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If an array is antisymmetric in every orthonormal basis and its contractions with every antisymmetric second-rank tensor are invariant scalars, it obeys the Cartesian second-rank tensor transformation law. The difference $D=R^{T}A\prime R-A$ is antisymmetric and orthogonal to every antisymmetric test matrix. Choosing the test matrix equal to $D$ gives $D:D=0$, hence $D=0$. The Frobenius inner product is nondegenerate on the antisymmetric subspace.

The Frobenius inner product of a symmetric matrix and an antisymmetric matrix is zero. Thus invariant contractions of an array with all symmetric second-rank tensors test only its symmetric part; an arbitrary basis-dependent antisymmetric part is invisible. A nonzero antisymmetric matrix in one basis and the zero matrix in another passes all such tests with scalar zero but is not a Cartesian second-rank tensor.

A Cartesian second-rank tensor has component matrix $T$ in each orthonormal basis, with transformation $T\prime =RT{R}^T$ under an orthogonal component change $R$. Both indices transform; this is why assigning arbitrary matrices independently in different bases does not define a tensor. The determinant, trace, and Frobenius inner product with another such tensor are unchanged by these basis transformations.

The convex cone of completely positive matrices:It is a closed convex cone and the dual cone of the copositive cone under the Frobenius inner product. The finite-sum definition imposes no closure by fiat; closedness of the completely positive cone supplies that fact.

On real symmetric matrices of size $n\ge 2$, the displayed pairing is symmetric because $\det (AC)=\det A\det C$. It is not bilinear: $B(I,2I)=2^n$, whereas linearity would require $2B(I,I)=2$. Its self-pairing is always nonnegative, but a nonzero singular matrix has $B(A,A)=0$, so positive definiteness fails as well. For $n=1$ it reduces to the ordinary real inner product. The Frobenius inner product $\mathrm{tr}(A^{T}C)$ supplies the appropriate bilinear, positive-definite geometry in every size.

The deviatoric strain is the trace-free part of the infinitesimal strain tensor. It removes the isotropic volume change. The decomposition is orthogonal for the Frobenius inner product, because a trace-free matrix has zero Frobenius inner product with $I$. ConsequentlyFor $\epsilon =\frac{1}{2}(\nabla v+(\nabla v{)}^{\mathsf{T}})$, its trace is $\mathrm{div}v$.

A tensor contraction over two paired indices. For real second-order tensors, it gives the displayed Frobenius inner product. For a fourth-order elasticity tensor and a second-order infinitesimal strain tensor, $(C:ϵ{)}_{ij}={\sum }_{kl}{C}_{ijkl}{ϵ}_{kl}$ remains a second-order tensor; double contraction need not always give a scalar.

The zero-boundary Korn inequality controls the full displacement gradient tensor by the infinitesimal strain tensor for $v\in H_0^1(\Omega ;{\mathbb{R}}^d)$. On a bounded Lipschitz domain, combining it with the Poincaré inequality makes a positive linear elasticity energy a coercive bilinear form in the first-order Sobolev norm. The zero Dirichlet boundary condition removes the otherwise uncontrolled rigid displacements.

For this zero-boundary version one can prove an exact identity. For a smooth compactly supported vector field, two applications of integration by parts giveExpanding the symmetric gradient and using density in the zero-boundary Sobolev space therefore yieldsIn three dimensions this also gives the stronger trace-free versionHere the matrix-valued L2 norm uses the Frobenius inner product. This explains why the trace-free form printed in the examination suffices for the usual Korn inequality estimate.

The infinitesimal strain tensor $\epsilon (v)=\frac{1}{2}(\nabla v+(\nabla v{)}^{\mathsf{T}})$ and the divergence are linear in $v$. Consequently the given $a$ is a symmetric bilinear form, andFor a matrix-valued function, $\Vert \cdot {\Vert }_0$ here uses the Frobenius inner product pointwise and then the L2 norm. All expressions are finite for $v\in H^1(\Omega ;{\mathbb{R}}^3)$.

