Regard $B$ as a linear map ${\mathbb{R}}^n\to {\mathbb{R}}^p$ and $A$ as a linear map ${\mathbb{R}}^p\to {\mathbb{R}}^m$ (the same argument holds over any field). ThenIts dimension cannot exceed either $\dim \mathrm{im}A$ or $\dim \mathrm{im}B$, since applying a linear map cannot increase dimension. The rank bound for a matrix product is thereforeThe bound $p$ is attained whenTheir product has an $I_p$ block and zeros elsewhere, so its rank of a matrix is $p$. Thus the bound depending only on the given matrix sizes is sharp.

The row rank of a matrix is the dimension of the linear span of its row vectors. Its column rank is the dimension of the linear span of its column vectors, equivalently the dimension of the image of the corresponding linear map. The equality of row rank and column rank states that these two numbers are equal for every matrix. Their common value is the rank of a matrix.

False. The real line with its usual metric is complete and is homeomorphic to $(0,1)$ with its usual metric, throughIts inverse is $h^{-1}(u)=\tan (\pi (u-1/2))$, so both maps are continuous. However, $(1/n{)}_{n\ge 2}$ is a Cauchy sequence in $(0,1)$ and has no limit there: its only possible real limit is zero. Hence $(0,1)$ is not a complete metric space. Completeness is a property of the metric, not just of its topology.

False. Give $X=\mathbb{R}$ its usual topology, let $Y$ be a one-point space, and take the constant continuous map $f:X\to Y$. The set $K=Y$ is compact, but $f^{-1}(K)=\mathbb{R}$ is not compact. For example, the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. Thus continuity alone does not preserve compactness under inverse images.

True. Let $\{V_{\alpha }\}$ be an open cover of $f(S)$ in its subspace topology. Write $V_{\alpha }=f(S)\cap W_{\alpha }$ with $W_{\alpha }$ open in $Y$. Since $f$ is continuous, the sets $S\cap f^{-1}(W_{\alpha })$ are open in $S$ and cover $S$. By compactness, finitely many cover $S$. Their corresponding $V_{\alpha }$ then cover $f(S)$, proving that the continuous image of a compact space is compact.

Define the Möbius transformationThe three prescribed distinct points go to $0,1,\infty$, respectively, and the specified cross-ratio is exactly $T(z_4)$.

To identify the preimage of the extended real line directly, write $T(z)=(az+b)/(cz+d)$. The condition $T(z)=\overline{T(z)}$ becomesAfter division by $2i$, this has the real form $A|z{|}^2+Bz+\overline{B}\overline{z}+C=0$, with $A,C$ real. If $A=0$ it is a line; otherwise completing the square gives a circle. It contains $z_1,z_2,z_3$, including the pole as a point mapped to infinity, so it is nondegenerate. Three distinct noncollinear points determine one circle; if the three are collinear, their unique generalized circle is their line.

Consequently $T(z_4)$ is real exactly when the fourth point is on the same generalized circle as the first three. This proves the real cross-ratio criterion for a generalized circle:

For the spherical circle, use a rotation to put the center at the north pole. Points at spherical distance $r$ have polar angle $r$ and admit the parametrizationIts speed is $|{\gamma }^{\prime }(\theta )|=\sin r$, so the arc length isFor small $r$, the Taylor series gives $C(r)=2\pi r-(\pi /3)r^3+O(r^5)$. Thus its circumference is smaller than the Euclidean circumference $2\pi r$ for the same intrinsic radius. Their ratio is $\sin r/r=1-r^2/6+O(r^4)$ and tends to one. The leading agreement expresses local Euclidean behavior; the cubic deficit reflects the positive Gaussian curvature of the unit sphere.

Hilbert's basis theorem: if $R$ is a commutative Noetherian ring with identity, then the polynomial ring $R[x]$ is Noetherian. Iteration gives the same assertion for every finite number of variables.

