the electric potential, the electric field is radial: $\mathbf{E}=E_r(r){\mathbf{e}}_r$ for $r>0$. On a sphere of radius $r$, Gauss's law therefore readsPut $R=a(1+\delta )$. For $r<a$, the enclosed electric charge is $4\pi {\rho }_0{r}^3/3$. In the spherical shell, the factor $s^2$ in the volume element cancels $s^{-2}$ in the charge density, soIt follows thatAt the origin the first formula gives $\mathbf{E}=0$ without needing a radial unit vector. At both shell interfaces the adjacent formulas agree: for every fixed $\delta >0$, there is no surface charge concentrated on either sphere, and the electric field is continuous there. No sign restriction on $h^{\prime }$ has been assumed, so the shell charge density and the radial electric field can be negative.

The space ${\mathbb{R}}^3$ is simply connected, and the electric field has zero curl. Hence it is a conservative vector field. More explicitly, Stokes theorem gives zero circulation around any closed loop, since such a loop bounds a spanning surface in ${\mathbb{R}}^3$. Therefore the line integral between two fixed endpoints is independent of the chosen path.

Fix a reference point ${\mathbf{r}}_0$ and define the electric potentialTaking a short straight terminal segment in each coordinate direction shows ${\partial }_j\phi =-E_j$. ConsequentlyThe electric potential is determined up to an additive constant; the later condition at infinity fixes that constant for the finite charge distribution in part (b).

For any bounded volume $V$ with piecewise smooth closed boundary and outward unit normal vector $\mathbf{n}$, Gauss's law in integral form isThus the outward electric flux equals the enclosed electric charge divided by the vacuum permittivity. To prove it, apply the divergence theorem to the electric field and use its given divergence:The classical proof assumes the requisite differentiability of the electric field. For piecewise smooth charge densities, one may apply the divergence theorem piece by piece and cancel the internal fluxes.

When $\alpha >n$, set $\mu =\sqrt{{\alpha }^2-n^2}>0$. The reduced ordinary differential equation is $F^{\prime \prime }-{\mu }^{2}F=0$. Enforcing the same initial conditions givesThe corresponding overdamped normal mode can also be writtenBoth rates in the exponential functions are negative because $0<\mu <\alpha$, so the hyperbolic sine factor does not prevent decay.

When $\alpha =n$, the reduced ordinary differential equation is $F^{\prime \prime }=0$. The initial conditions $F(0)=0$, $F^{\prime }(0)=1$ giveThis is the critically damped normal mode, also obtained by taking $\nu \to 0$ in $\sin (\nu t)/\nu$.

Substitute the separated solution $y=e^{-\alpha t}F(t)\sin (nx)$ into the damped wave equation. Its time derivatives givewhereas $y_{xx}=-n^2{e}^{-\alpha t}F\sin (nx)$. ThusThe last two conditions follow from $y(x,0)=0$ and $y_t(x,0)=\sin (nx)$. For $0<\alpha <n$, let $\nu =\sqrt{n^2-{\alpha }^2}>0$. The simple harmonic motion equation has solution of a differential equationThis is an underdamped normal mode: its oscillations have the decaying factor $e^{-\alpha t}$.

Use the same difference $w$ and wave energy as in part (i). The damped wave equation now gives the energy dissipation identity for a linearly damped stringBecause $E(0)=0$ and $E(t)\ge 0$, the energy remains zero. Hence $w_t=w_x=0$, and the homogeneous Dirichlet boundary conditions force $w=0$. Therefore the solution of a differential equation is unique also for $\boxed{\alpha >0}$; positive damping can only decrease this energy.

Let $w$ be the difference of two sufficiently regular solutions of the wave equation with the prescribed data. By linearity, $w$ has zero initial displacement, zero initial velocity, and homogeneous Dirichlet boundary conditions. Define its wave energy byFor $\alpha =0$, differentiation under the integral sign and integration by parts yieldThe boundary term vanishes because $w=0$ at both fixed endpoints implies $w_t=0$ there. The zero initial data give $E(0)=0$, so $w_t=w_x=0$ for all times. Thus $w$ is spatially constant, and its zero boundary values give $\boxed{w=0}$. This proves uniqueness for the undamped wave equation.