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Let $H$ be the $r$-uniform hypergraph and use strong saturation of a uniform hypergraph in its immediate-completion sense: each missing edge $e$ lies in an $(r+t)$-set $T$ whose other $r$-subsets are already edges of $H$. The argument allows existing hypergraph cliques. Take $n\ge r+t$, the usual parameter range; if $r\le n<r+t$, strong saturation forces every possible edge to be present, and the inequality is immediate whenever its binomial terms are defined. The case $t=0$ has the trivial lower bound zero.

For $t\ge 1$, choose distinct real numbers $z_1,\dots ,z_n$ and form the $t\times n$ Vandermonde matrix $M$ whose $i$th column is $(1,z_i,\dots ,z_i^{t-1}{)}^T$. Every $t$ columns are linearly independent by the Vandermonde determinant. Set $U=\ker M\subseteq {\mathbb{R}}^n$, so $\dim U=n-t$, and work in the quotient vector spaceHere the inclusion of ${\bigwedge }^{r}U$ is injective: extend a basis of $U$ to a basis of ${\mathbb{R}}^n$ and use the resulting wedge bases of the exterior powers. Write $w_e$ for the image in $W$ of the coordinate wedge indexed by an $r$-set $e$. All these images span $W$.

For any $(r+t)$-set $T$, let $V_T$ be its coordinate subspace. The restriction of $M$ has rank $t$, so $U\cap V_T$ has dimension $r$. Wedge a basis of this intersection to obtain a nonzero elementEvery coefficient $c_e$ is nonzero. Indeed, projection of $U\cap V_T$ onto the $e$ coordinates is an isomorphism: a vector in its kernel is supported on $T\setminus e$, whose $t$ columns of $M$ are independent. The coefficient is the determinant of that projection in the chosen bases. Passing to $W$ therefore gives a linear dependence

If $e$ is missing from $H$, take its strong-saturation witness $T$. The dependence expresses $w_e$ in the span of the $w_f$ for the other edges $f\subseteq T$, all of which are present. Thus the vectors indexed by the present edges span all of $W$. The exterior-power quotient bound for strong hypergraph saturation follows:The bound is sharp: include precisely the $r$-sets meeting one fixed $t$-set $S$. A missing edge $e$ is disjoint from $S$, and every other $r$-subset of $e\cup S$ meets $S$, so adding $e$ immediately completes a hypergraph clique of order $r+t$. The number of included edges is exactly the displayed bound.

On the smooth-function space, the product rule gives $ABf=(xf{)}^{\prime }=f+x{f}^{\prime }$ while $BAf=x{f}^{\prime }$. Hence $\boxed{[A,B]=I}$, and both maps preserve the stated smooth space.

For general operators with this commutator relation, use $[A,BC]=[A,B]C+B[A,C]$. Induction gives $[A,B^i]=i{B}^{i-1}$ for $i\ge 1$. Since $Ay=0$,Every displayed vector lies in $W$, so $W$ is invariant under $A$, as well as under $B$. Iterating the lowering formula gives $A^k{B}^{i}y=i!B^{i-k}y/(i-k)!$ for $k\le i$, and zero for $k>i$.

Suppose a nontrivial finite linear dependence has highest nonzero term $c_n{B}^{n}y$. Applying $A^n$ kills all lower terms and yields $c_{n}n!y=0$. The field is real, $n!\ne 0$, and $y\ne 0$, so this contradicts $c_n\ne 0$. Thus the entire sequence is linearly independent. This identity commutator cyclic ladder also proves that such a representation with a nonzero vector in the kernel of $A$ cannot be finite dimensional.

In an oriented orthonormal coordinate system, the dot product and cross product areThe scalar triple product is $a\cdot (b\times c)$; equivalently it is the determinant of the matrix whose columns are $a,b,c$. The three vectors have linear independence when

Let $U=(a_1{a}_2{a}_3)$ and $V=(b_1{b}_2{b}_3)$ be column matrices. Since $(U^{T}V{)}_{ij}=a_i\cdot b_j$, we have $S=U^{T}V$. Multiplicativity of the determinant and its invariance under matrix transpose yieldA square matrix has maximal matrix rank exactly when its determinant is nonzero, equivalently when its columns have linear independence. A product of two real numbers is nonzero exactly when both factors are nonzero. Hence $\boxed{\mathrm{rank}S=3⟺(a_i{)}_{i=1}^3\text{ and }(b_j{)}_{j=1}^3\text{ are both independent}}$.

