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Let $[L:F]=n<\infty$. The intermediate field $K$ is an $F$-linear linear subspace of $L$, so the finite-dimensional linear subspace theorem gives $[K:F]\le n$. To handle the other field extension, take an $F$-basis $v_1,\dots ,v_n$ of $L$. Every $\ell \in L$ is a sum $\sum f_i{v}_i$ with $f_i\in F\subseteq K$, so the same list spans $L$ as a $K$-vector space. It need not remain linearly independent, but spanning suffices to give $[L:K]\le n$. Hence both intermediate extensions are finite.

The degree of a field extension is the dimension of the larger field as a vector space over the smaller one:The field extension is a finite field extension precisely when this dimension is finite, equivalently when finitely many elements of $K$ span $K$ over $F$. Finiteness here concerns the dimension, not the number of elements of the fields; for example, $\mathbb{C}$ has degree two over $\mathbb{R}$ although both fields are infinite.

Let $[L:F]=n<\infty$. The intermediate field $K$ is an $F$-linear linear subspace of $L$, so the finite-dimensional linear subspace theorem gives $[K:F]\le n$. To handle the other field extension, take an $F$-basis $v_1,\dots ,v_n$ of $L$. Every $\ell \in L$ is a sum $\sum f_i{v}_i$ with $f_i\in F\subseteq K$, so the same list spans $L$ as a $K$-vector space. It need not remain linearly independent, but spanning suffices to give $[L:K]\le n$. Hence both intermediate extensions are finite.

A cubic spline is a twice continuously differentiable function whose restriction to each knot interval is a polynomial of degree at most three. Its second derivative is continuous and piecewise linear. Natural endpoint conditions are $C^{\prime \prime }(x_0)=C^{\prime \prime }(x_n)=0$.

Linearity of both the piecewise-polynomial and smoothness conditions, and of the endpoint equations, makes the natural splines a vector space $V$. A continuous piecewise linear second derivative with zero endpoint values is determined freely by its $n-1$ interior nodal values. Integrating twice adds two arbitrary affine constants. HenceThis count includes the case of one interval.

An interpolating spline has $S(x_i)=f_i$ at every node. Set $e=T-S$, so $e$ vanishes at all nodes. On each open knot interval $S^{(3)}$ is constant. Two integrations by parts giveThe terms containing $e$ vanish at nodes. When summed, the interior $e^{\prime }{S}^{\prime \prime }$ terms cancel because $e^{\prime }$ and $S^{\prime \prime }$ are continuous. The two outer terms vanish by the natural endpoint conditions. ThusIntegrating the given square identity now yields the sharper energy decompositionIn particular the natural interpolating spline minimizes the squared-curvature integral.

If $T$ is another natural interpolant, the same minimizing argument in both directions makes the two energies equal. The final nonnegative integral then vanishes, so $T^{\prime \prime }=S^{\prime \prime }$ everywhere by continuity. Their difference is affine and zero at all knots, including the two distinct endpoints, so $T=S$. This proves uniqueness.

It also supplies existence if needed: nodal evaluation is a linear map from the $(n+1)$-dimensional space $V$ to ${\mathbb{R}}^{n+1}$. Its kernel is zero by the same argument with zero data and the zero competitor. It is therefore surjective, so every data vector has a natural spline interpolant.

Regard $K$ as a vector space over its prime subfield ${\mathbb{F}}_p$. Since $K$ is finite, a maximal linearly independent list terminates and is a finite basis, of some dimension $n\ge 1$. Each field element has a unique coordinate vector in ${\mathbb{F}}_p^n$, so

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 $d\ge 1$.

The degree of a field extension $[K:F]$ is the dimension of $K$ as a vector space over $F$. The tower law states that for finite extensions $F\subseteq E\subseteq K$,If the left side is prime, one factor is one. Degree one forces equality of the corresponding fields, so there is no proper intermediate field.

Let $K$ have 32 elements. Its prime field is ${\mathbb{F}}_2$, and its dimension over that field is five, since a dimension-$d$ vector space over ${\mathbb{F}}_2$ has $2^d$ elements. Every element outside ${\mathbb{F}}_2$ has minimal polynomial of degree five: its generated intermediate field has degree greater than one and dividing five.

