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The algebraic dual space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all real linear functionals. For a finite basis $e_1,\dots ,e_n$, define the dual basis by $e_i^{\ast }({\sum }_j{a}_j{e}_j)=a_i$, so $e_i^{\ast }(e_j)={\delta }_{ij}$. Every linear functional satisfiesThis equality follows by evaluating both sides on each vector of the basis. It proves spanning directly. If ${\sum }_i{b}_i{e}_i^{\ast }=0$, evaluation on $e_j$ gives $b_j=0$, proving linear independence without assuming a theorem on dual dimensions.

For a vector subspace $U$, its annihilator of a vector subspace isExtend a basis $e_1,\dots ,e_r$ of $U$ to a basis of $V$. Then $U^{\circ }$ has basis $e_{r+1}^{\ast },\dots ,e_n^{\ast }$, so $\boxed{\dim U^{\circ }=n-\dim U}$.

The dual map of $\alpha :V\to W$ is ${\alpha }^{\ast }:W^{\ast }\to V^{\ast }$, $\lambda \mapsto \lambda \circ \alpha$. In the corresponding dual bases its matrix is the transpose of the matrix for $\alpha$. A matrix and its transpose have the same rank, giving $\mathrm{rank}{\alpha }^{\ast }=\mathrm{rank}\alpha$. Directly,Thus $\boxed{\ker {\alpha }^{\ast }=(\mathrm{im}\alpha {)}^{\circ }}$. Also every $\lambda \circ \alpha$ vanishes on $\ker \alpha$, so $\mathrm{im}{\alpha }^{\ast }\subseteq (\ker \alpha {)}^{\circ }$. Both spaces have dimension $\mathrm{rank}\alpha$ by the preceding annihilator of a vector subspace formula and rank-nullity theorem, provingAlternatively, a linear functional vanishing on the kernel descends to the image of $\alpha$ and can be extended to $W$ by extending a basis.

For the infinite-dimensional polynomial ring, the coefficient functionals $L_i$ are linearly independent: evaluate a finite relation on each monomial. However, they do not span the algebraic dual space. The linear functional $p\mapsto p(1)$ takes value one on every monomial, while any finite combination of the $L_i$ vanishes on every sufficiently high-degree monomial. Thus the $L_i$ do not form a basis of $V^{\ast }$. An infinite formal sum of coefficient functionals is not a finite linear combination.

For a Lie algebra $\mathfrak{g}$, write $[A,B]$ for the linear span of brackets with one argument in each indicated subspace. The three definitions areThese are respectively an Abelian Lie algebra, a solvable Lie algebra and a nilpotent Lie algebra. The second and third sequences are the derived series of a Lie algebra and the lower central series of a Lie algebra.

An Abelian Lie algebra has ${\gamma }_2=0$, and a nilpotent Lie algebra is a solvable Lie algebra: induction gives ${\mathfrak{g}}^{(j)}\subseteq {\gamma }_{j+1}$. Thus all the implications are generated byNone of the reverse implications holds. The Heisenberg Lie algebra with basis $x,y,z$ and $[x,y]=z$, all other basic brackets zero, is nonabelian but has ${\gamma }_2=\mathbb{C}z$, ${\gamma }_3=0$. The two-dimensional affine Lie algebra of the line with $[h,e]=e$ has ${\mathfrak{g}}^{(1)}=\mathbb{C}e$ and ${\mathfrak{g}}^{(2)}=0$, but ${\gamma }_j=\mathbb{C}e$ for every $j\ge 2$. It is solvable and neither nilpotent nor abelian. These two examples answer all six ordered-pair comparisons.

The Lie theorem says that a finite-dimensional Lie algebra representation of a finite-dimensional solvable Lie algebra over an algebraically closed field of characteristic zero has a common eigenvector whenever its representation space is nonzero. Equivalently it admits an invariant complete flag, or simultaneous upper triangularization. Here the field may be taken to be $\mathbb{C}$; these hypotheses are essential.

To prove the common-eigenvector assertion, induct on $\dim \mathfrak{g}$. The zero algebra is immediate. Since $\mathfrak{g}$ is nonzero and solvable, its derived algebra is proper. Choose a codimension-one Lie algebra ideal $\mathfrak{h}$ containing it, and write $\mathfrak{g}=\mathfrak{h}\oplus \mathbb{C}x$. The Lie algebra $\mathfrak{h}$ is solvable, so induction supplies $v\ne 0$ and a linear functional $\lambda :\mathfrak{h}\to \mathbb{C}$ with $hv=\lambda (h)v$.

