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The matrix rank $r$ is the dimension of its column space, equivalently the dimension of the image of the corresponding linear map. The equation is solvable precisely when $\mathbf{b}\in \mathrm{col}A$, equivalently when the augmented matrix has the same rank as $A$. If ${\mathbf{x}}_0$ is one solution, every solution is ${\mathbf{x}}_0+\mathbf{u}$ with $\mathbf{u}\in \ker A$, since subtracting any two solutions gives a kernel element. The rank-nullity theorem gives $\dim \ker A=3-r$.

Consequently, for rank three there is one solution for every $\mathbf{b}$, namely $A^{-1}\mathbf{b}$. For rank two there are no solutions outside the column plane, and an affine solution space of a linear equation of dimension one when consistent. For rank one, consistency requires $\mathbf{b}$ to lie on the column line and the solution set is an affine plane of dimension two. For rank zero, $A=0$: the solution set is all of ${\mathbb{R}}^3$ if $\mathbf{b}=0$, and empty otherwise. These descriptions include homogeneous systems, whose solution spaces pass through the origin.

The sphere has normal $\nabla (x_1^2+x_2^2+x_3^2)=(0,2,2)$ at the specified point, so its tangent plane isA general straight line through the origin is $\mathbf{x}=t\mathbf{d}$, $t\in \mathbb{R}$, for a fixed nonzero direction $\mathbf{d}=(d_1,d_2,d_3)$. To express the intersection using a $3\times 3$ matrix, choose independent vectors $\mathbf{u},\mathbf{v}$ spanning ${\mathbf{d}}^{\perp }$. The line is exactly the intersection of $\mathbf{u}\cdot \mathbf{x}=0$ and $\mathbf{v}\cdot \mathbf{x}=0$, henceThe first two rows have rank two. The third is independent of them exactly when $(0,1,1)\cdot \mathbf{d}=d_2+d_3\ne 0$. In that case the rank is three and the unique intersection isGeometrically the line crosses the tangent plane transversely. If $d_2+d_3=0$, the matrix rank is two: its third row is a linear combination of the first two, but the corresponding right-hand-side combination would be zero rather than two. The system is inconsistent and there is no intersection. The line is parallel to the tangent plane and cannot lie in it, since it contains the origin while the plane does not. Thus no choice of a genuine direction produces an infinite intersection in this particular problem.

Suppose, towards a contradiction, that $|a_n{z}^n|\le M$ for every $n$, where $|z|>R$. Choose a radius $\rho$ strictly between $R$ and $|z|$, and take any $w$ with $|w|=\rho$. ThenThe geometric series on the right converges because its ratio lies in $[0,1)$. The comparison test for series therefore gives absolute convergence of the power series at $w$, contrary to the defining property of its radius of convergence. Thus $|a_n{z}^n|$ is unbounded in $n$ whenever $|z|>R$. This is the bounded power-series terms force interior absolute convergence principle. The case $R=\infty$ has no points outside its convergence disk and the assertion is vacuous there.

Every possible radius of convergence is realized by an explicit power series:For the first, consecutive absolute terms have ratio $(n+1)|z|$, so at every $z\ne 0$ they eventually grow and do not tend to zero; at zero the series is defined. The middle geometric series converges absolutely exactly for $|z|<R$, and outside that disk its terms grow. The last has consecutive-term ratio $|z|/(n+1)\to 0$, so the ratio test gives absolute convergence for every finite $z$. These arguments use the necessary zero-term condition for convergence, the comparison test for series, and absolute convergence of a geometric series with ratio less than one; they do not assume convergence on the boundary.

A real number $s$ is the supremum of a nonempty set $A\subset \mathbb{R}$ if every $a\in A$ satisfies $a\le s$ and no smaller real number is an upper bound. Equivalently, for every $\eta >0$ there is $a\in A$ with $s-\eta <a\le s$. A greatest element, or maximum, additionally belongs to $A$.

Let $s=\sup A$ and suppose $s\notin A$. Fix $\epsilon >0$. The approximation property gives at least one element of $A$ in $(s-\epsilon ,s)$. If there were only finitely many, they would have a largest value $m<s$. Every other element of $A$ is at most $s-\epsilon$, so $\max \{m,s-\epsilon \}<s$ would be an upper bound for the whole set. This contradicts the definition of the supremum. Therefore there are infinitely many such elements, the accumulation below an unattained supremum.

For a counterexample when the supremum is attained, take $A=\{0\}$. Its supremum and greatest element are both zero, while every interval $(-\epsilon ,0)$ contains no elements of $A$.

