A map $T:V\to W$ between vector spaces over the same field $\mathbb{F}$ is a linear map precisely when it preserves vector addition and scalar multiplication:Equivalently, $T(au+bv)=aT(u)+bT(v)$ for all $u,v\in V$ and $a,b\in \mathbb{F}$.

A vector space over a field $\mathbb{F}$ is a set $V$ with vector addition $V\times V\to V$ and scalar multiplication $\mathbb{F}\times V\to V$ satisfying the following vector space axioms. Under vector addition, $V$ is an abelian group: addition is associative and commutative, there is a zero vector $0_V$, and each $v$ has an additive inverse $-v$. For all $a,b\in \mathbb{F}$ and $u,v\in V$,The operations are defined for every indicated pair and take their values in $V$.

Write $A=I_3+N$, whereThe binomial theorem for commuting matrices applies because $I_{3}N=N{I}_3$. All terms of degree at least three vanish, so for every positive integer $n$,The same expression directly gives $A$ when $n=1$, so no small exponent is excluded.

If $A$ is a strictly upper triangular matrix, then $A_{ij}=0$ unless $j-i\ge 1$. We prove by mathematical induction thatThe assertion for $k=1$ is the definition. In the matrix multiplication formulaa nonzero summand would require $\ell -i\ge k$ and $j-\ell \ge 1$, hence $j-i\ge k+1$. This proves the induction step. Since $j-i\le n-1$ for all entries, $\boxed{A^n=0}$. In particular, every strictly upper triangular matrix is a nilpotent matrix.

Perform the elementary row operations $R_2\leftarrow R_2+2R_1$ and $R_3\leftarrow R_3-R_1$, then $R_2\leftarrow R_2/2$, $R_1\leftarrow R_1-R_2$, and finally $R_1\leftarrow R_1+R_3$, $R_2\leftarrow R_2-3R_3$:Consequently the reduced row echelon form is $I_3$. Applying the same operations to the augmented column $b=(b_1,b_2,b_3{)}^T$ givesThus every $b\in {\mathbb{R}}^3$ admits exactly one solution. This follows because the elementary row operations preserve the solution set of the augmented linear system, and the reduced coefficient matrix is the identity matrix.

For an $m\times n$ matrix, the corresponding elementary matrices are $m\times m$, because the operation is implemented by left matrix multiplication. Apply the desired elementary row operation to the identity matrix $I_m$. Writing $E_{ij}$ for the matrix unit with its only nonzero entry, $1$, in position $(i,j)$ gives:Indeed, these matrix products interchange rows $i,j$, scale row $i$, and add $c$ times row $j$ to row $i$, respectively. The displayed inverses undo precisely those operations. Algebraically, $E_{ij}^2=0$ when $i\ne j$, so $(I_m+c{E}_{ij})(I_m-c{E}_{ij})=I_m$; the other two inverse identities follow immediately from their row actions. Thus every elementary matrix is an invertible matrix.

The three elementary row operations on an $m\times n$ real matrix are:

Each operation changes whole rows, so it acts on every column by the same rule. Each can be undone by an elementary row operation.

Eliminate the two pure second derivatives by taking $\alpha ,\beta$ to be the roots of $z^2+2Bz+C=0$. They are distinct real numbers exactly whenThus the required function is $H(B)=B^2$. The mixed coefficient becomes $4(C-B^2)\ne 0$, so the transformed equation is $y_{\xi \eta }=0$. Integrating once in each coordinate giveswith arbitrary sufficiently smooth functions $F,G$. This is the characteristic form of a strictly hyperbolic partial differential equation.

The multivariable chain rule gives ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ and ${\partial }_t=\alpha {\partial }_{\xi }+\beta {\partial }_{\eta }$. Substitution into the differential equation yieldsThe coordinate transformation is invertible exactly when $\alpha \ne \beta$.

At $x=L/2$, the integration interval first moves through positive arguments, then crosses zero, and finally contains all of $[-L,L]$. The result isThe final plateau is the integral of the entire initial velocity profile divided by $2c$.

At $ct=L/2$, the interval of integration has half-width $L/2$. Evaluating the cumulative integral in part a gives the even functionIt has value $L^2/(8c)$ at zero and maxima $L^2/(4c)$ at $x=\pm L/2$, then falls to zero at the two wavefronts.

Take the boundary pieces in the orderOn the circular arc $C_1$, use $(x,y,z)=(2\cos \theta ,2\sin \theta ,0)$ for $0\le \theta \le \pi /2$. Then $\mathbf{F}=(0,0,1)$ and $dz=0$, so its line integral is zero.