Consider the finite-dimensional cyclic subspace $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Until the first linear dependence, these powers form a basis. The identityand $[h,x]\in \mathfrak{h}$ show by induction, simultaneously for all $h\in \mathfrak{h}$, that $W$ is $\mathfrak{h}$-invariant and that $h$ is upper triangular on it with every diagonal entry $\lambda (h)$. It is also $x$-invariant by construction. HenceThe trace of a commutator is zero, and characteristic zero gives $\lambda ([x,h])=0$.

The simultaneous eigenspace $E_{\lambda }=\{w:hw=\lambda (h)w\text{ for every }h\in \mathfrak{h}\}$ is nonzero and $x$-invariant, sinceOver an algebraically closed field, $x{|}_{E_{\lambda }}$ has an eigenvector, which is therefore a common eigenvector for all of $\mathfrak{g}$. This finishes induction. Apply the same assertion to the quotient representation by its invariant line, and then to successive quotients. A basis adapted to the resulting complete flag gives the stated simultaneous triangularization of a Lie algebra representation, completing the proof of the Lie theorem.

A nilpotent Lie algebra is a solvable Lie algebra, so the inclusion $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ satisfies the Lie theorem and is upper triangular in a suitable basis, for finite-dimensional complex $V$.

This is insufficient to prove the Engel theorem. Its matrix version starts with a Lie subalgebra of nilpotent endomorphisms and concludes that they are simultaneously strictly upper triangular; its abstract version concludes nilpotence from nilpotence of every Adjoint representation endomorphism. Merely upper triangular matrices can have nonzero diagonal entries: the one-dimensional algebra $\mathbb{C}{I}_V$ is an Abelian Lie algebra and a nilpotent Lie algebra, but acts by a nonnilpotent identity matrix. This is nilpotent Lie algebras need not act nilpotently. Moreover, in the abstract Engel theorem nilpotence of the algebra is a conclusion, so assuming it first to invoke the Lie theorem would be circular. Abstract nilpotence and nilpotence of each representing matrix are different conditions.

The Riesz representation theorem states that every bounded linear functional $L$ on a real or complex Hilbert space $H$ is represented by a unique $h\in H$:For the complex case take the inner product to be linear in its first argument.

If $L=0$, choose $h=0$. Otherwise its kernel $K$ is a closed linear subspace. Choose $x$ with $L(x)\ne 0$ and let $z=x-P_{K}x$, using the orthogonal projection. Then $z\ne 0$, $z\perp K$, and $L(z)=L(x)\ne 0$. For every $v$,Therefore take $h=L(z)z/\Vert z{\Vert }^2$ in the real case, and $h=\overline{L(z)}z/\Vert z{\Vert }^2$ in the complex case. The conjugate in the latter formula compensates for conjugate linearity in the second argument.

The Cauchy-Schwarz inequality gives $|L(v)|\le \Vert v\Vert \Vert h\Vert$, and evaluation at $v=h/\Vert h\Vert$ when $h\ne 0$ gives equality of the norms. If two vectors represent $L$, their difference is orthogonal to every vector, including itself, hence zero. This proves all assertions of the Riesz representation theorem.

The real Lax-Milgram theorem applies to a Hilbert space $H$ and a bounded bilinear form $a$ satisfyingFor every bounded linear functional $L$ there is a unique $u\in H$ withSymmetry of the bilinear form is not required.

By the Riesz representation theorem, write $a(u,v)=⟨Au,v⟩$ and $L(v)=⟨g,v⟩$. The operator $A$ is linear and bounded, with $\Vert A\Vert \le M$. The coercive bilinear form bound and the Cauchy-Schwarz inequality implyHence $A$ is injective. Its range is closed: if $A{u}_n$ converges, this last inequality applied to differences makes $u_n$ a Cauchy sequence, and its limit maps to the proposed range limit. If $w$ is in the orthogonal complement of the range, then $a(u,w)=0$ for all $u$; taking $u=w$ and using coercivity gives $w=0$. The range is thus dense as well as closed, so it is all of $H$. Solve $Au=g$ uniquely; the displayed lower bound gives the asserted estimate.

For complex Hilbert spaces the same proof works for a bounded sesquilinear form, linear in the first argument, with $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }^2$. In the convention $a(u,v)=L(v)$, $L$ must then be a bounded conjugate-linear functional represented as $⟨g,v⟩$.

