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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.

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.

For a convex cone $D$ in an inner product space, use the nonnegative-pairing conventionHere the pairing is the Frobenius inner product on real symmetric matrices. Let $C_0$ be the conic hull of the nonnegative rank-one matrices $x{x}^T$, and let $C=\overline{C_0}$. For $A\in K$, $⟨A,x{x}^T{⟩}_F=x^{T}Ax\ge 0$. This extends to conic combinations, and by continuity to their limits. Hence $C\subseteq K^{\ast }$.

For the converse, if $B\notin C$, separation from a closed convex cone supplies a symmetric $H$ withIn particular $x^{T}Hx\ge 0$ for every $x\ge 0$, so $H\in K$. The negative pairing then excludes $B$ from $K^{\ast }$. ThereforeThe same generator test gives $C^{\ast }=K$. This is the duality of copositive and completely positive cones; the next argument removes the closure.

The Krivine rounding constant has $t=c_K\pi /2=\log (1+\sqrt{2})<\pi /2$. Since $|X_{ij}|\le 1$, the principal real inverse sine satisfieson every cross-block entry. The matrix $A$ has zero diagonal blocks, so those cross blocks are the only contributors to its Frobenius inner product with $\arcsin [Y]$. Explicitly,No analogous identity is needed for the diagonal blocks involving $\sinh$.

Combining with the preceding expectation formula and optimality of $X$ gives the bipartite sign rounding boundAt least one feasible rounded outcome attains at least this expectation. This is an expectation-based approximation guarantee for the specific bipartite sign problem; no claim is made that $c_K$ is the largest possible constant. The inequalities remain valid when the optimal objective is zero.

If an array specified in each orthonormal basis has invariant Frobenius inner product with every Cartesian second-rank tensor, it transforms as such a tensor. For a component change $R$, invariance says $(R^{T}A\prime R-A):B=0$ for every matrix $B$. Nondegeneracy of the Frobenius inner product forces the difference to vanish. Testing every tensor, rather than a single selected tensor, is essential.