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The generator acts only in the $ct,z$ plane, with its square equal to the identity on that two-dimensional subspace. Its exponential isFor the vector in (c), this giveswhile $x^{\prime }=x$ and $y^{\prime }=y$. Since the vector is future-null, its spatial magnitude is $c{t}^{\prime }$. ThereforeThe transverse azimuth does not change. These are exactly the direction and scale obtained from the spinor congruence. In particular $c{t}^{\prime }=\mathcal{R}ct$ and $\tan ({\theta }^{\prime }/2)=e^{-\kappa }\tan (\theta /2)$. The azimuth-sign qualification in (c) is compatible with this check because a boost along $z$ leaves both transverse coordinates unchanged.

Interpret the two displayed exchange components as belonging to the common nearest-neighbour sum with coefficient $-J$. Write $H_f=|h|$ and, for $h\ne 0$, $s=\mathrm{sgn}h$. The source explicitly fixes $J>0$ but does not separately constrain $h$ in this question; keeping $H_f$ covers either field direction.

For classical vectors of length $S$, define ${\mathbf{q}}_j=(S_j^x,S_j^y)$. The Cauchy-Schwarz inequality gives ${\mathbf{q}}_j\cdot {\mathbf{q}}_{j+1}\le |{\mathbf{q}}_j||{\mathbf{q}}_{j+1}|\le S^2$, and $h{S}_j^x\le H_{f}S$. Each term is therefore bounded below, soUniform alignment along the field saturates every bound:Equality requires all planar vectors to be parallel with length $S$, and the nonzero field then fixes their direction. At $h=0$, every uniformly oriented vector in the $xy$ plane is a classical ground state:A nonuniform configuration cannot improve these simultaneously saturated bond bounds, including on the periodic ring.

A phase transition is a qualitative change of thermodynamic phase as control parameters such as temperature, pressure or chemical potential are varied. A genuine equilibrium singularity is a nonanalyticity of the appropriate thermodynamic potential in the thermodynamic limit; finite systems generally show rounded crossovers instead.

An order parameter is a macroscopic quantity that distinguishes the phases. It often vanishes in a symmetric disordered phase and becomes nonzero after spontaneous symmetry breaking, although density-driven transitions need not have that simple symmetry interpretation. It can be a scalar, vector or tensor. At a first-order phase transition an appropriate first derivative of the free energy is discontinuous, and the order parameter commonly jumps; at a continuous phase transition the order parameter can approach zero continuously while a susceptibility or a higher derivative becomes singular.

Three examples are:

These examples also illustrate why an order parameter must reflect the physical distinction between phases rather than merely be any quantity that changes with temperature.

For a linear map $T:V\to W$, its image of a linear map and kernel of a linear map areTheir dimensions define the rank of a linear map and nullity:The image of a linear map measures which output vectors can occur, while the kernel of a linear map measures which input vectors are sent to zero.

Take the linear mapIts kernel of a linear map consists of all $(x,0)$, and its image of a linear map is the same line:Their set intersection is nonzero and their sum is only one-dimensional. In particular, $(0,1)$ cannot be written as a sum of one vector from each of these vector subspaces. The nonzero set intersection also prevents an internal direct sum of vector spaces. ThusThe matrix is a nonzero nilpotent matrix; its failure of diagonalizability explains why the preceding result does not apply.

Here $v_3=2v_1-v_2$, so the image of a linear map is the linear span of $v_1,v_2$. These two vectors are linearly independent: if $a{v}_1+b{v}_2=0$, the $e_1$ and $e_2$ coordinates give $a+b=0$ and $a-b=0$, hence $a=b=0$. Consequently the rank of a linear map is two. By the rank-nullity theorem, the nullity is $3-2=1$:More explicitly, the relation $-2v_1+v_2+v_3=0$ gives the kernel of a linear map

The printed word “eigenvalues” must read “eigenvectors”: eigenvalues are real numbers, whereas a basis of ${\mathbb{R}}^n$ consists of vectors. The intended assertion isThis wording error is present in the original scan.^[ref]

If $X^{-1}AX=D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ for an invertible matrix $X$, write $x_i$ for its columns. The matrix identity $AX=XD$ says $A{x}_i={\lambda }_i{x}_i$. The columns of an invertible matrix are a basis, hence they give the required basis of eigenvectors.

Conversely, put a basis of real eigenvectors $x_1,\dots ,x_n$ into the columns of $X$, with corresponding eigenvalues ${\lambda }_i$. The basis property makes $X$ an invertible matrix. The equations $A{x}_i={\lambda }_i{x}_i$ give $AX=XD$, so $X^{-1}AX=D$. This is precisely diagonalizability over the real numbers.

If $x\in \mathrm{Null}(A)$, the $i$th entry of $Ax=0$ says $r_i\cdot x=0$. Any $y\in \mathrm{Row}(A)$ is a linear combination $y={\sum }_i{\alpha }_i{r}_i$, so bilinearity of the dot product givesHenceThis proves the orthogonality for every pair of vectors in the two vector subspaces. In fact the null space equals the orthogonal complement of the row space, since being orthogonal to every row is exactly the system $Ax=0$.