Use the Hausdorff convention for a locally convex space: a real vector space $X$ is equipped with a separating family $\mathcal{P}$ of seminorms. Thus each $p$ is nonnegative, subadditive and satisfies $p(tx)=|t|p(x)$, and for every $x\ne 0$ some $p\in \mathcal{P}$ has $p(x)>0$. The topology has a neighborhood basis at $a\in X$ consisting ofEquivalently, it is the coarsest vector-space topology making all these seminorms continuous. The separating condition is precisely what makes this topology Hausdorff. If the Hausdorff requirement were omitted, continuous linear functionals could not distinguish points in the common kernel of the seminorms.

The continuous dual space $X^{\ast }$ consists of all continuous linear functionals $X\to \mathbb{R}$. We first prove the needed Hahn-Banach theorem. Let $p$ be a real-valued sublinear function on $X$, meaning $p(a+b)\le p(a)+p(b)$ and $p(ta)=tp(a)$ for $t\ge 0$, and let $g$ be linear on a vector subspace $M$, with $g(m)\le p(m)$. To extend across $v\notin M$, writeThe necessary bounds on $c$ areThey are compatible becauseTaking one variable equal to zero also shows that $L,U$ are finite. Choose $c\in [L,U]$. The upper bound proves domination when $t>0$, after dividing $m+tv$ by $t$; the lower bound proves it when $t<0$, after dividing by $-t$. Thus $\tilde{g}\le p$ on $M+\mathbb{R}v$. This is the one-dimensional dominated extension of a real linear functional.

Order all dominated extensions of $g$ by extension of their domains. A chain has an upper bound obtained by taking the union of the domains and linear functionals. Zorn's lemma gives a maximal extension, and the one-dimensional construction shows that its domain must be all of $X$. We have therefore proved the real dominated-extension theorem. In particular, when $p$ is a seminorm, domination at both $x$ and $-x$ gives $|f(x)|\le p(x)$.

For $x_0\ne 0$, choose a continuous seminorm $p$ with $p(x_0)>0$, and define $g(t{x}_0)=tp(x_0)$ on its one-dimensional span. Then $|g(t{x}_0)|\le p(t{x}_0)$. Extend by the theorem just proved to $f$ with $|f(x)|\le p(x)$. This bound makes $f$ continuous, and $f(x_0)=p(x_0)>0$. Applying this to the difference of two distinct points proves that $X^{\ast }$ separates the points of $X$, the continuous-dual separation theorem for Hausdorff locally convex spaces.

For separation from a closed linear subspace, choose a basic balanced neighborhood $V$ of zero such that $(x_0+V)\cap Y=\varnothing$. WriteThen $q(x_0-y)\ge 1$ for every $y\in Y$. On $Y+\mathbb{R}{x}_0$, define $g(y+t{x}_0)=t$, which is well-defined since $x_0\notin Y$. For $t\ne 0$,for $t=0$ the inequality is immediate. The proved extension theorem gives a continuous $f$ dominated by $q$, withThis proves separation of a point from a closed linear subspace without using any unproved Hahn-Banach extension or separation result.

Let $q:X^{\ast }\to X^{\ast }/Y$ be the quotient map. Since $Y$ is norm closed and has finite codimension, the quotient is a finite-dimensional normed space. Choose a basis of its continuous dual and compose its coordinate linear functionals with $q$. This gives ${\varphi }_1,\dots ,{\varphi }_n\in X^{\ast \ast }$ such thatIf $Jx\in F$, it vanishes on $Y$. Evaluation at $x$ is weak-star continuous, and $Y$ is weak-star dense, so it vanishes on all of $X^{\ast }$. The dual norm formula gives $x=0$. Hence $F\cap JX=\{0\}$.

Assume $X\ne \{0\}$; the zero space is trivially normed by any positive constant. We claim thatIf not, there would be $x_j\in S_X$ and ${\psi }_j\in F$ with $\Vert J{x}_j-{\psi }_j\Vert \to 0$. The sequence ${\psi }_j$ is bounded. Finite-dimensionality of $F$ gives a norm-convergent subsequence with limit $\psi \in F$. Then $J{x}_j\to \psi$, and $JX$ is norm closed because $X$ is Banach. Thus $\psi \in F\cap JX$ and $\Vert \psi \Vert =1$, a contradiction. This is positive distance between a unit sphere and a disjoint finite-dimensional subspace.