Expand the quadratic term at $\theta v+(1-\theta )w$. The linearity of $\ell$ cancels its contribution to the difference, givingFor $0\le \theta \le 1$, this is nonnegative. Thereforeso $J$ is a convex functional on the whole Sobolev space. No boundary condition is needed for this conclusion. The argument uses nonnegativity of the quadratic term; uniqueness on the zero-boundary Sobolev space will instead use the Korn inequality in part (b).

Write $V=H_0^1(\Omega ;{\mathbb{R}}^3)$. For any $v\in V$ and any real $t$, $u+tv$ is admissible. Expanding the functional givesSince the left side is nonnegative for positive and negative $t$, its derivative at zero vanishes. Thus the first variation gives the weak formulation

To identify its differential equation, define the stress tensorThis is a symmetric matrix field. The Frobenius inner product of a symmetric matrix with a skew-symmetric matrix is zero, so $\sigma (u):\nabla v=\sigma (u):\epsilon (v)$. Since $\mathrm{tr}\epsilon (v)=\mathrm{div}v$, we obtainFor smooth compactly supported test functions, integration by parts in each row givesHence $-\mathrm{div}\sigma (u)=f$ as a distributional identity, that is,Membership in $V$ supplies the zero Sobolev trace. This identifies precisely the required weak solution of the linear elasticity problem, with row-wise divergence.

For uniqueness, use the exact form of the Korn inequality supplied in the paper. Put $E=\epsilon (v)$ and $d=\mathrm{tr}E=\mathrm{div}v$. The deviatoric strain is $D=E-\frac{1}{3}dI$. Because $\mathrm{tr}D=0$ and $I:I=3$, orthogonality for the Frobenius inner product givesIf $C_K$ is the supplied Korn inequality constant and $K$ the permitted Poincaré inequality constant, thenConsequently,If $u_1,u_2$ are two weak solutions, their difference $d_u\in V$ satisfies $a(d_u,v)=0$ for every $v\in V$. Choosing $v=d_u$ and using this coercive bilinear form estimate forces $d_u=0$. The minimizer therefore gives the unique weak solution.

For later use, $|\epsilon (v){|}_F\le |\nabla v{|}_F$ and $|\mathrm{div}v|\le \sqrt{3}|\nabla v{|}_F$ also give the bounded bilinear form and linear functional estimatesTogether with the coercive bilinear form estimate and the Lax-Milgram theorem, these would establish existence independently of the assumed minimizer.

The cyclic property of the trace gives $\mathrm{tr}(AB)=\mathrm{tr}(BA)$, which is C1. The linearity of trace and matrix multiplication gives C2 and C3. Finally symmetry of $A$ giveswith equality exactly when every entry is zero. Thus C4 also holds. This is the Frobenius inner product restricted to the real vector space of symmetric matrices.

Put $m=p^a$, with $p$ a prime number, and $s=m-1$. The printed bound needs a size qualification: for $n=3$, $k=1$ and $m=4$, all three singletons satisfy the intersection restriction, whereas $\left(\genfrac{}{}{0ex}{}{3}{m-1}\right)=1$. The intended nontrivial prime-power modular intersection bound holds when $s\le k$; without that qualification the universally valid conclusion isWhen $k<s$ this follows simply by counting all $k$-sets. We prove the sharper intended form for $s\le k$.

Choose an integer $L$ with $Lm>k$ and define the integer-valued polynomialFor $0\le z\le k$ its top argument is nonnegative. The bottom argument has all its first $a$ base-$p$ digits equal to $p-1$. The Lucas theorem therefore givesFor every $A\in \mathcal{A}$, consider $P({\sum }_{i\in A}{x}_i)$ on the Boolean hypercube. Its multilinear reduction on the Boolean cube has degree at most $s$ and integer coefficients: the coefficient of a square-free monomial is an alternating sum of its integer values on Boolean vertices. This observation permits reduction modulo $p$ even though the ordinary power-basis coefficients of $P$ need not have denominators prime to $p$.

