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Set $N=2^n$, $a=N^{-1/2}=\sin t$ and $c=\cos t$, with $N>1$. The ordered vectors $|w⟩,|s⟩$ are normalized but nonorthogonal, since $⟨w|s⟩=a$. Their coordinate Gram matrix is thereforeAn operator matrix in this basis is obtained from the coefficients in the images of its basis vectors, not by treating those vectors as orthogonal measurement outcomes. Directly applying the two reflection operators givesTheir columns are consequentlyThese are physical unitary operators even though the coordinate matrices are not unitary in the ordinary Euclidean coordinate norm: they preserve $S$, so $U^{†}SU=S$.

The Grover iteration in a nonorthogonal basis isIt has determinant one and trace $2-4a^2=2\cos 2t$, hence eigenvalues $e^{\pm 2it}$. For example, the invertible matrix of eigenvectorssatisfiesThis diagonalization is exact for $N>1$; no large-$n$ approximation is needed.

To extract an actual probability using the Born rule, introduce the orthonormal unmarked superposition $|b⟩=(|s⟩-a|w⟩)/c$. In the ordered basis $|w⟩,|b⟩$, the same iteration is the rotationAfter $r$ iterations,Equivalently the original coordinates are $[\sin (2rt)/c,\cos ((2r+1)t)/c{]}^T$. The overlap with $|w⟩$ is the first coordinate plus $a$ times the second, recovering the same amplitude. Squaring just the first coordinate would be incorrect in a nonorthogonal basis.

For large $n$, $t\simeq N^{-1/2}$, and dropping the initial angle $t$ gives the requested approximation $P_r\simeq {\sin }^2(2rt)$. The first near-unit success occurs at the integer closest to $\pi /(4t)-1/2$:One can choose $r_{\mathrm{opt}}=\lfloor \pi /(4t)\rfloor$, with either neighbor equivalent in a tie. Further iterations overshoot and reduce success rather than monotonically improving it. The special $N=1$ problem needs no search and the two-vector coordinate construction becomes singular.

For an unstructured search, a reversible Boolean quantum oracle implements a phase change on the marked entry, while the diffusion reflection about a uniform state is known. Repeating the pair of reflections amplifies the marked amplitude and measurement yields a candidate that can be checked by another oracle query. The bounded-error query count is $O(\sqrt{N})$ for one marked item, compared with classical $\Theta (N)$ queries. It is a quadratic oracle-query speedup: state preparation, the cost of implementing the oracle and access to a classical database still matter. The method does not reveal an unknown marked value without the promised query access.^[ref]

Use the thin singular value decomposition $a=Ud{V}^{†}$, with $U^{†}U=I_N$, $V$ unitary and $d=\mathrm{diag}(d_k)$ nonnegative. Expanding its entries givesThe columns of $U$ and of $V^{\ast }$ are orthonormal, establishing the Schmidt decomposition. Bob's basis uses the complex-conjugated columns of $V$. Omitting that conjugation would not reproduce the coefficient matrix. The squared Schmidt coefficients are the eigenvalues of $a^{†}a$; equivalently $d^2=V^{†}{a}^{†}aV$ and $d=U^{†}aV$. The two reduced density matrices are $a{a}^{†}$ and $(a^{†}a{)}^T$, which have the same nonzero eigenvalues.

The printed coefficient $0.41231$ slightly rounds a normalized state: each squared norm is $0.9999995361$. Normalize before calculating an entropy. For the two states, the resulting Gram matrices are approximatelyFor a trace-one two-dimensional Hermitian matrix with diagonal entry $u$ and off-diagonal entry $c$, its eigenvalues areSort them decreasingly and take square roots:The Von Neumann entropy is $S=-{\lambda }_0{\log }_2{\lambda }_0-{\lambda }_1{\log }_2{\lambda }_1$. $\Psi$ is more entangled: its Schmidt weights are closer to equal and its reduced-state entropy is larger. Neither state is a product state or maximally entangled on its two-dimensional support.

By Nielsen's pure-state conversion theorem, deterministic LOCC conversion requires the input vector of squared Schmidt coefficients to be majorized by the output vector. Here $0.699941<0.894238$ and both sums are one, so $\lambda (\Psi )\prec \lambda (\Phi )$. Therefore $\Psi \to \Phi$ is possible deterministically; $\Phi \to \Psi$ is not. Entanglement can be discarded by local operations but cannot increase deterministically for these pure inputs. Probabilistic concentration is a separate question and is allowed below.^[ref]

There is a dimensional qualification in the proposed measurement. Write $P=|0_A^{\prime }⟩⟨0_A^{\prime }|+|1_A^{\prime }⟩⟨1_A^{\prime }|$. The two supplied measurement operators obeyThus their effects form a POVM on the occupied support, where $P$ is the identity. Alice's full space is three-dimensional, so these two operators alone do not sum to $I_3$. A full-space completion is $M_3=|2_A^{\prime }⟩⟨2_A^{\prime }|$. It has zero probability on either input, so it changes none of the requested outcomes. Strictly, the POVM consists of the effects $E_r=M_r^{†}{M}_r$; the specified $M_r$ additionally fix a quantum instrument and therefore the conditional states.