Fix $x\in S_X$, identifying it with $Jx$, and put $\delta =\mathrm{dist}(x,F)\ge d$. On $F+\mathbb{R}x$, define $h(\varphi +tx)=t\delta$. The distance definition givesThe Hahn-Banach theorem extends it to $\Psi \in X^{\ast \ast \ast }$ with $\Vert \Psi \Vert \le 1$, $\Psi {|}_F=0$ and $\Psi (x)=\delta$.

Apply part (ii) with the Banach space $X^{\ast }$, its finite-dimensional dual vector subspace $E=F+\mathbb{R}x\subseteq X^{\ast \ast }$, and the bidual space element $\Psi \in X^{\ast \ast \ast }$. For every $\eta >0$ it supplies $y_{\eta }\in X^{\ast }$ satisfyingThus $y_{\eta }\in Y$. Normalize by $1+\eta$ and let $\eta \downarrow 0$ to obtainHomogeneity gives, for all $x\in X$,Consequently a finite-codimensional weak-star dense dual subspace is norming, with $c=d$. Since $Y$ is a real linear vector subspace and its ball is symmetric, the same supremum is obtained if an absolute value is inserted. The argument uses near-unit interpolation and a limiting supremum, not an assertion that the supremum is attained.

Consider the restriction operatorIt is onto: otherwise its range, a vector subspace of the finite-dimensional vector space $E^{\ast }$, would have a nonzero annihilator in $E^{\ast \ast }=E$. That annihilator would be an $e\in E$ vanishing on all of $X$, so $e=0$, a contradiction. Moreover $T$ is open. To see this directly, choose preimages of a basis of $E^{\ast }$; they define a linear right inverse $S:E^{\ast }\to X$, continuous because its domain is finite-dimensional. Small changes of an image can then be lifted by small changes using $S$.

Put $r=1+\epsilon$ and $C=T(\{x:\Vert x\Vert <r\})$. It is open and convex in $E^{\ast }$. If $a\notin C$, the Hahn-Banach separation theorem gives a nonzero real linear functional on $E^{\ast }$, represented by some $e\in E$, such thatBut $a(e)=x^{\ast \ast }(e)\le \Vert e\Vert$, contradicting $r>1$. Therefore $a\in C$, and the finite-dimensional interpolation form of Goldstine's theorem yieldsIf $E=\{0\}$ one simply takes $x=0$.

To recover the Goldstine theorem, start with $x^{\ast \ast }\in B_{X^{\ast \ast }}$ and finitely many tests $e_1,\dots ,e_m\in X^{\ast }$. Apply the result to their span with a small parameter $\delta >0$. The resulting $x$ need not lie in $B_X$, but $v=x/(1+\delta )$ does. For every test,Choosing $\delta$ sufficiently small puts $Jv$ in any prescribed basic weak-star neighborhood. This proves the asserted weak-star density of the closed unit ball. The reverse inclusion follows because $B_{X^{\ast \ast }}$ is weak-star closed, being the intersection of the conditions $|x^{\ast \ast }(e)|\le \Vert e\Vert$ for $e\in X^{\ast }$.

Suppose the weak-star closure of $Z$ were a proper linear vector subspace of $X^{\ast }$. The Hahn-Banach separation theorem would give a nonzero weak-star continuous linear functional vanishing on it. A continuous dual of a weak-star topology consists precisely of evaluations at points of $X$: continuity bounds the linear functional by finitely many evaluations, so it factors through their finite-dimensional coordinate map and is a linear combination of them. Thus some $0\ne x\in X$ would satisfy $z(x)=0$ for all $z\in Z$. The norming inequality would imply $c\Vert x\Vert \le 0$, a contradiction. Every norming subspace of $X^{\ast }$ is weak-star dense.

For the infinite-codimension example, takeHere $X$ is the absolutely summable sequence space and $c_0$ is the space of sequences converging to zero, a norm-closed vector subspace of ${\ell }^{\infty }$. The dual identification is $y(x)={\sum }_n{x}_n{y}_n$: a bounded sequence defines a linear functional of norm $\Vert y{\Vert }_{\infty }$, and every linear functional on ${\ell }^1$ has this form by evaluating on the coordinate vectors and using density of finitely supported sequences.

For $x\in {\ell }^1$, use the finitely supported sequence $y^{(N)}$ whose first $N$ entries are $\mathrm{sgn}(x_n)$ and whose remaining entries vanish. It belongs to $c_0$, has norm at most one, andThe reverse inequality follows from $|y(x)|\le \Vert y{\Vert }_{\infty }\Vert x{\Vert }_1$. Hence $c_0$ is 1-norming for ${\ell }^1$, an instance of vanishing sequences norm the summable sequence space.