On $C_2$, use $x=0$, $z=4-y^2$, with $y$ decreasing from two to zero. Since $dx=0$ and $F_y=0$, $\mathbf{F}\cdot d\mathbf{r}=dz$, giving four. On $C_3$, use $y=0$, $z=4-x^2$, with $x$ increasing from zero to two. Again $\mathbf{F}\cdot d\mathbf{r}=dz$, giving minus four. The total is $\boxed{0+4-4=0}$, agreeing with Stokes theorem.

The curl is $\nabla \times \mathbf{F}=(x,y,-2z)$. For the upward orientation of the paraboloid, the vector surface area element is $(2x,2y,1)dxdy$. Therefore Stokes theorem givesThe upward normal gives the prescribed anticlockwise boundary orientation viewed from above.

For $f(t)=\sin 2t$, the positive-$t$ branch is above the negative one, so the orientation in part i is outward. ConsequentlyThe integral follows from ${\sin }^{2}2t=(1-\cos 4t)/2$.

Parametrize the surface of revolution byAn oriented vector surface area element isUse the vector field $\mathbf{F}=(x/2,y/2,0)$, whose divergence is one. If this parametrization has the outward orientation on an enclosing surface, the divergence theorem givesThe second equality is integration by parts, using the endpoint values. For example this outward convention applies when the positive-$t$ branch lies above the negative-$t$ branch, $f(t)>f(-t)$ for $0<t<\pi /2$.

The stated $C^1$ regularity and endpoint conditions alone do not guarantee an embedded enclosing surface: branches can cross or coincide. Without an appropriate orientation assumption the flux formula is an algebraic signed volume. If “inside” is defined as the region between the two graph branches for each radius, its nonnegative geometric volume is insteadThis explicitly handles a reversal or change in the order of the two branches.

Take $v=|\mathbf{x}{|}^2$ in Green's second identity. Since ${\nabla }^{2}u=0$, ${\nabla }^{2}v=6$, $v=1$ on the unit sphere, and ${\partial }_{n}v=2$ there,The final term vanishes by the divergence theorem, because it equals ${\iiint }_B{\nabla }^{2}udV=0$. ThereforeWe assume sufficient boundary regularity for the stated boundary values and integrals, or use the corresponding trace/limit formulation. This is the equality of volume and surface averages for a harmonic function, since the unit ball and sphere have volume $4\pi /3$ and area $4\pi$.

Apply the divergence theorem to $\mathbf{F}=u\nabla v-v\nabla u$. The product rule givesbecause the two gradient inner products cancel. Also $\mathbf{F}\cdot \mathbf{n}=u{\partial }_{n}v-v{\partial }_{n}u$. ThusThis is Green's second identity, under the usual regularity assumptions permitting these derivatives and boundary integrals.

For a bounded region $R\subseteq {\mathbb{R}}^3$ with piecewise smooth boundary and a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of its closure, the divergence theorem sayswhere $\mathbf{n}$ is the outward unit normal vector. The left side integrates divergence over the volume; the right side is the outward flux.

The truncated Fourier cosine series gives the better approximation on the closed interval, in the senses of uniform convergence and the asymptotic L2 norm error. To make the comparison precise, let $C_N$ and $S_N$ retain the frequencies $1\le n\le N$, with the constant term also included in $C_N$.

The cosine Fourier coefficients are $O(n^{-2})$. In fact, absolute convergence gives the uniform tail boundFor the Fourier sine series, every Fourier partial sum satisfies $S_N(0)=S_N(\pi )=0$. Thusso there is no uniform convergence to $f$ on the closed interval. Its odd extension has jump discontinuities, producing the Gibbs phenomenon near the endpoints; its Fourier coefficients decrease only like $1/n$.

The advantage persists in the L2 norm, which ignores individual endpoint values. By Parseval's identity and Fourier orthogonality,The nonzero odd cosine coefficients and both parity classes of sine coefficients also give matching lower bounds of these orders by comparison with the corresponding p-series. Hence their L2 norm errors decrease respectively like $N^{-3/2}$ and $N^{-1/2}$. This is a comparison of approximation quality in specified norms, not a claim that one finite sum has smaller error at every individual interior point.

For the Fourier cosine series, integration by parts givesand, for $n\ge 1$,Only odd Fourier harmonics remain, so

For the Fourier sine series, another integration by parts givesConsequentlyAt $x=0$ and $x=\pi$, the Fourier sine series instead has value zero.

To describe the requested graphs precisely, take $t\in [-\pi ,\pi )$ congruent to $x$ modulo $2\pi$. The Fourier cosine series converges to the continuous triangular periodic extensionIts minima occur at the even multiples of $\pi$, with value $1$, and its maxima at the odd multiples, with value $1+\pi$. The Fourier sine series converges toThe jump discontinuities at even multiples of $\pi$ go from $-1$ to $1$; those at odd multiples go from $1+\pi$ to $-(1+\pi )$. Each jump is assigned its midpoint, zero, by the Fourier series.