In the appropriate Schmidt basis for either input, write $|\psi ⟩=\sqrt{{\lambda }_0}|00⟩+\sqrt{{\lambda }_1}|11⟩$. The two outcomes have probabilitiesand normalized conditional statesThese formulas give both outcomes for each state by substituting its row of the table. For example,The two inputs have different Schmidt bases; the primed measurement basis is adapted separately to the state, as the question specifies, not one common unexamined laboratory basis.

For the $M_1$ outcome to be a maximally entangled state, its two amplitudes must be equal. HenceThe requested numerical results areThus $\Psi$ has the larger success probability, consistent with its greater initial entanglement. The $M_2$ branch cannot be balanced in the allowed angle range for either nonmaximal input: it would require $\tan \theta =\sqrt{{\lambda }_0/{\lambda }_1}>1$.

The success probabilities belong to this particular binary Schmidt filter. They are not the optimum over every possible quantum instrument. The diagonal success operator $\mathrm{diag}(\sqrt{{\lambda }_1/{\lambda }_0},1)$ instead gives the standard single-copy entanglement concentration probability $2{\lambda }_1$. Postselection can increase entanglement in the retained branch, while the outcome-averaged entanglement cannot increase under LOCC.

Let $P$ be the $3$-by-$3$ evaluation matrix $P_{i,\ell +1}=p_{\ell }({\lambda }_i)$. The kernel evaluation matrix is the Gram matrix $P{P}^T$, so multiplicativity of the determinant givesBecause the polynomials are monic of consecutive degrees, their coefficient matrix relative to $1,x,x^2$ is a triangular matrix with unit diagonal. It has determinant one, so the Vandermonde determinant invariance under monic basis change givesSquaring removes the ordering sign. Thus the Gaussian unitary ensemble joint probability density function has precisely the kernel formNo orthogonality of the polynomials is needed for this step.

Both $L$ and $R$ have $m$ linearly independent columns. The Gram matrix $R^{\top }R$ is invertible becauseConsequently $R^{\top }$ is a surjective linear map: for any $y\in {\mathbb{R}}^m$, the choice $x=R(R^{\top }R{)}^{-1}y$ satisfies $R^{\top }x=y$. HenceThe latter image of a linear map has dimension $m$, giving $\mathrm{rank}B=m$. If $m=0$, $B=0$ and the conclusion is immediate. This argument explains why the product retains its full possible matrix rank; a general product of two matrices need not do so.

For a positive-definite weighted inner product on a compact interval, a least-squares polynomial is the orthogonal projection of a function onto the finite-dimensional space ${\Pi }_n$ of polynomials of degree at most $n$. In a basis ${\varphi }_0,\dots ,{\varphi }_n$, its coefficients solve the normal equations $Gc=d$, where $G_{ij}=({\varphi }_i,{\varphi }_j)$ and $d_i=(u,{\varphi }_i)$. The Gram matrix is positive definite: $z^{T}Gz=\Vert \sum z_i{\varphi }_i{\Vert }^2>0$ for $z\ne 0$. Thus the coefficients exist and are unique. The residual is orthogonal to every candidate polynomial, giving $\Vert u-q{\Vert }^2=\Vert u-p_n{\Vert }^2+\Vert p_n-q{\Vert }^2$ and proving optimality. An orthogonal basis makes these equations diagonal, as in a least-squares polynomial in an orthogonal-polynomial basis.

A sum of squares polynomial is a real polynomial expressible as $p={\sum }_i{q}_i^2$ for real polynomials $q_i$. In the homogeneous degree-$2m$ case the summands may be taken homogeneous of degree $m$. Every such polynomial is nonnegative on real inputs; the converse fails for multivariate polynomials in general.

A real polynomial $p$ is a sum of squares polynomial if $p={\sum }_{\ell }{q}_{\ell }^2$ for real polynomials $q_{\ell }$. Such a representation certifies global nonnegativity. If $m(z)$ is the vector of relevant monomials, the identity $p(z)=m(z{)}^{T}Gm(z)$ with a positive semidefinite matrix $G$ encodes a sum of squares through a Gram matrix. Equating coefficients is linear in $G$, so finding a certificate is a semidefinite program.