To prove infinite codimension, take disjoint infinite setsTheir indicator sequences have linearly independent classes in ${\ell }^{\infty }/c_0$. Indeed, a finite combination has the constant value $a_j$ on $A_j$; if it tends to zero, every $a_j$ must vanish. Thus the quotient is infinite-dimensional.

The Riesz-Markov-Kakutani representation theorem says that a positive linear functional on $C(K)$ has the form $f\mapsto {\int }_{K}fd\nu$ for a unique finite positive regular Borel measure $\nu$, with norm $\nu (K)$. Its complex form says that every bounded complex linear functional is represented by a unique finite regular complex measure $\mu$, andThe extension from positive to arbitrary linear functionals follows by positive/negative decomposition of real linear functionals and then real/imaginary decomposition. The last quantity is the total variation norm of a measure, so this identifies the dual isometrically.

An extreme point $e$ of a convex set $C$ cannot be written as $e=ta+(1-t)b$ with $0<t<1$ and distinct $a,b\in C$. The Krein-Milman theorem, applied with the underlying real locally convex weak-star topology, states

We next prove Milman's converse to the Krein-Milman theorem. Suppose an extreme point $e$ were outside the weak-star closure of $S$. A basic weak-star neighborhood of $e$ disjoint from $S$ uses finitely many real coordinates: real and imaginary parts of evaluations at elements of $X$. Its complement is the union of finitely many closed half-spacesIntersect these with $C$, discard empty intersections, and call the resulting compact convex sets $C_1,\dots ,C_m$. They cover $S$, and none contains $e$.

The convex hull of their union is compact. Every point in it can be written ${\sum }_{i=1}^m{t}_i{c}_i$ with $(t_i)$ in the finite simplex and $c_i\in C_i$, by combining terms from the same convex set. The map from the simplex times ${\prod }_i{C}_i$ to that sum is weak-star continuous, so its image is compact and closed. It therefore contains ${\overline{\mathrm{conv}S}}^{w^{\ast }}=C$, in particular $e$. But extremality forces every $c_i$ having a positive coefficient in a representation of $e$ to equal $e$, contradicting $e\notin C_i$. Hence

Assume $K\ne \varnothing$. We claim that the extreme points of the dual unit ball of C(K) areA measure of norm less than one is not extreme, since it admits a small nonzero perturbation $\pm \eta {\delta }_t$ within the ball. For a measure $\mu$ of norm one, suppose $|\mu |$ is not a point mass. There is a Borel set $E$ with $0<a=|\mu |(E)<1$: if the measure support has two points, choose disjoint neighborhoods of positive mass; a regular probability measure supported at just one point is the corresponding point mass. Thenis a convex combination of distinct norm-one measures. Thus an extreme measure must have variation concentrated at one point, and must be $\alpha {\delta }_t$ with $|\alpha |=1$.

Conversely, if $\alpha {\delta }_t=(\nu +\eta )/2$ with $\Vert \nu \Vert ,\Vert \eta \Vert \le 1$, thenEquality throughout forces both measures to have all their variation at $t$, and forces their phases to be $\alpha$. Hence $\nu =\eta =\alpha {\delta }_t$, proving extremality. If $K=\varnothing$, the dual ball is $\{0\}$ and its sole extreme point is $0$.

Every finite Borel measure on $[0,1]$ is regular, so the given $\mu$ belongs to the dual unit ball of $C[0,1]$. Apply Banach-Alaoglu theorem and Krein-Milman theorem to that ball. It is the weak-star closed convex hull of these phased point masses. The unit ball of the finite-dimensional vector subspace $F$ is norm compact. Choose a finite $\epsilon /4$-net $f_1,\dots ,f_m$ in it. There is a convex combinationwhose integrals differ from those of $\mu$ by less than $\epsilon /2$ on every $f_j$. Put $t_i=a_i{\alpha }_i$. Then ${\sum }_i|t_i|=1$ and $\Vert \nu \Vert \le 1$. For any $f$ in the unit ball of $F$, choose $f_j$ with $\Vert f-f_j\Vert <\epsilon /4$. The linear functional $L_{\mu }-L_{\nu }$ has norm at most two, soScaling yields the requested estimate for every $f\in F$. This is atomic approximation on finite-dimensional spaces of continuous functions. Repeated nodes are allowed: keeping their individual terms preserves the exact sum of coefficient magnitudes even if their phases cancel. If $F=\{0\}$, one point with coefficient one suffices.