The reduced row echelon form givesChoose the free variables $x_3=s,x_4=t$. ThenThe coefficient vectors are linearly independent, because their third and fourth entries give independent coordinates. ConsequentlySubstitution in the original matrix verifies $Ax=0$ for both basis vectors. The rank-nullity theorem supplies the independent dimension check $2+2=4$.

With column vectors, entry $P_{ij}$ is the fraction moving from region $j$ to region $i$. The first column is therefore $(1/2,1/4,1/4{)}^{\mathsf{T}}$, the second is $(1/2,1/2,0{)}^{\mathsf{T}}$, and the third is $(1/2,0,1/2{)}^{\mathsf{T}}$. HenceEach column sums to one, as required for this convention for a stochastic matrix. Multiplying by $P$ conserves the total resources.

For a square matrix $M$, an eigenvalue–eigenvector pair is a scalar $\lambda$ and a nonzero vector $v$ satisfyingThus the linear map represented by $M$ scales this particular direction by $\lambda$. The nonzero condition excludes the zero vector, which would otherwise satisfy the displayed equation for every scalar.

The other eigenvalues are $1/2$ and $0$. Their magnitudes are strictly less than one. All three eigenvalues are distinct, so their eigenvectors form a basis and the matrix is diagonalizable. Every component outside the stationary direction therefore decays under repeated application of $P$.

For an explicit justification, chooseThen $Pw=w/2$, $Pu=0$, andThe zero-eigenvalue component disappears after one step. For every $n\ge 1$,Thus the shares converge to $v$ and the resource vector converges to $10v$, since the conserved total is ten. For any other initial resource vector of total $T$, the same eigenvector argument gives the limit $Tv$.

For a real $m\times n$ matrix $A$, the row space is the linear span of its rows in ${\mathbb{R}}^n$, and the column space is the linear span of its columns in ${\mathbb{R}}^m$. Its null space consists of the input vectors sent to zero:The rank of a matrix isThe equality of row and column dimensions is the row-rank/column-rank theorem. Each displayed space is a linear subspace of the indicated ambient vector space.

Write the columns of $A$ as $a_1,\dots ,a_n$ and the coordinates of the vector $x$ as $x_1,\dots ,x_n$. By the definition of matrix multiplication,Indeed, its $i$th coordinate is ${\sum }_{j=1}^n{A}_{ij}{x}_j$, exactly the $i$th coordinate of this linear combination. Since the column space contains every linear combination of the columns, the claim holds for every vector $x$.

The right-hand vector is in the column space, becauseConsequently the linear system is consistent. A concrete solution isAs a compatibility check, the original rows satisfy $r_1+11r_2-5r_3=0$, and the right-hand entries satisfy the same relation $1+11\cdot 4-5\cdot 9=0$. There are infinitely many solutions, obtained by adding any vector in the null space to the displayed solution.

The reduced row echelon form yieldsTake the free variables in a linear system to be $s=x_2$, $t=x_4$, $u=x_5$. ThenThusEach listed vector lies in the null space, and every solution has the displayed form. Their coordinates in positions 2, 4 and 5 prove linear independence. The dimension also follows from the rank-nullity theorem: $5-2=3$.

Solve two triangular linear systems, rather than forming an inverse matrix:For this LU decomposition, forward substitution gives $y_1=b_1$, $y_2=b_2+2y_1$ and $y_3=b_3-3y_1+y_2$. Then back substitution givesAll diagonal entries used as divisors are nonzero. The same LU decomposition can be reused for many right-hand vectors, with only these two triangular solves repeated.

For a square matrix $M$, an eigenvalue–eigenvector pair consists of a scalar $\lambda$ and a nonzero vector $v$ satisfyingThe linear map represented by $M$ therefore scales $v$ by $\lambda$. The exclusion of the zero vector matters: otherwise every scalar would satisfy the equation trivially.

The reduced row echelon form givesTaking the free variables as $x_2=s$ and $x_4=5t$, the null space isThe two displayed vectors are linearly independent, as their second and fourth coordinates show, and each satisfies $Ax=0$. ThusThis also agrees with the rank-nullity theorem: $4-2=2$.

Source gap / TODO: the original printed page 4, confirmed visually, omits the unknown vector. It literally equates a $3\times 3$ matrix with a $3\times 1$ vector, so it does not define a well-formed system of linear equations. The following answer is conditional on the usual intended interpretation $A\mathbf{x}=(1,k^2,k{)}^T$; the missing factor is not present in the source.

Under that interpretation, the rows satisfy $r_1-2r_2+r_3=0$. Applying this linear relation to the right-hand side gives the necessary consistency conditionIt is sufficient too: the upper-left $2\times 2$ minor has determinant $4-9=-5\ne 0$, so the first two rows are linearly independent. The third row is $2r_2-r_1$, and imposes precisely the condition already found. ThereforeFor completeness, setting $x_3=t$ givesand direct substitution satisfies all three linear equations exactly for those two values of $k$. An unconditional answer to the malformed printed equality awaits a source correction.