Every element of $K$ satisfies $x^{32}=x$, by the multiplicative group of a field of order 31 for nonzero elements. Hence $X^{32}-X$ splits completely over $K$ and has no repeated roots, since its derivative is minus one. Conversely any irreducible polynomial $q$ over ${\mathbb{F}}_2$ gives a quotient ring ${\mathbb{F}}_2[X]/(q)$ with 32 elements. Its defining element satisfies $X^{32}-X$, so $q$ divides that polynomial and therefore has all five distinct roots in $K$.

The 30 elements of $K\setminus {\mathbb{F}}_2$ are thus partitioned into disjoint five-element root sets of the irreducible polynomials. Their number isNo classification of the six individual polynomials is needed; the tower degree and separability already count them all.

Any nonzero vector in ${\mathbb{F}}_2^3$ extends to a basis. Sending one such basis to another gives an element of the general linear group, so its group action on the seven nonzero vectors is a transitive group action. Counting successive independent columns givesBy the orbit-stabilizer theorem, the stabilizer $H$ of a chosen nonzero vector $v$ has order $168/7=24$.

There are three two-dimensional linear subspaces containing $v$, corresponding to the three one-dimensional linear subspaces of ${\mathbb{F}}_2^3/⟨v⟩$. The stabilizer $H$ acts transitively on them: extend a basis of either plane beginning with $v$ to a basis of the whole vector space. Thus the stabilizer of a flag $⟨v⟩<P$ has order $24/3=8$.

If $Q\le G$ has order eight, each orbit of a group action of $Q$ on the seven nonzero vectors has size a power of two. Since seven is odd, at least one orbit of a group action has size one. Thus $Q$ fixes a nonzero vector $v$. Its group action on the three planes containing $v$ likewise has a fixed plane $P$. Therefore $Q$ is contained in, and by order equals, the stabilizer of that flag. The group action of $G$ is transitive on these flags, so all such subgroups are conjugate subgroups.

Distinct flags give distinct stabilizers. In a basis adapted to a flag, its order-eight stabilizer consists ofIts only common fixed nonzero vector is $e_1$. Its only invariant plane is $⟨e_1,e_2⟩$: an invariant plane is the kernel of a linear map of a nonzero linear functional, and over ${\mathbb{F}}_2$ that linear functional must be fixed; the displayed matrices have only the fixed nonzero linear functional $e_3^{\ast }$. Hence each subgroup specifies its flag uniquely. There are seven choices of $v$ and three choices of $P$, giving

Choose a basis for the first subspace of a flag, extend it to a basis for the second, and continue until obtaining a basis of the whole vector space. Do the same for the target flag. The unique linear map sending the first adapted basis to the second is invertible and sends every flagged subspace to its counterpart. Hence the action is transitive.

The finite field $K$ is a finite-dimensional vector space over its prime subfield $F$ of size $p$. A basis has some positive length $n$, and every element is uniquely specified by $n$ coefficients in $F$. There are $p$ choices per coefficient, so $\boxed{|K|=p^n}$ and $[K:F]=n$.

Regard $K$ as a vector space over $F$. Since $K$ is finite, a maximal linearly independent list has finite length $d$ and is a basis. Every element is uniquely a linear combination of its $d$ basis vectors, with $q$ choices for each coefficient. Hence $\boxed{|K|=q^d}$, with $d=[K:F]$. This includes $d=1$ when the extension is trivial.

No field has six elements. A finite field has positive characteristic: the sequence $0,1,2\cdot 1,\dots$ must repeat. By part (a), this characteristic is a prime number $p$. Translation by $1$ partitions the field into cycles of length $p$, since returning to a starting point after $k$ steps is equivalent to $k1=0$. Thus $p$ divides six and $p=2$ or $p=3$.