A rational matrix $A=(a_1\cdots a_n)$ is partition regular over the positive integers if every finite colouring of $\mathbb{N}$ admits a vector $x\in {\mathbb{N}}^n$ whose coordinates all have the same colour and satisfy $Ax=0$. Equal coordinates are allowed. Rado's theorem says that this holds exactly when the columns can be partitioned into nonempty blocks $I_0,\dots ,I_t$ withThis is the columns condition. Clearing an overall denominator lets us prove necessity for an integer matrix.

Suppose the columns condition fails. There are only finitely many ordered partitions of the finite column index set. For each ordered partition, choose a failing block: its sum $b_j$ is outside the rational span of preceding columns, with that span interpreted as zero for the first block. Finite-dimensional linear algebra gives a rational linear functional vanishing on that span but not on $b_j$. For completeness, take a basis of the span, append $b_j$, extend to a basis of the column space, and prescribe values zero on the first basis vectors and one on $b_j$. Clearing the functional's denominators gives an integer row functional $\ell$ with $\ell (b_j)\ne 0$.

Choose one prime $p$ larger than the absolute values of all these finitely many nonzero integers $\ell (b_j)$. Colour each positive integer by its first nonzero base-$p$ digit:Assume a monochromatic solution $x$ exists, and partition its coordinates into blocks of equal P-adic valuation, ordered by increasing valuations $v_0<\cdots <v_t$. Their first nonzero digit is a common $d\ne 0(\mathrm{mod}p)$. For this ordered partition take its chosen failing block $I_j$ and functional $\ell$. Apply $\ell$ to ${\sum }_i{a}_i{x}_i=0$. All preceding-block terms disappear exactly. Divide the remaining integer equation by $p^{v_j}$ and reduce modulo $p$. Later blocks disappear, while the current block givesThis contradicts the prime's choice. Thus the colouring has no monochromatic solution, proving necessity. This is the finite separating-functional proof of the columns condition; it needs neither a limiting argument nor a bound on the valuations themselves.

For sufficiency, suppose the blocks satisfy the columns condition. Choose rational coefficients $q_{ji}$, for $i$ in preceding blocks, such thatLet $c$ be a positive integer clearing their denominators, and choose an integer $P\ge 1$ bounding all $|c{q}_{ji}|$. Use the allowed monochromatic m-p-c set theorem with $t+1$ generators. We use the triangular convention in which the resulting positive integer set contains every numberin one colour. This is an m-p-c set, with the generator indices reversed if the alternative lower-triangular convention is used. The permitted theorem applies to either convention, since it applies to every finite number of generators. Its set lies in $\mathbb{N}$, so all of the displayed combinations are positive.

For $i\in I_j$, defineEach coordinate belongs to that monochromatic set. Summing the column contributions and collecting coefficients of $u_k$ givesFor $k=0$ the inner sum is empty and $b_0=0$. This proves sufficiency by a Rado solution inside an m-p-c set, and completes the theorem.

For the requested application, chooseThe matrix, in the coordinate order $(x,y,z,w)$, has columnsThe block $I_0=\{x,z,w\}$ has zero sum. The remaining column satisfies $a_y=-a_z/3+2a_w$, so it is in the span of the first block. By Rado's theorem, positive monochromatic $x,y,z,w$ satisfy the system. In particular,One can see the positivity directly in the same triangular construction: a monochromatic set containing $3u+\lambda v$ for $|\lambda |\le 6$, together with $3v$, givesAll four are positive because they belong to that positive m-p-c set. This is a partition-regular system forcing an ordered Schur relation.

Let $e_1,\dots ,e_n$ be the standard basis and define $a_j=f(e_j)$. By linearity,Thus every linear functional has this form. Its representing vector is unique because evaluating $a\cdot x=b\cdot x$ at each $e_j$ gives $a_j=b_j$.

For the prescribed kernel of a linear map, the third listed vector is the first minus the second, so it adds no independent constraint. The first two requireTaking $a_1=s$ and $a_4=t$ gives $a_2=2s-2t$, $a_3=-3s+3t$. Consequently all possible representing vectors areConversely, every such vector is perpendicular to each of the three prescribed vectors, so its linear functional vanishes on them. This also includes the zero linear functional.