For real vectors $x_i$ in a Euclidean inner product space, their Gram matrix $G_{ij}=x_i^T{x}_j$ satisfies the displayed identity by expanding the squared norm. Thus every Gram matrix is a positive semidefinite matrix, and it is a positive-definite matrix exactly when the vector family is linearly independent. The complex version replaces transposes by conjugate transposes and uses $c^{\ast }Gc=\Vert {\sum }_i{c}_i{x}_i{\Vert }^2$.

The Specht module $S^{(n-1,1)}$ is the augmentation subspace of the natural permutation module $F^n$, with the standard tabloid bilinear form. Its orthogonal complement in $F^n$ is the constant-vector line. That line lies in the augmentation subspace exactly when the field characteristic divides $n$, giving the displayed radical. Equivalently, the basis $v_i-v_n$ has Gram matrix $I_{n-1}+J_{n-1}$. The James submodule theorem makes every nonzero radical quotient simple, with dimension $n-2$ when $p\mid n$ and $n-1$ otherwise. The case $n=2,p=2$ gives the zero quotient rather than a nonzero simple module.

The Gram matrix of the canonical height pairing of rational points gives a robust independence certificate. For two points, set $a=\hat{h}(P)$, $b=\hat{h}(Q)$ and $c=(\hat{h}(P+Q)-a-b)/2$. If bounds from rounded measurements imply $a>0$ and $ab-c^2>0$, the matrix is positive-definite. Any integer relation $rP+sQ=O$ would give $a{r}^2+2crs+b{s}^2=0$, forcing $r=s=0$. Thus the points generate a free abelian group of rank two. Checking an interval lower bound for the determinant, rather than just its approximate central value, makes the certificate insensitive to rounding. It certifies independence, not that the points are a basis of the entire Mordell-Weil group.

For coefficient column $q=(q_0,\dots ,q_d{)}^T$, $M_{ij}=q_i\overline{q_j}$ is a Hermitian matrix and a positive semidefinite matrix because $x^{\ast }Mx=|q^{\ast }x{|}^2$. Expanding $|q(z){|}^2$ on the unit circle gives coefficient $p_k={\sum }_{i-j=k}{M}_{ij}$. Thus $M$ has matrix rank one when $q\ne 0$ and zero otherwise. General feasible Gram matrices need not have matrix rank one.

In a real finite-dimensional inner product space, let two bases satisfy $⟨u_i,w_j⟩={\delta }_{ij}$. Taking inner products with $u_i$ in $v={\sum }_j{b}_j{w}_j$ gives $b_i=⟨u_i,v⟩$. If also $v={\sum }_j{a}_j{u}_j$, then $b_i={\sum }_j⟨u_i,u_j⟩a_j$. Thus the Gram matrix $G_U$ converts the original coordinates into reciprocal basis coordinates, and $G_W=G_U^{-1}$ converts back. With Euclidean Cartesian column basis matrices $U,W$, reciprocity is $U^{T}W=I$, hence $W=U^{-T}$ and $W^{-1}U=U^{T}U=G_U$. This is why dot products with nonorthogonal basis vectors are coordinates in the reciprocal basis, rather than coordinates in the original basis.

For normalized linearly independent pure states, their Gram matrix $G_{ij}=⟨{\psi }_i|{\psi }_j⟩$ is positive definite. The displayed reciprocal vectors satisfy $⟨{\chi }_i|{\psi }_j⟩={\delta }_{ij}$. The operator $R={\sum }_i|{\chi }_i⟩⟨{\chi }_i|$ has nonzero eigenvalues equal to those of $G^{-1}$: use the polar factorization of the matrix whose columns are ${\psi }_i$. Choosing $0<ϵ\le {\lambda }_{\min }(G)$ gives the POVM effects $E_i$ and $E_{?}=I-ϵR$. Each conclusive outcome has probability $ϵ$ on its own state and zero on every other input. The inconclusive effect preserves positivity, so the construction is a physical measurement in quantum mechanics.

Let the left and right fermions transform in $D\oplus \epsilon$, where $D$ is the two-dimensional irreducible representation of the symmetric group $S_3$ and $\epsilon$ is the sign representation. Let the scalar field be a trivial representation. In matching orthonormal bases, invariance of ${\bar{\psi }}_L\lambda {\psi }_R$ says that $\lambda$ commutes with $D\oplus \epsilon$. The Schur lemma forces each diagonal irreducible block to be scalar and both mixed blocks to vanish, giving the displayed form. Thus after a vacuum expectation value $h$, the physical fermion mass matrix has singular values $|ha|,|ha|,|hb|$: unbroken $S_3$ enforces a degenerate doublet. In a common nonorthonormal basis with invariant Gram matrix $K$, the doublet coefficient is $aK$; canonical normalization recovers $a{I}_2$.