The same argument works in every dimension. Set $U_n=(c_1\cdots c_n)$ and $V_n=(d_1\cdots d_n)$. Thenso $\boxed{\mathrm{rank}T=n⟺(c_i{)}_{i=1}^n\text{ and }(d_j{)}_{j=1}^n\text{ are both independent}}$.

For the last construction, choose two standard basis vectors $e_1,e_2$ in ${\mathbb{R}}^n$ and take $c_j=e_1+j{e}_2$ for $j=1,\dots ,n$. If $\alpha c_i+\beta c_j=0$ with $i\ne j$, its first two coordinates give $\alpha +\beta =0$ and $i\alpha +j\beta =0$, hence $\alpha =\beta =0$. Every pair therefore has linear independence. Every triple lies in the two-dimensional linear span of $e_1,e_2$ and has linear dependence. More explicitly, for distinct $i,j,k$,a nontrivial dependence. Thus $\boxed{c_j=e_1+j{e}_2(1\le j\le n)\text{ works for every }n>2}$.

Put $f=[k_L:k]$. For an unramified extension the ramification index is one and $[L:K]=f$. The residue $\bar{\alpha }$ has degree $f$ over $k$. Thus $1,\alpha ,\dots ,{\alpha }^{f-1}$ are linearly independent over $K$: a nonzero linear dependence can be divided by a coefficient of least valuation and reduced to a nonzero relation among the reduced powers. Hence $[K(\alpha ):K]\ge f=[L:K]$, and $\alpha$ generates $L$.

The same argument gives the integral statement, which is stronger than merely generating the field. WriteIf $m={\min }_j{v}_K(c_j)$, divide by ${\pi }_K^m$. Since ${\pi }_K$ remains a uniformizer in the unramified extension and the reduced powers form a basis of $k_L/k$, the resulting sum has nonzero residue. Therefore $v_L(x)=m$. In particular $x$ is integral if and only if every coefficient is integral. This proves unramified residue generator integral basis:

A set is linearly dependent if some finite selection of distinct vectors admits a relation ${\sum }_j{c}_j{v}_j=0$ with coefficients not all zero. For a finite set this is just such a nontrivial relation using its vectors.

Prove dependence of $n+1$ vectors in ${\mathbb{R}}^n$ by induction, starting with the zero-dimensional case. If all first coordinates vanish, select $n$ of the vectors and apply induction in ${\mathbb{R}}^{n-1}$. Otherwise relabel so the last vector has first coordinate $a\ne 0$. For $1\le j\le n$, put $w_j=v_j-(v_{j,1}/a)v_{n+1}$. These $n$ vectors have zero first coordinate, so induction supplies ${\sum }_{j=1}^n{c}_j{w}_j=0$, with some $c_j\ne 0$. Expanding gives a nontrivial linear dependence among the original vectors. The same proof works over any field.

Now express vectors of $V$ in coordinates relative to its given $n$-element basis. Any other basis has at most $n$ elements: more would contain $n+1$ linearly dependent vectors. If it has $m<n$ elements, expressing the original basis in its coordinates makes the original $n$ vectors linearly dependent, a contradiction. Thus every basis has exactly $n$ elements.

Finally, the functions $\sin (jx)$, $j=1,2,\dots$, are bounded and continuous. Any finite relation among them can be multiplied by $\sin (rx)$ and integrated over $[0,2\pi ]$. The orthogonality identityforces every coefficient to vanish. Arbitrarily large linearly independent finite sets exist, so the vector space of bounded continuous functions is infinite-dimensional.

The Steinitz exchange lemma says that if $u_1,\dots ,u_r$ are linearly independent and $v_1,\dots ,v_s$ span the same vector space, then $r\le s$ and, after reordering the $v_j$, the list $u_1,\dots ,u_r,v_{r+1},\dots ,v_s$ still spans it.

To prove this, suppose $u_1,\dots ,u_k$ have already replaced $k$ of the spanning vectors. Express $u_{k+1}$ in that spanning list. At least one coefficient of an unreplaced $v_j$ is nonzero: otherwise $u_{k+1}$ lies in the span of $u_1,\dots ,u_k$, contradicting linear independence. Solve this expression for that $v_j$ and replace it by $u_{k+1}$. The new list still spans. If all $s$ vectors were replaced before the $u_i$ were exhausted, the next $u_i$ would be dependent on its predecessors. Hence $r\le s$, and the replacement can continue through $u_r$.

