There are $p^n$ possible translation vectors. An invertible matrix is specified by an ordered basis of ${\mathbb{F}}_p^n$. Its first column can be any nonzero vector, giving $p^n-1$ choices. After choosing $i$ independent columns, their span has $p^i$ elements, so the next column has $p^n-p^i$ choices. Therefore

Every affine transformation factors asso $G=NH$. A translation fixing zero has zero translation vector, so $H\cap N=\{(I,0)\}$. Together with normality from part (c), these are the internal semidirect product conditions. ThusThe reconstruction data are the additive group $F^n$, the linear group ${\mathrm{GL}}_n(F)$, and its standard action $A:c\mapsto Ac$ on translations. In pair notation $(b,A)$, the product is $(b,A)(d,C)=(b+Ad,AC)$, equivalent to the formula in part (b).

Embed ${\mathrm{GL}}_n(F)$ by $A\mapsto (A,0)$. The multiplication and inverse formulas show that this is an injective group homomorphism, so its image $H$ is a subgroup. The translations form $N=\{(I,c):c\in F^n\}$, with $(I,c)(I,d)=(I,c+d)$ and inverse $(I,-c)$. Hence $N$ is isomorphic to the additive group of $F^n$.

For an arbitrary affine transformation,Conjugation therefore preserves $N$, proving that $N$ is normal in $G$.

Use composition from right to left. Applying $(C,d)$ and then $(A,b)$ givessoThe product remains an affine transformation because $AC$ is invertible. Composition of functions is associative. The identity is $(I,0)$, and solving $w=Av+b$ gives the inverseThus these transformations form the affine group. The representation by $(A,b)$ is unique: evaluating at zero determines $b$, and then evaluating on the standard basis determines $A$.

With $N◃G$, the internal semidirect product conditions are $G=NH$ and $N\cap H=\{1\}$. Every element then has a unique expression $nh$. The extra data needed to reconstruct the group are the conjugation actionOn the set $N\times H$, the external semidirect product $N{⋊}_{\varphi }H$ has multiplicationThe map $(n,h)\mapsto nh$ identifies it with $G$. A direct product results when this action is trivial.

For example, $S_3=A_3⋊⟨(12)⟩\cong C_3⋊C_2$. The two subgroups intersect trivially and their orders multiply to six, so they generate $S_3$. Conjugation by the transposition inverts the three-cycle. This action is nontrivial, and $S_3$ is nonabelian whereas $C_3\times C_2$ is abelian, so this is not their direct product.

For $\alpha \ne 0$, Lagrange's theorem for finite groups in $K^{\times }$, whose order is $q^m-1$, gives ${\alpha }^{q^m-1}=1$. The zero element satisfies the equation too, so every $\alpha \in K$ has ${\alpha }^{q^m}=\alpha$.

Now let $f\in F[X]$ be irreducible of degree $m$. The quotient $F[X]/(f)$ is a field with $q^m$ elements, by its basis $1,\overline{X},\dots ,{\overline{X}}^{m-1}$. Its element $\overline{X}$ therefore satisfies ${\overline{X}}^{q^m}-\overline{X}=0$. This means precisely that $X^{q^m}-X$ lies in the ideal $(f)$, giving

By the allowed finite multiplicative-subgroup theorem, the group $K^{\times }$ is cyclic. Let $z$ generate it. Every nonzero element of $K$ is a power of $z$, hence belongs to $F(z)$; zero belongs as well. Since the reverse inclusion is automatic,Thus a finite extension of a finite field is a simple field extension.

Write $Q=q^m$ and take a splitting field $L$ of $X^Q-X$ over $F$. Its formal derivative is $-1$, since the characteristic divides $Q$, so it has $Q$ distinct roots in $L$. Let $K$ be their set. The Frobenius endomorphism identity givesThus $K$ contains $0,1$ and is closed under addition, negatives and multiplication. If $u\ne 0$ belongs to $K$, then $(u^{-1}{)}^Q=(u^Q{)}^{-1}=u^{-1}$, so it is closed under inverses too. Hence $K$ is a field with exactly $Q$ elements. Every $c\in F$ satisfies $c^q=c$ by Lagrange's theorem for finite groups applied to $F^{\times }$, together with the zero case; iterating gives $c^Q=c$. Therefore

The extension $K/F$ is a vector space over $F$. Since $K$ is finite, a maximal linearly independent subset is finite and spans it; let its size be $d$. Relative to this basis, every element of $K$ has a unique list of $d$ coefficients in $F$. There are $q$ choices for each coefficient, soThe inclusion $F\subseteq K$ makes $d\ge 1$.

Apply successive Householder reflections supported on smaller trailing coordinate sets. At step $k=1,\dots ,n-2$, let $x$ be the entries in rows $k+1,\dots ,n$ of column $k$ of the current symmetric matrix. If $x\ne 0$, part (c) gives an orthogonal reflection $H$ taking it to $c{e}_1$; if $x=0$, use the identity. Embed this as $Q_k=\mathrm{diag}(I_k,H)$ and replace the current matrix by $Q_k^{T}A{Q}_k$.

The first $k$ columns are unchanged by right multiplication. Left multiplication turns the tail of column $k$ into a vector supported only in row $k+1$. Previously treated columns have zero entries throughout this trailing coordinate set, so their zeros are preserved. Symmetry makes the matching row zeros appear as well. After the final step all entries below the first subdiagonal, and by symmetry above the first superdiagonal, vanish.