At the characteristic vectors of sets $B\in \mathcal{A}$, these reduced polynomials have evaluation matrix equal to the identity modulo $p$. Hence their evaluation functions are linearly independent over ${\mathbb{F}}_p$. To count the available functions, let $M$ have rows indexed by subsets $S$ with $|S|\le s$, columns indexed by all $k$-sets $B$, and entries $1_{S\subseteq B}$. Over the rational numbers, for $|S|=j$,Thus the matrix rank over the rational numbers is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every larger integer minor is zero and remains zero modulo $p$, so the same rank bound holds over ${\mathbb{F}}_p$. This low-degree evaluation rank on a uniform layer proves

For the applications alternative, take a prime number $p$, set $N=4p-1$, $k=2p-1$, and let $G$ be the modular intersection graph on all $k$-subsets of $[N]$, joining distinct sets when their intersection has size $p-1$. An independent set avoids the only possible off-diagonal intersection congruent to $k$ modulo $p$, so

First embed $A$ as $v_A=1_A/\sqrt{2p}$ in Euclidean space. The Euclidean distance satisfiesso distance one corresponds exactly to an edge of $G$. Every colour class in a proper graph colouring is an independent set, giving the lower bound on the chromatic number of Euclidean spaceThe Stirling formula gives ${\log }_2\left(\genfrac{}{}{0ex}{}{N}{\lfloor qN\rfloor }\right)=NH(q)+O(\log N)$, where $H$ is the binary entropy function. Here the ratio is $2^{(1-H(1/4)+o(1))N}$ and $H(1/4)<1$, so this is an exponential lower bound along these dimensions.

Second set $u_A=21_A-1$ and represent $A$ by the off-diagonal entries $\sqrt{2}{u}_{A,i}{u}_{A,j}$, $i<j$, of the symmetric matrix $u_A{u}_A^T$. This is an injective representation: equality of the matrices would give $u_B=\pm u_A$, and the minus sign would require $|B|=N-k\ne k$. The common diagonal can be omitted without changing distances. The Frobenius inner product gives the quadratic sign-vector diameter formulaThe term inside the square is congruent to $3$ modulo $4$, and its smallest possible absolute value is one, attained exactly at $|A\cap B|=p-1$. Such pairs exist. Consequently the diameter pairs of this finite set in dimension $D=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ are precisely the edges of $G$. Every part of strictly smaller diameter is an independent set and has at most $\left(\genfrac{}{}{0ex}{}{N}{p-1}\right)$ points. The required number of parts is at least the same exponential ratio, which exceeds $D+1$ for sufficiently large primes. Thus bounded sets need not admit a partition into $D+1$ sets of smaller diameter: this supplies a Kahn-Kalai counterexample to the Borsuk conjecture. Both applications use the prime case of the proved prime-power result.

The Frankl-Wilson theorem is a striking example of the polynomial method in combinatorics: an intersection restriction creates a family of linearly independent low-degree polynomials, whose dimension bounds the set family. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}$ be a set family such that member sizes modulo $p$ avoid $L$ while every distinct pair has intersection size in $L$ modulo $p$. Its general modular form givesIf all members have the same size $k$ and $s\le k$, the uniform refinement gives $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$.

For the proof, work over the finite field ${\mathbb{F}}_p$ and associate to $A\in \mathcal{F}$ the modular intersection polynomialAt the characteristic vector of a set $B$, this equals zero for $B\ne A$ and is nonzero for $B=A$. A linear relation among the evaluation functions therefore has every coefficient zero. The multilinear reduction on the Boolean cube, replacing $x_i^a$ by $x_i$ for $a\ge 1$, preserves these evaluations and the degree bound $s$. There are ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ possible square-free monomials, proving the general form.