For real symmetric $A$, replace sign-vector lifts $x{x}^T$ by all matrices in the elliptope:Every sign vector gives a feasible rank-one matrix with the same objective, so the relaxed maximum is an upper bound. The omitted rank-one condition is substantive: an arbitrary feasible Gram matrix need not come from signs.

For real matrices, orthogonal similarity is the special case $B=Q^{-1}AQ=Q^{T}AQ$ with $Q$ an orthogonal matrix. Then a symmetric matrix remains symmetric, because $B^T=Q^T{A}^{T}Q=B$. The expression $QA{Q}^T$ describes the reverse conversion and has the same property. An orthonormal basis retains the ordinary dot product formula for its coordinate vectors.

An arbitrary change of basis need not preserve entrywise symmetry. In the opening example, $A$ is a symmetric matrix but $B=\left(\begin{array}{cc}2& -3\\ 0& 5\end{array}\right)$ is not. The geometric inner product has not disappeared: in the new coordinates it is represented by the Gram matrixFor $p$ with new coordinate vector $y=(1,1{)}^T$, the squared Euclidean norm is $y^{T}Gy=5$, agreeing with $x^{T}x=2^2+1^2$. The naive coordinate sum $y^{T}y=2$ would use the wrong inner product.

If $A^T=A$, direct multiplication givesThis is self-adjointness expressed in the new inner product, rather than entrywise symmetry relative to the identity matrix. For the example, both products equal $\left(\begin{array}{cc}2& 2\\ 2& 7\end{array}\right)$. Thus arbitrary matrix similarity preserves the linear operator, while its geometric interpretation requires transforming the Gram matrix as well. Matrix congruence supplies the corresponding rule for a bilinear form and should be distinguished from matrix similarity.

With the nonnegative Laplace-Beltrami operator, a frequency $w$ in the dual lattice gives the eigenfunction $e^{2\pi i⟨w,x⟩}$ and eigenvalue $4{\pi }^2|w{|}^2$. Its multiplicity is the number of dual vectors with that norm. Fourier series prove completeness. In dimension two the shortest vector, shortest independent vector and covolume determine a reduced Gram matrix, proving spectral rigidity.

Let $b_i=N{e}_i$ be the interpolating cardinal basis and set ${\Gamma }_{ij}={\int }_a^b{b}_i^{\prime \prime }{b}_j^{\prime \prime }dx$. Linearity gives $J(N\mathbf{v})={\mathbf{v}}^T\Gamma \mathbf{v}$, and this Gram matrix is a positive semidefinite matrix. Its null space consists precisely of vectors $(\alpha +\beta x_i{)}_i$, by the null-space statement for the second derivative roughness penalty. Thus its rank is $n-2$, including the zero matrix when $n=2$.

An order-$k$ B-spline is supported on $[t_i,t_{i+k}]$. Below this interval its truncated-power knot data agree with a degree-$k-1$ polynomial, annihilated by the order-$k$ divided difference; above it all data vanish. Therefore its ordinary or mixed Gram matrix has entries zero when $|i-j|\ge k$. The half-bandwidth is $k-1$, and for strictly increasing knots this bound is sharp.

If two simple roots have angle $5\pi /6$, normalize them to unit vectors $u,v$ with $(u,v)=-\sqrt{3}/2$. For any other unit simple root $w$, set $r=(u,w)\le 0$, $s=(v,w)\le 0$. Its three-root Gram matrix has determinant $1/4-r^2-s^2-\sqrt{3}rs$. Any nonzero $r$ or $s$ has absolute value at least $1/2$, contradicting positive definiteness. Thus all other simple roots are orthogonal to $u,v$. An irreducible root system containing this triple bond must be the rank-two G2 root system.

The vector-length multiset of a rank-two Euclidean lattice determines it up to an orthogonal map. Let $a$ be the shortest nonzero length. Removing two copies of each positive integer multiple of $a$ removes one primitive lattice line; the shortest remaining length $b$ is the shortest independent-vector length. Those two vectors form a lattice basis: otherwise a nonzero point in the centred basis parallelogram would be an independent lattice vector of length less than $b$. Lattice-point density determines the covolume $A$. The basis Gram matrix then has diagonal entries $a^2,b^2$ and absolute off-diagonal entry $\sqrt{a^2{b}^2-A^2}$. This proves two-dimensional spectral rigidity for a flat torus through its dual lattice.