For two bases of sizes $r,s$, apply the Steinitz exchange lemma in both directions to get $r\le s\le r$. Thus every basis has the same size. Given a linearly independent set and any finite basis, the replacement procedure produces a finite spanning list containing that set. Successively delete any remaining vector dependent on the others; none of the prescribed independent vectors need be deleted, since one can instead delete a dependent supplementary vector. Equivalently, greedily retain only supplementary vectors outside the span already built. The resulting list is a basis containing the prescribed set.

For the three-dimensional example, the columns have matrixTherefore the three vectors are a basis. For the four-dimensional example,This is a nontrivial linear dependence, so the four vectors are not a basis. The distinction is an instance of the adjacent sums of a cyclic basis criterion: alternating coefficients close consistently around an even cycle, but not an odd one.

If $x=0$, the generated subspace is zero and invariant. Otherwise, let $r$ be the first index for which $x,Tx,\dots ,T^{r}x$ are linearly dependent. Such an index exists with $1\le r\le n$, since there are $n+1$ vectors. The preceding vectors are independent, so the coefficient of $T^{r}x$ in this first relation cannot vanish. HenceThe linear operator $T$ sends each of the first $r-1$ generators to the next and sends the last into their span. This proves that their span is an invariant subspace. Every later iterate lies in the same subspace by induction, so it equals the span of the first $n$ iterates. Therefore the indicated subspace is $T$-invariant. This proof of a cyclic subspace uses only finite dimension and linear dependence.

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.

First $w\in {\ell }^2$, since ${\sum }_{n\ge 1}{4}^{-n}=1/3$. Any finite linear dependence among $w$ and the $e_m$ has the form $aw+{\sum }_{m\in F}{b}_m{e}_m=0$ for a finite index set $F$. At an index outside $F$, its coordinate equals $a{2}^{-n}$, so $a=0$. Each remaining coordinate then gives $b_m=0$. Therefore the set is linearly independent, and the prescribed values define a unique linear map on its algebraic span.

Take the tailsThey belong to $V$, satisfy $f(v_N)=1$, and haveThus $v_N\to 0$ while $f(v_N)$ does not approach $f(0)=0$. The map is not continuous, and part (b) also shows it is unbounded. This is a discontinuous functional detected by vanishing sequence tails, illustrating why the finite-dimensional conclusion in part (a) cannot be extended to arbitrary normed spaces.

Put $q=(1,1,1,1{)}^{\mathsf{T}}$ and arrange the four vectors as the columns of $A=(x-1)I+q{q}^{\mathsf{T}}$. On the one-dimensional vector subspace spanned by $q$, the eigenvalue is $x+3$. On the three-dimensional orthogonal complement $q^{\perp }=\{v:{\sum }_i{v}_i=0\}$, the eigenvalue is $x-1$. Thusand the vectors fail to be a basis exactly when $x=1$ or $x=-3$.

For $x=1$, all four columns equal $q$. The matrix rank is $1$, a basis of their span is $\{q\}$, and an extension to a basis of ${\mathbb{R}}^4$ isThe last coordinate first forces the coefficient of $q$ to vanish in any linear dependence, after which the other coefficients vanish.

For $x=-3$, the matrix rank is $3$ and the span is exactly $q^{\perp }$. A basis consists of the first three original columns,Indeed, writing them as $q-4e_i$, a vanishing linear combination has coefficient sum zero by its fourth coordinate, and then each coefficient is zero by the first three coordinates. Adjoin $q$ to extend this basis to ${\mathbb{R}}^4$, since $q\notin q^{\perp }$.

The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

Expand $\det (A+tB)$ by multilinearity of the determinant in its columns. The coefficient of $t^j$ is a sum of determinants obtained by choosing $j$ columns from $B$ and the remaining columns from $A$. Since the columns of $B$ span a space of dimension $k$, every choice of more than $k$ columns from $B$ is linearly dependent. All coefficients of $t^j$ for $j>k$ therefore vanish, so

If $k=n$, then $B$ is invertible and the coefficient of $t^n$ is $\det B\ne 0$. Hence the degree is exactly $n$.

For a zero-polynomial example with $k=1$, takeThen $\mathrm{rank}B=1$, while $\det (A+tB)=0$ for every $t$.

If the two regressors are strongly associated, the data may provide little support for varying one while holding the other fixed. Multicollinearity makes the separate slope estimates unstable, and exact linear dependence makes them unidentifiable. A conditional slope may also be an extrapolation beyond combinations actually observed.