For uniform members, the low-degree evaluation rank on a uniform layer supplies the sharper dimension bound. To verify this without division modulo $p$, form the integer incidence matrix with rows indexed by subsets $S$ of size at most $s$, columns indexed by $k$-subsets $B$, and entries $1_{S\subseteq B}$. Over the rational numbers, a row with $|S|=j$ is ${\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)}^{-1}$ times the sum of all size-$s$ rows containing $S$. Its matrix rank is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every larger integer minor is consequently zero and remains zero modulo $p$, giving the same bound over ${\mathbb{F}}_p$. The nonzero diagonal evaluation matrix of the $P_A$ then forces $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$.

The theorem, proved by Frankl and Wilson in 1981, extends the theme of the Ray-Chaudhuri–Wilson theorem from ordinary intersection sizes to residues. Modular restrictions can be weaker than exact intersection restrictions yet still force unexpectedly small families. Here are three consequences from one explicit modular intersection graph.

Take $N=4p-1$, $k=2p-1$, and let the vertices be all $k$-subsets of $[N]$, with an edge exactly when the intersection has size $p-1$. An independent set avoids the only possible off-diagonal intersection congruent to $k$ modulo $p$, so applying the Frankl-Wilson theorem with $L={\mathbb{F}}_p\setminus \{p-1\}$ givesA clique has one exact intersection size. Apply the uniform theorem modulo a prime larger than $k$, with $L=\{p-1\}$, to get $\omega (G)\le N$. Since there are $v=\left(\genfrac{}{}{0ex}{}{N}{k}\right)$ vertices, this gives the explicit asymmetric Ramsey number bound $R(N+1,B+1)>v$. It illustrates how algebraic constructions produce graphs with simultaneously restricted cliques and independent sets.

For a geometric application, map each vertex to $x_A=1_A/\sqrt{2p}$ in ${\mathbb{R}}^N$. Then $|x_A-x_B{|}^2=(k-|A\cap B|)/p$, so edges have unit Euclidean distance. Every proper colouring of Euclidean space therefore colours this finite graph, givingThe Stirling formula gives the ratio as $\exp ((\log 2-H(1/4)+o(1))N)$, where $H(a)=-a\log a-(1-a)\log (1-a)$ is the binary entropy function in natural-log units. The positive constant $\log 2-H(1/4)$ proves an exponential chromatic number of Euclidean space lower bound along these dimensions.

Finally, a quadratic embedding converts this same forbidden intersection into a forbidden diameter pair. Set $u_A=21_A-1$ and $Q_A=u_A{u}_A^T$. The quadratic sign-vector diameter formula follows from the Frobenius inner product:For distinct members the smallest possible absolute inner product is one, attained exactly when the intersection is $p-1$. Thus maximum-distance pairs are exactly edges of $G$. The matrices have common diagonal; using their off-diagonal coordinates, multiplied by $\sqrt{2}$, realizes them isometrically in $D=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ dimensions. They are distinct because equal outer products require $u_A=\pm u_B$, and complements have size $2p$, not $k$.

Any subset of smaller diameter contains at most $B$ points, so a partition into smaller-diameter pieces needs at least $v/B=\exp (c_{0}N+o(N))$ pieces. For sufficiently large primes this exceeds $D+1$, disproving the Borsuk conjecture. This is the mechanism behind the Kahn-Kalai counterexample to the Borsuk conjecture: the number of required pieces grows exponentially in the original set-system parameter while dimension grows quadratically. Their 1993 paper establishes the resulting high-dimensional counterexample.

Writing $X=\left(\begin{array}{cc}x& y\\ z& w\end{array}\right)$ and using the Frobenius inner product, the three linear equations areThe first two differ by twice the third; equivalently $A-B=2C$. Thus there are only two independent constraints. Substituting $z=-w$ gives $y=2w-x$, so every element of the vector subspace has the unique expressionThe two coefficient matrices satisfy all three constraints. Their linear independence follows by looking first at entry $(1,1)$ and then at entry $(2,2)$. Therefore a basis for this orthogonal complement is

The matrix trace calculation in (i) givesA sum of nonnegative real numbers is zero exactly when each is zero. Thus $\boxed{\mathrm{tr}(A^{T}A)=0⟺A=0}$, so the real Frobenius inner product is positive-definite.