Expanding the cross product component by component gives its cross-product matrix:It is a real skew-symmetric matrix and annihilates $\mathbf{n}$. By the vector triple product,so $A^2=\mathbf{n}{\mathbf{n}}^T-I$ for a unit vector $\mathbf{n}$. Choose a unit vector $\mathbf{u}\perp \mathbf{n}$ and put $\mathbf{v}=\mathbf{n}\times \mathbf{u}$. Then $(\mathbf{u},\mathbf{v},\mathbf{n})$ is an orthonormal basis and $A\mathbf{u}=\mathbf{v}$, $A\mathbf{v}=-\mathbf{u}$, $A\mathbf{n}=0$. Thus the characteristic polynomial is $\lambda ({\lambda }^2+1)$ and the cross-product matrix spectrum isOver the complex numbers, corresponding eigenvectors are $\mathbf{n}$, $\mathbf{u}-i\mathbf{v}$, and $\mathbf{u}+i\mathbf{v}$. Only the zero eigenvalue has real eigenvectors.

For the specified third coordinate axis, direct multiplication givesIts zero eigenspace is $\mathrm{span}\{(0,0,1)\}$, and its $-1$ eigenspace is the whole plane $\mathrm{span}\{(1,0,0),(0,1,0)\}$. Hence the eigenvalues of $A^2$ are $-1,-1,0$, with every nonzero vector in the appropriate eigenspace an eigenvector.

Rolle theorem states that if $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there is $c\in (a,b)$ with $f^{\prime }(c)=0$.

If $f$ is constant, any interior point works. Otherwise the extreme value theorem gives an attained maximum and minimum on the interval, and at least one of them differs from the common endpoint value. That extremum must occur at an interior point $c$. At an interior maximum, for small positive $h$ the difference quotient $[f(c+h)-f(c)]/h$ is nonpositive, while for negative $h$ it is nonnegative. Existence of the common derivative forces it to be both nonpositive and nonnegative, hence zero. The same argument with inequalities reversed applies to an interior minimum. This proves Rolle theorem, including the derivative-at-an-extremum step rather than assuming it.

Now let the nonconstant polynomial $p$ have distinct real roots $r_1<\cdots <r_k$, with multiplicities $m_1,\dots ,m_k$ adding to its degree $n$. If $p(x)=(x-r_j{)}^{m_j}{q}_j(x)$ and $q_j(r_j)\ne 0$, differentiation givesThe bracket is nonzero at $r_j$, so its multiplicity as a root of $p^{\prime }$ is exactly $m_j-1$. These roots already contribute ${\sum }_j(m_j-1)=n-k$ real roots counted with multiplicity. Rolle theorem supplies an additional root of $p^{\prime }$ in each of the $k-1$ disjoint intervals $(r_j,r_{j+1})$. These additional roots are distinct from each other and from all the $r_j$.

We have therefore exhibited at least $(n-k)+(k-1)=n-1$ real roots counted with multiplicity. The derivative has degree $n-1$, so there is no room for any additional nonreal roots. Every root of $p^{\prime }$ is real. This is the Rolle root count with multiplicities; for $n=1$ the nonzero constant derivative has no roots, as expected.

For failure of the converse, takeThe derivative has only the real root zero, of multiplicity two. But $p(x)=(x+1)(x^2-x+1)$ and the quadratic factor has roots $(1\pm i\sqrt{3})/2$, which are nonreal. Thus real-rootedness of the derivative does not imply real-rootedness of the original polynomial.

To prove the required zero-existence statement directly, let $S=\{x\in [a,b]:f(x)\le 0\}$ and $c=\sup S$. This set is nonempty and bounded. Continuity and the strict signs at the endpoints give a right neighbourhood of $a$ with negative values and a left neighbourhood of $b$ with positive values; consequently $a<c<b$. If $f(c)<0$, continuity would give points of $S$ to the right of $c$, contradicting its upper-bound property. If $f(c)>0$, continuity would give an interval immediately to the left of $c$ containing no points of $S$, contradicting the approximation property of its supremum. Hence $f(c)=0$ for some $c\in (a,b)$, proving this form of the intermediate value theorem.

For the half-interval claim, put $H(x)=g(x+1/2)-g(x)$ on $[0,1/2]$. It is a continuous function andIf either endpoint value is zero, it supplies the answer. Otherwise the signs are opposite, so the intermediate value theorem, applied also to $-H$ if necessary, gives an interior zero. Thus $g(c+1/2)=g(c)$ for some $c\in [0,1/2]$.

For general positive integer $n$, define $H_n(x)=g(x+1/n)-g(x)$ on $[0,(n-1)/n]$. At its grid points,If a sampled value is zero, it is the desired point. If none is zero, not all sampled values can have the same sign, so two have opposite signs; continuity gives a zero between their grid points. This proves the horizontal chord of reciprocal-integer length:For $n=1$ the interval is the single point zero, which works by the endpoint equality itself. The proof does not require a differentiable function or a periodic extension.

For a bounded real function on $[a,b]$, take a partition $a=t_0<\cdots <t_m=b$. Its lower and upper Darboux sums areThe function is Riemann integrable when ${\sup }_{P}L(f,P)={\inf }_{P}U(f,P)$; the common value is its Riemann integral. Equivalently, for every $\epsilon >0$ there is a partition with $U(f,P)-L(f,P)<\epsilon$. A continuous function is bounded and Riemann integrable on each compact interval, so all the following integrals exist.