Here is a bounded-degree coefficient proof of Hilbert basis theorem. Let $I$ be any ideal of $R[x]$. For $n\ge 0$, define $J_n\subseteq R$ to be the coefficients of $x^n$ in the polynomials of $I$ of degree at most $n$, including the zero coefficient. Closure under addition and scalar multiplication makes $J_n$ an ideal of $R$. Multiplication of a polynomial by $x$ shows $J_n\subseteq J_{n+1}$. By the ascending chain condition, there is $N$ such that $J_N=J_{N+1}=\cdots$.

Each $J_n$ for $0\le n\le N$ is finitely generated. Choose generators $a_{n,1},\dots ,a_{n,r_n}$ and corresponding polynomials $f_{n,j}\in I$ of degree at most $n$ whose coefficient of $x^n$ is $a_{n,j}$. There are only finitely many chosen polynomials. We prove they generate $I$ by induction on degree.

Take a nonzero $f\in I$ of degree $d$ and leading coefficient $a$. If $d\le N$, express $a={\sum }_j{c}_j{a}_{d,j}$ and subtract ${\sum }_j{c}_j{f}_{d,j}$. If $d>N$, use $a\in J_d=J_N$, express $a={\sum }_j{c}_j{a}_{N,j}$, and subtract ${\sum }_j{c}_j{x}^{d-N}{f}_{N,j}$. In both cases the leading coefficient cancels, leaving a polynomial in $I$ of smaller degree. For degree zero the remainder is zero, and the induction completes the claim.

Thus every ideal of $R[x]$ is finitely generated, proving the Hilbert basis theorem. The proof uses no cancellation of nonzero elements of $R$, so it remains valid when $R$ has zero divisors.

A field, for example $\mathbb{Q}$, is a Noetherian ring: its only ideals are $(0)$ and the whole field, both finitely generated.

For a non-Noetherian example, take $R=\mathbb{Q}[x_1,x_2,\dots ]$, the polynomial ring in countably many variables, each individual polynomial using only finitely many variables. The ascending chain of idealsis strictly increasing. Indeed, if $x_{n+1}$ belonged to $(x_1,\dots ,x_n)$, setting the first $n$ variables to zero would give the impossible equality $x_{n+1}=0$. The chain never stabilizes, contradicting the ascending chain condition. Thus $\mathbb{Q}$ is Noetherian, while $\mathbb{Q}[x_1,x_2,\dots ]$ is not.

For $d$ variables, integrate over the product of $d$ unit circles:For multi-indices $\alpha ,\beta \in {\mathbb{Z}}_{\ge 0}^d$, applying one-variable Fourier orthogonality in each variable givesIf $f={\sum }_{\alpha }{a}_{\alpha }{z}^{\alpha }$, its squared norm is ${\sum }_{\alpha }|a_{\alpha }{|}^2$, a finite sum. The form is therefore a positive definite Hermitian form andThis is the monomial orthonormal basis on the unit torus.

For nonnegative integers $m,n$, Fourier orthogonality givesEvery polynomial is a finite linear combination of these monomials, and distinct monomials are linearly independent. ThereforeThis is an algebraic basis: no infinite linear combinations are needed in the polynomial vector space.

The integral is linear in its first argument and conjugate-linear in its second, and complex conjugation of the integrand shows $⟨g,f⟩=\overline{⟨f,g⟩}$. Thus it is a Hermitian form.

If $f(z)={\sum }_{j=0}^N{a}_j{z}^j$, Fourier orthogonality givesThis is nonnegative and vanishes only when every coefficient vanishes, that is, only for the zero polynomial. Hence the Hermitian form is positive definite.

If $\alpha =\beta$, $\psi =-\alpha (x-y{)}^2/2$. Its contours are the parallel lines $x-y=d$. On each such line,For $d\ne 0$ this is an unbounded straight streamline. Every point on $x=y$ has zero velocity, so that diagonal is a line of equilibria, rather than one moving streamline.