Write $\alpha =\mathrm{tr}(A^{T}A)$, $\beta =\mathrm{tr}(B^{T}B)$ and $\gamma =\mathrm{tr}(A^{T}B)$. If $B\ne 0$, then $\beta >0$, and applying the same nonnegativity to $A-tB$ givesTaking $t=\gamma /\beta$ yields $\alpha -{\gamma }^2/\beta \ge 0$. If $B=0$, the claimed inequality is immediate. Hence the Cauchy-Schwarz inequality here isWhen $B\ne 0$, equality holds exactly when $\mathrm{tr}((A-(\gamma /\beta )B{)}^T(A-(\gamma /\beta )B))=0$, hence exactly when $A=(\gamma /\beta )B$. Together with the case $B=0$, the complete equality condition is $\boxed{A,B\text{ are linearly dependent}}$, including the case that either is zero.

The matrix trace is the sum of the diagonal entries:Using the definitions of matrix multiplication and matrix transpose,This is the symmetry of the real Frobenius inner product.

False. Tests against symmetric second-rank tensors see only the symmetric part $A_s=(A+A^T)/2$, because the Frobenius inner product of a symmetric matrix and an antisymmetric matrix is zero:An arbitrary basis-dependent antisymmetric part is invisible to these tests. For example, prescribein one basis, and $A^{\prime }=0$ in another rotated basis, choosing antisymmetric arrays in every remaining basis. Every contraction with any symmetric second-rank tensor is the invariant scalar zero. But an invertible rotation cannot transform this nonzero matrix to zero, so the array is not a Cartesian second-rank tensor. This is the blindness of symmetric contraction tests to antisymmetric arrays.

True. In all four tests, an array is understood to have components specified in each rotated orthonormal basis; being a scalar means that the contraction is invariant under those changes. Write the Frobenius inner product as $A:B={\sum }_{ij}{A}_{ij}{B}_{ij}$. Every test Cartesian second-rank tensor obeys $B^{\prime }=RB{R}^T$. The assumption saysChoosing the elementary matrices as $B$, or choosing $B=R^T{A}^{\prime }R-A$, gives $R^T{A}^{\prime }R=A$. Thereforewhich is the required Cartesian second-rank tensor transformation law. This is the scalar contraction test for a Cartesian tensor.

True. Since $A$ and $A^{\prime }$ are antisymmetric, the matrix $D=R^T{A}^{\prime }R-A$ is antisymmetric. Invariance of contractions with every antisymmetric second-rank tensor gives $D:B=0$ for every antisymmetric test matrix $B$. Such a matrix can be freely prescribed in one basis and then transformed to define a valid test tensor. Choose $B=D$ in that basis. Thenso $D=0$ and $A^{\prime }=RA{R}^T$. Together with the given antisymmetry, this proves that $A$ is an antisymmetric second-rank tensor. The antisymmetric contraction test for an antisymmetric tensor works because the Frobenius inner product is nondegenerate on the antisymmetric subspace.

For each fixed $x$ in the nonnegative orthant, the map $A\mapsto x^{T}Ax$ is a continuous linear function on the vector space of real symmetric matrices. Thusis an intersection of closed half-spaces, using the Frobenius inner product $⟨A,B{⟩}_F=\mathrm{tr}(AB)$ on this space. Arbitrary intersections of closed sets are closed, so $K$ is closed.

If $A,B\in K$ and $s,t\ge 0$, then $x^T(sA+tB)x=s{x}^{T}Ax+t{x}^{T}Bx\ge 0$ for every $x\ge 0$. Therefore $sA+tB\in K$, including the zero matrix when $s=t=0$. HenceThe copositive cone is defined in the real space of symmetric matrices; symmetry is needed later to recover every matrix entry from these quadratic forms.