Furthermore, a slope in a linear regression model is not automatically a causal effect. A causal interpretation needs appropriate design or assumptions excluding confounding; independence of the errors across observations and zero mean alone do not establish those assumptions. The coefficients describe the specified linear mean relation, but changing a regressor by intervention can have a different interpretation.

Assume $\alpha <\beta$ and abbreviateThe integral of the nonnegative square expands to $A+2\lambda B+{\lambda }^{2}C\ge 0$ for every real $\lambda$. If $C>0$, completing the square, or choosing $\lambda =-B/C$, gives $A-B^2/C\ge 0$. Hence the Cauchy-Schwarz inequality isIf $C=0$, continuity and nonnegativity imply $g=0$ throughout the interval, so the same inequality is immediate.

When $C>0$, equality holds exactly when $\int (f-(B/C)g{)}^{2}dx=0$. The integrand is continuous and nonnegative, so it must vanish everywhere, giving $f=(B/C)g$. Including the case of a zero function, equality is equivalent to linear dependence of $f$ and $g$ as elements of the real vector space of continuous functions on the interval.

No. Every vector in the plane can be written asThe two linearly independent vectors in part (ii) therefore form a vector space basis for this vector subspace. Its dimension is $2$, so any three vectors in it have linear dependence.

The nonzero direction vectors now have linear dependence, so the lines are parallel or coincident. Subtract from $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ its orthogonal projection along ${\mathbf{v}}_1$; the remaining component is the shortest joining displacement. The distance between parallel lines isThe second expression follows from the relation between the cross product and the perpendicular component. It includes distance zero for coincident lines.

No. Every one of the three vectors has equal first and second Cartesian coordinates, so their linear span is contained in the plane $x=y$. That linear subspace has dimension two and cannot equal ${\mathbb{R}}^3$.

More explicitly, call the three given vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ in their listed order. ThenThis nontrivial linear relation proves linear dependence, whereas a basis must be linearly independent. Being unit vectors alone does not ensure that they form a basis.

Take a normalized vector in the required common operator domains. Expanding the squared norm and using self-adjointness givesThus its final term is $\hslash t⟨C⟩$. Set $a=\Vert A\psi {\Vert }^2$, $b=\Vert B\psi {\Vert }^2$, $c=⟨C⟩$. If $b>0$, minimizing the nonnegative quadratic at $t_{\ast }=-\hslash c/(2b)$ givesEquality holds precisely when the minimum norm vanishes, that is $(A-i{t}_{\ast }B)\psi =0$. If $b=0$, however, then $B\psi =0$ and $c=0$, so equality is automatic even when $A\psi \ne 0$. The printed equality criterion needs this zero-vector exception. For example $A=I$, $B=0$, $C=0$ saturates the inequality but admits no such real $t$. The exact general criterion is linear dependence of $A\psi ,B\psi$ together with vanishing real part of their inner product, as in the uncentered second-moment commutator inequality.

Apply the same calculation to $A-⟨A⟩I$, $B-⟨B⟩I$, whose commutator is unchanged. This gives the Robertson uncertainty principleIf $\psi$ is an eigenvector of $A$, its dispersion is zero; finite dispersion of $B$ then forces $\boxed{⟨C⟩=0}$.

In the position representation, $X\psi =x\psi$ and $P\psi =-i\hslash {\psi }^{\prime }$. The product rule gives $[P,X]\psi =-i\hslash \psi$, hence $\boxed{[P,X]=-i\hslash I}$. Since $V(X)$ commutes with $X$,For an energy eigenstate, $⟨[H,X]⟩=E⟨X⟩-E⟨X⟩=0$, so $\boxed{⟨P⟩=0}$. For the oscillator $[H,P]=i\hslash m{\omega }^{2}X$, and the same argument gives $\boxed{⟨X⟩=0}$. These expectation identities require the stated finite moments and commutator domains; formal position eigenvectors are not normalizable exceptions.

Apply the Robertson uncertainty principle to each angular-momentum pair:If $\psi$ is an eigenvector of $L_3$, then $\Delta L_3=0$. The inequalities for $(L_2,L_3)$ and $(L_3,L_1)$ force $\boxed{⟨L_1⟩=⟨L_2⟩=0}$, assuming the relevant moments exist.