The product $Q=Q_1\cdots Q_{n-2}$ is orthogonal, andRestricting each transformation to its trailing coordinates is what preserves earlier zero patterns.

Let $r=\Vert x\Vert >0$. Choose $c=-r$ if $x_1\ge 0$ and $c=r$ if $x_1<0$, and set $v=x-c{e}_1$. This opposite-sign choice ensures $v\ne 0$. Because $c^2=r^2$,ThereforeThe denominator is positive, including when $x$ already lies on the first coordinate axis.

If $x\in S_v$, then $v^{T}x=0$ and the defining formula immediately gives $H_{v}x=x$. For arbitrary $x$,Thus the Householder reflection fixes the hyperplane perpendicular to $v$ pointwise and reverses the component along its normal.

Let $P=v{v}^T/(v^{T}v)$, which is a nonzero symmetric orthogonal projection: $P^T=P$ and $P^2=P$. Then $H_{v,\alpha }=I-\alpha P$ is symmetric andSince $P\ne 0$, this equals the identity exactly when $\alpha (\alpha -2)=0$. Thus

Every derivative of order $n+1$ of $\sin (2x)$ has absolute value at most $2^{n+1}$. The equally spaced nodes are symmetric about $1/2$. Pair each node $a\le 1/2$ with $1-a$; for $x\in [0,1]$,If an unpaired middle node remains, its factor satisfies $|x-1/2|\le 1/2$. Consequently the complete nodal product is at most $2^{-(n+1)}$. The error formula actually gives the stronger estimateThe first bound tends to zero, proving uniform convergence on the whole interval. There is no appeal to a general claim that equispaced interpolation always converges; the derivative and symmetric-product bounds are particular to this analytic function and these nodes.

If $x$ is a node the error and the product both vanish, and any interior $\xi$ works. Otherwise put $\omega (t)={\prod }_{i=0}^n(t-x_i)$ and $C=(f(x)-p_n(x))/\omega (x)$. The functionhas $n+2$ distinct zeros, at the nodes and at $x$. Repeated Rolle's theorem gives a point $\xi$ between the extreme zeros, and therefore in $(a,b)$, where $R^{(n+1)}(\xi )=0$. Since $p_n^{(n+1)}=0$ and ${\omega }^{(n+1)}=(n+1)!$, one gets $C=f^{(n+1)}(\xi )/(n+1)!$. Substitution gives

Multiplication by the displayed Vandermonde matrix maps the coefficient vector of a polynomial of degree at most $n$ to its vector of values at the nodes. Part (a) proves that every vector of node values is attained uniquely, so this square linear map is bijective. Therefore $A_n$ is invertible. Equivalently, a kernel vector would define a polynomial of degree at most $n$ with $n+1$ distinct roots and would have to vanish.

Interpolation means $p_n(x_i)=f(x_i)$ for every node. Define the Lagrange basis polynomialsDistinctness of the nodes makes every denominator nonzero, and ${\ell }_i(x_j)={\delta }_{ij}$. Thusis a polynomial of degree at most $n$ with the required values. If two such polynomials existed, their difference would have $n+1$ distinct roots while having degree at most $n$; the polynomial root bound makes it identically zero. Hence existence and uniqueness hold. The actual degree can be smaller than $n$, despite the conventional name.

For the one-parameter fit to ten grouped categories, the usual reference degrees of freedom are $10-1-1=8$. The table minimum is about $8.29$, far below the supplied critical value $15.51$, so do not reject the fitted log-series model at the five-percent level.

This standard grouped likelihood-ratio calibration uses the grouped maximum, approximately $0.93$, as the preceding answer explains. If the ungrouped estimate from the total abundance is inserted into a grouped statistic instead, the usual automatic loss of exactly one chi-squared degree of freedom is not justified by that same grouped-likelihood argument. The problem's table-based decision should therefore be kept distinct from that alternative procedure.

The displayed table has ten categories, including one pooled upper-tail category. With the parameter fixed, no model parameter is fitted and the appropriate reference degrees of freedom are $10-1=9$. The supplied statistic is $15.59$, below the stated five-percent critical value $16.92$. Hence do not reject the specified log-series model at the five-percent level. Using eight degrees of freedom here would incorrectly charge a fitted parameter to a fully specified null.

Put $D^{\ast }=D-A$, with $A=(X^{T}X{)}^{-1}{X}^T$. Then $D^{\ast }X=0$, and therefore $D^{\ast }{A}^T=D^{\ast }X(X^{T}X{)}^{-1}=0$; its transpose also vanishes. Expanding givesHenceThe difference from the least-squares covariance is positive semidefinite, since $v^T{D}^{\ast }{D}^{\ast T}v=\Vert D^{\ast T}v{\Vert }^2\ge 0$. In particular its $i$th diagonal entry is ${\sigma }^2{\sum }_j(D_{ij}^{\ast }{)}^2$, proving every stated variance inequality. For estimating ${\theta }_1$, prefer least squares: among linear unbiased estimators it has minimum variance, and therefore minimum mean squared error in that class. A competitor can tie if the first row of $D^{\ast }$ is zero. This is the Gauss-Markov theorem; it does not assert optimality against all biased or nonlinear estimators.