Rotate coordinates to $s=(x+y)/\sqrt{2}$ and $t=(x-y)/\sqrt{2}$. The stream function becomesFor $\alpha <\beta$, both coefficients are negative, so every nonzero contour is a bounded ellipse surrounding the stationary origin. For $\alpha >\beta$, the form is indefinite: nonzero contours are unbounded hyperbolas, and the zero contour consists of straight separatrices. Consequently all nonstationary streamlines are bounded exactly when $\alpha <\beta$. The origin is itself a stationary bounded trajectory in every case; the assertion about boundedness concerns the nonstationary streamlines. The equality case is handled separately below. These are the linear planar strain and rotation streamlines.

The velocity divergence is ${\partial }_x{u}_x+{\partial }_y{u}_y=\alpha -\alpha =0$, satisfying the constant-density incompressible flow condition. Choose the stream function convention $(u_x,u_y)=({\psi }_y,-{\psi }_x)$. Integration then givesIndeed, ${\psi }_y=\alpha x-\beta y$ and $-{\psi }_x=\beta x-\alpha y$. Streamlines are contours of this function. The vector field is also compatible with steady inviscid momentum balance without body forces: $(\mathbf{u}\cdot \nabla )\mathbf{u}=({\alpha }^2-{\beta }^2)(x,y)$ is balanced by pressure $p=p_0-\rho ({\alpha }^2-{\beta }^2)(x^2+y^2)/2$.

The vorticity is the curl of the velocity, perpendicular to the flow plane:The planar rotation rate is half this scalar vorticity, not the vorticity itself.

The uniform prior and likelihood give the posterior densitya Beta distribution with parameters $K+1,n-K+1$. Under squared-error loss for $\varphi$, the posterior expected loss is $\mathbb{E}[{\varphi }^2\mid K]-2a\mathbb{E}[\varphi \mid K]+a^2$, minimized uniquely at the posterior mean of $\varphi$, not at the product of the two posterior mean factors.

Using the supplied beta integral,Thus the Bayes estimate isThis remains positive even on the endpoint samples, unlike the extended MLE; it integrates over the posterior uncertainty in $\theta$.

Independence gives $\mathbb{E}[X_1(1-X_2)]=\theta (1-\theta )=\varphi$, proving unbiasedness. Conditional on $K=k$, all binary sample strings with $k$ ones are equally likely. ThereforeThe Rao-Blackwell theorem, or the tower property and total variance decomposition directly, yields the improved estimatorIt is unbiased by conditional expectation and is a function of the requested MLE. Its variance is strictly smaller, sinceFor $0<\theta <1$, the event $K=1$ has positive probability. On it, $X_1(1-X_2)$ has probability $1/n$ of being one, strictly between zero and one for $n\ge 2$. Thus the final expectation is strictly positive. This proves strict variance reduction, not merely a weak inequality.

Let $K={\sum }_i{X}_i$ and $\bar{X}=K/n$. The likelihood is ${\theta }^K(1-\theta {)}^{n-K}$, so for $0<K<n$ its unique maximizer is $\hat{\theta }=K/n$. By invariance of maximum likelihood estimation,For $K=0$ or $n$, the strict parameter interval $(0,1)$ has no attained likelihood maximum; the usual extended estimate takes the limiting endpoint and gives $\hat{\varphi }=0$. Thus the displayed estimator is the conventional extended MLE on those samples, an important boundary qualification. This is the first estimator in estimating Bernoulli variance.

At the cone's vertex height $h$, the first part gives $U(h)=U_0\sqrt{1-2gh/U_0^2}$ and $A(h)U(h)=A_0{U}_0$. Insert these into the assumed local impingement force from part (ii). Static support requires upward force equal to $Mg$, givingEquivalently,For a nonnegative support height the ratio in parentheses must not exceed one, and the prescribed $h\ll U_0^2/(2g)$ requires that ratio to be close to one. If the weight exceeds the lift available at the nozzle, this model has no supported height above the nozzle.