By the spectral theorem a symmetric positive-definite matrix $C$ has $C=Q\mathrm{diag}(c_i)Q^{\mathsf{T}}$, with $Q$ orthogonal and $c_i>0$. Define $S=Q\mathrm{diag}(\sqrt{c_i})Q^{\mathsf{T}}$. It is symmetric positive-definite and $S^2=C$. For uniqueness, any symmetric positive-definite square root $T$ commutes with $C=T^2$, hence preserves each eigenspace of $C$. On its $c$-eigenspace, $T$ is symmetric positive-definite and all its eigenvalues have square $c$, so all equal $\sqrt{c}$. Therefore $T=S$, including repeated eigenspaces.

For $\det F>0$, putThen $R^{\mathsf{T}}R=I$ and $\det R=\det F/\det U>0$, so $R\in SO(3)$; also $F=RU=VR$. Any other right polar decomposition has $F^{\mathsf{T}}F=U^2$, so uniqueness of the positive root fixes $U$ and then $R$. The left factor is consequently fixed too.

The principal stretches $v_i$ are the positive eigenvalues of $U$, equivalently the singular values of $F$. For a unit vector $N={\sum }_i{c}_i{N}_i$ in an orthonormal eigenbasis of $U$,Thus $\boxed{\underset{|N|=1}{\max }|FN|=v_{\max }}$, attained precisely by the unit vectors in the eigenspace of $U$ belonging to $v_{\max }$. With a simple largest eigenvalue these are its two opposite unit eigenvectors; degeneracy allows every unit combination within that eigenspace.

For the simple shear,A line with nonzero direction $N$ keeps its length exactly whenHence $\boxed{N_2=0\text{or}2N_1+\gamma N_2=0}$, with $N_3$ arbitrary. Every segment in the cube parallel to one of these directions preserves length; there is no restriction on its starting point other than remaining in the cube.

The two in-plane stretches are $v_{\pm }=(\sqrt{4+{\gamma }^2}\pm \gamma )/2$, and the third is one. A maximizing direction for $\gamma >0$ is $N\parallel (1,v_{+},0)$, since the $C$ eigenvalue equation gives $N_2/N_1=(v_{+}^2-1)/\gamma =v_{+}$. Therefore $\boxed{N\to \pm (1,1,0)/\sqrt{2}}$ as $\gamma \downarrow 0$. At exactly zero shear all directions tie; the limit refers to the selected maximizing direction for positive shear. These are the length-preserving directions and principal directions of simple shear.

On each half-space the deformation is affine. Continuity on the separating plane makes the two affine constants equal and gives $(A-B)t=0$ for every tangent vector $t\perp N$. Decomposing any $x$ into its tangential part and $(x\cdot N)N$ givesThis is the Hadamard jump condition. Since $B=A-a\otimes N=A(I-A^{-1}a\otimes N)$, the matrix determinant lemma givesBoth determinants are positive, hence $\boxed{A^{-1}a\cdot N<1}$.

Write $a_1={\eta }_1$, $a_2={\eta }_2$, $S=a_1^2+a_2^2$ and $\Delta =a_2^2-a_1^2\ne 0$. The two proposed angles have $c=2a_1{a}_2/S$ and $s=\kappa \Delta /S$, and $c^2+s^2=1$. Therefore the displayed rotations lie in $SO(3)$. Direct multiplication givesThis is a rank-one connection for each sign. With a unit normal one may choose $N=(1,\kappa ,0)/\sqrt{2}$ and jump vector $a=\sqrt{2}\Delta (-a_2,\kappa a_1,0)/S$. The two rotations are distinct because $\Delta \ne 0$; the permitted at-most-two result therefore proves these are all solutions. Flipping both $a$ and $N$ changes no connection.

It remains to test the material plane. Because $y(0)=0$, the affine constants vanish. Let $t=(-p_2,p_1,0)$, so the plane is spanned by $t$ and $e_3$. Both deformation gradients send $e_3$ to $a_1{e}_3$. If the plane is not the twin plane, it cuts the two half-spaces; its two image halves are coplanar exactly when the in-plane vectors $Bt$ and $At$ are parallel. Put $d=(-a_2,\kappa a_1)$ and $n=(1,\kappa )$ in their two-dimensional sections. ThenIts prefactor is nonzero, so coplanarity requires $p_1=\kappa p_2$, precisely $p\parallel (1,\kappa ,0)\parallel N$. Conversely, on that twin plane $(A-B)x=0$, so the two affine maps agree and its image is a plane. Thus $\boxed{\Pi \text{ remains plane}⟺p\parallel N}$. All other allowed planes acquire a kink along their intersection with the twin. This is the planarity test across a cubic-to-tetragonal twin.