Set $h(x)=x^{-1}{\int }_0^{x}f(t)dt$ for $x>0$. Strict decrease gives $f(x)<f(t)<f(0)$ for $0<t<x$. Integrating, with strict inequalities because the difference is positive on interior subintervals, givesThe intermediate value theorem produces $g(x)\in (0,x)$ with $f(g(x))=h(x)$, and strict monotonicity makes this point unique. Thus there is exactly one such $g(x)$ in $(0,x)$.

For the given exponential function, $h(x)=(1-e^{-x})/x$, soThe preceding strict bounds already prove $0<g(x)<x$; no limiting value at zero needs to be included in its domain.

For differentiability, a continuous strictly monotone function has a continuous inverse on its range. If it is differentiable at an interior point $t$ with $f^{\prime }(t)\ne 0$, the inverse is differentiable at $y=f(t)$, with derivative $1/f^{\prime }(t)$. Indeed, setting $u=f^{-1}(y^{\prime })$, inverse continuity gives $u\to t$ as $y^{\prime }\to y$, andThis one-dimensional inverse result needs no assumption that $f^{\prime }$ is continuous. It is enough here that $f$ is differentiable and $f^{\prime }<0$ at every positive argument.

The fundamental theorem of calculus and the quotient rule giveNow $g=f^{-1}\circ h$ and $g(x)>0$, so the inverse derivative formula applies at every relevant point. The monotone mean-value point of an integral average satisfiesBoth numerator and denominator are negative. For the exponential example, this also agrees with $g^{\prime }(x)=1/x-1/(e^x-1)>0$, since $e^x>1+x$ for $x>0$.

Write $S_{xx}={\sum }_i{x}_i^2$. When $S_{xx}>0$, minimizing ${\sum }_i(Y_i-b{x}_i{)}^2$ gives the strictly convex quadratic normal equation $b{S}_{xx}={\sum }_i{x}_i{Y}_i$. The least-squares estimator is thereforeThis is linear regression through the origin, with no fitted intercept. For $n\ge 2$, let $\mathrm{SSE}={\sum }_i(Y_i-\hat{\beta }{x}_i{)}^2$ and $s^2=\mathrm{SSE}/(n-1)$. An orthonormal change of coordinates in the Gaussian vector $Y$ separates the unit design direction $x/\sqrt{S_{xx}}$ from its $n-1$ orthogonal residual directions. The first coordinate is normal with mean $\beta \sqrt{S_{xx}}$ and variance ${\sigma }^2$; the remaining coordinates are independent centered normals with that variance. Consequentlyand these variables are independent. Thus under the null, the Student t test for regression through the origin statistic isFor a prescribed significance level $\eta$, reject the two-sided null when $|T|>t_{n-1,1-\eta /2}$. The two-sided p-value is $2[1-F_{t_{n-1}}(|T|)]$. There is one fitted coefficient, so the variance degrees of freedom are $n-1$, not $n-2$.

The assumptions $S_{xx}>0$ and $n\ge 2$ matter. If all $x_i=0$, the distribution contains no information about $\beta$ and no informative test is possible. With one nonzero design observation there is no residual variance degree of freedom, so the stated unknown-variance t test cannot be constructed.

Taking the curl of the Euler equation gives the vorticity equation for an inviscid incompressible flow without body forces:The given velocity has divergence $-1+0+1=0$ and curl $(0,0,\omega (t))$. Its vorticity is spatially uniform, so its advective derivative vanishes. The stretching term is $\omega {\partial }_{z}u=(0,0,\omega )$. ThusThe signed vertical component is ${\omega }_0{e}^t$; its magnitude is $|{\omega }_0|e^t$, with the source's magnitude convention taking ${\omega }_0\ge 0$. As an additional consistency check, after imposing this equation the material acceleration is $(x,0,z-1)$, so a pressure $p=-\rho [x^2+(z-1{)}^2]/2+p_0(t)$ satisfies the full Euler equation.

The four-momentum is $(E/c,p_x,p_y,p_z)$. It transforms by the same Lorentz boost as other four-vectors. For an observer moving at $v$ in the positive $x$ direction, its time and longitudinal components transform asSubstituting the given momentum components proves all three displayed transformation relations.

A photon has zero rest mass, so the relativistic energy-momentum relation gives $E=cp$ on the positive-energy branch. Lorentz invariance preserves $E^2-c^2{p}^2=0$, and $E^{\prime }=\gamma E(1-(v/c)\cos \theta )>0$ for $|v|<c$. Therefore $E^{\prime }=c{p}^{\prime }$. Dividing the transformed longitudinal momentum by $p^{\prime }=E^{\prime }/c$ gives the relativistic aberration formulaAs $v\to c^{-}$, for every fixed $\theta \ne 0$ the ratio tends to $-1$: photon directions are concentrated along the negative $x$ direction in the observer's frame. The exactly forward photon is exceptional: if $\theta =0$, then $\cos {\theta }^{\prime }=1$ for every $v<c$, and its energy tends to zero. For other fixed angles $E^{\prime }$ diverges. The limiting angular behavior is not uniform near $\theta =0$, and $v=c$ itself is not an admissible inertial observer velocity.