The multiples of $1$ form a copy of the prime field ${\mathbb{F}}_p$. For completeness, the inverse of a nonzero residue $a$ lies among these multiples: the integer Bezout identity $ab+cp=1$ gives $(a1)(b1)=1$. The entire field is a vector space over this subfield. A maximal linearly independent set is finite and spans, so it is a basis of some positive size $d$. Unique basis coordinates give exactly $p^d$ elements. This proves finite field cardinality is a prime power here. But six is neither a power of two nor a power of three, a contradiction.

Yes. An $\mathbb{F}$-module over a field is a vector space. Start with a finite generating list. If it is linearly dependent, one vector has a nonzero coefficient in a linear relation; dividing by that coefficient expresses the vector using the others, so it can be removed without changing the span. Repeating terminates with a linearly independent generating list, which is a basis and therefore a basis of a module.

Thus every finitely generated module over a field is a free module. For the zero module, the empty list is the required basis of a module.

Use part (a) to put the four projective points atA quadratic polynomial vanishing at the first three has no square terms, and vanishing at the fourth givesThese form a two-dimensional vector space. Choose the two membersTheir coefficient triples are independent and both satisfy the relation. Their symmetric bilinear form matrices are of the formFor $(A,B,C)=(1,1,-2)$ and $(1,-2,1)$, both determinants are $-1/2$, so both selected conics are nonsingular. Any coefficient triple with zero sum is uniquely a linear combination of these two. Hence every conic through the four points has matrixUndoing the projective transformation carries the entire family back to the original four points, preserving nonsingularity. This is a pencil of conics through four points.

Apply the regular level set theorem to $F(A)=\det A-1$ on the $n^2$-dimensional vector space of real matrices. At a point with determinant one, part (b) gives $D{F}_A(H)=\mathrm{tr}(A^{-1}H)$. Choosing $H=A/n$ gives $D{F}_A(H)=1$, so this derivative is surjective.

Hence the special linear group $\mathrm{SL}(n,\mathbb{R})$ is a smooth submanifold of dimension $n^2-1$. At the identity,This is the special linear Lie algebra. When $n=1$, the group is a single point and the tangent space is zero, consistent with the same formula.

Use the six-dimensional vector space of real symmetric $3\times 3$ matrices as the target, rather than all nine matrix entries independently. DefineIts derivative isThe given $J$ is symmetric with $J^2=I$. Every $A\in G$ is invertible, since taking determinants of $A^{T}JA=J$ gives $(\det A{)}^2=1$. For any symmetric target $S$, chooseThen $A^{T}JH=\frac{1}{2}JJS=S/2$, and its transpose is also $S/2$. Thus $D{F}_A(H)=S$ and the derivative is surjective at every point of $G$.

The regular level set theorem now givesIt also proves smooth embeddedness in the entire ambient vector space of matrices. The target symmetry is what gives the correct codimension.

The curve-based tangent space at $x$ consists of velocities ${\gamma }^{\prime }(0)$ of smooth curves $\gamma :(-\epsilon ,\epsilon )\to M$ satisfying $\gamma (0)=x$. For an arbitrary subset these velocities need not form a vector space; for a smooth submanifold they do, and this is its usual tangent space. This definition suffices for the direct computation at the identity.

For a real $m\times n$ matrix $A$, the row space is the linear span of its rows in ${\mathbb{R}}^n$, and the column space is the linear span of its columns in ${\mathbb{R}}^m$. Its null space consists of the input vectors sent to zero:The rank of a matrix isThe equality of row and column dimensions is the row-rank/column-rank theorem. Each displayed space is a linear subspace of the indicated ambient vector space.

False. In the real vector space $\mathbb{R}$, both $\{1\}$ and $\{2\}$ are bases: each is linearly independent and spans the whole space. A basis need not be a unique set. What is fixed for a finite-dimensional vector space is the number of basis elements, its dimension.

The rank-nullity theorem states $\dim V=\dim \ker T+\dim \mathrm{im}T$ for a linear map from a finite-dimensional vector space $V$. Here the domain has dimension four and the image has dimension two. Thus $\dim \ker T=4-2=2$. For example, a kernel basis is $\{(-2,-1,1,0),(-1,-2,0,1)\}$, as follows from the reduced equations.