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$[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$, $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, $[L_i,S_j]=0$, and $J=L+S$ obeys the same angular-momentum algebra. The commuting set $H_0,L^2,S^2,L_z,S_z$ therefore has the stated simultaneous basis.

$L\cdot S=\frac{1}{2}(J^2-L^2-S^2)$. Hence the coupled $|n\ell sj{j}_z⟩$ basis diagonalizes the spin-orbit term.

A piecewise-linear hat function is a continuous piecewise linear function that selects one node of a finite element mesh: it is one there and zero at every other node. For an interior node $x_i$, its exact formula on a possibly nonuniform mesh isIts support of a function is $[x_{i-1},x_{i+1}]$. It rises on the left interval and falls on the right, producing its hat-shaped graph. The interior hat functions form a basis of the continuous $P_1$ conforming finite element space with homogeneous Dirichlet boundary conditions. Chen's author notes define this nodal basis and illustrate it in one and two dimensions.^[ref] The nodal property ${\varphi }_i(x_j)={\delta }_{ij}$ makes the coefficients directly readable: if $v_h={\sum }_i{c}_i{\varphi }_i$, then $v_h(x_i)=c_i$. Unlike the global Lagrange cardinal polynomials, each of these functions is confined to two neighboring elements.

For example, with $x_j=j/4$, the middle hat function is ${\varphi }_2(x)=4x-1$ on $[1/4,1/2]$, $3-4x$ on $[1/2,3/4]$, and zero elsewhere. Hence ${\varphi }_2(3/8)=1/2$, while ${\varphi }_1(3/8)=1/2$ and ${\varphi }_3(3/8)=0$. The finite element solution of the one-dimensional Poisson equation therefore has $u_h(3/8)=(3/32)/2+(1/8)/2=7/64$.

Local assembly makes this locality computational. On an element $[a,b]$ of length $d=b-a$, the two local basis functions are $N_L=(b-x)/d$ and $N_R=(x-a)/d$. For the Poisson equation with unit forcing, their local stiffness matrix and load vector areAdd these four matrix entries and two loads to the rows and columns belonging to the element's endpoints, then impose the prescribed boundary values. An interior node receives contributions from its two neighboring elements. Writing $d_L=x_i-x_{i-1}$ and $d_R=x_{i+1}-x_i$ giveswith off-diagonal entries included only for interior neighbors. On a uniform mesh these reduce to $2/h$, $-1/h$ and $h$. Nonadjacent hat functions have no derivative overlap of positive length, so all distant entries vanish. This produces a sparse matrix with only three diagonals: the global coupling is built entirely from local two-node interactions.

For arbitrary positive integers $e,f$, the displayed extension has degree $ef$, ramification index $e$, and residue degree $f$. If $|t|=p^{-1}$, its unique extended absolute value on a field is $|\sum a_n{u}^n|=p^{-\min \{n:a_n\ne 0\}/e}$. A basis consists of ${\gamma }_i{u}^j$, where the ${\gamma }_i$ form an basis of ${\mathbb{F}}_{p^f}$ and $0\le j<e$. Regrouping a Laurent series by its exponent modulo $e$ proves the basis assertion. When $p$ divides $e$ the extension can be inseparable; the valuation calculation still applies.

The prime field of a field $K$ is its smallest subfield: the arithmetic forced upon us by the identity element $1_K$. Every subfield must contain $1_K$, its repeated sums, their additive inverses, and the multiplicative inverses of nonzero results. In characteristic zero these operations produce a copy of the rational numbers $\mathbb{Q}$. In positive characteristic of a field $p$, they produce a copy of ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$, with addition and multiplication given by modular arithmetic.

For example, inside the field with four elements, repeated addition of $1$ gives $0,1,0,1,\dots$. Thus its prime field is $\{0,1\}$, and the other two elements cannot be reached by repeatedly adding $1$. This is why a finite field need not consist entirely of ordinary integer residues: a larger finite field extends the arithmetic of its prime field with additional elements.

The positive characteristic of a field must be a prime number. If its least vanishing multiple of $1_K$ were $rs$ with $1<r,s<rs$, then $(r{1}_K)(s{1}_K)=0$ would exhibit zero divisors. The absence of zero divisors rules this out. The prime field supplies the scalars when viewing $K$ as a vector space; its basis then explains the possible sizes of a finite field.

A reduced QR decomposition of an $m\times k$ matrix over the real numbers with full column rank, $m\ge k$, writes $A=QR$, where $Q$ is $m\times k$ with orthonormal columns and $R$ is a $k\times k$ upper triangular matrix with positive diagonal. Applying the Gram-Schmidt process to the columns $v_j$ of $A$ constructs $Q=(q_1\cdots q_k)$ and records the coefficientsThis last column identity is exactly the matrix multiplication $A=QR$. The upper triangular matrix records that the $j$th input uses only the first $j$ new directions, while its positive diagonal records the norms of the nonzero orthogonal projection residuals. Since $Q^{T}Q=I$, one also has $R=Q^{T}A$; $Q^T$ denotes the transpose and $I$ the identity matrix. Over the complex numbers, use the conjugate transpose $Q^{\ast }$ in these identities.

For the three-dimensional Gram-Schmidt process example, the exact matrices areFor example, the last column reconstructs asThe diagonal entries are approximately $1.414214,1.224745,1.154701$. Each is a norm of the new direction before it was scaled to length one.

For a linear least-squares problem, the orthogonal projection of $b$ onto the column space of $A$ is $Q{Q}^{T}b$. Hence the minimizing coefficients solve the linear system $Rx=Q^{T}b$ by back substitution. Let $A_2=(v_1{v}_2)$ consist of the first two columns of the example; its reduced factors are $Q_2=(q_1{q}_2)$ and the leading $2\times 2$ block $R_2$ of $R$. For $b=(0,0,1{)}^T$, the linear system $R_{2}x=Q_2^{T}b$ isThus $A_{2}x=(1/3,-1/3,2/3{)}^T$, agreeing with the orthogonal projection computed above, and the minimum Euclidean norm of $A_{2}x-b$ is $1/\sqrt{3}$. The QR decomposition separates building an orthonormal basis from recovering coefficients in the original basis.^[ref]

For a field $k$ and $q\in k^{\times }$, the quantum plane is $k_q[X,Y]=k⟨X,Y⟩/(YX-qXY)$. Its ordered monomials $X^i{Y}^j$, $i,j\ge 0$, are a basis, and their multiplication is $X^i{Y}^j{X}^u{Y}^v=q^{ju}{X}^{i+u}{Y}^{j+v}$. Leading monomials show it is a noncommutative domain. Some sources interchange $q$ with $q^{-1}$ in the defining relation.

A quantum superposition is a linear combination of quantum state vectors in a complex Hilbert space. In a specified orthonormal basis, a normalized pure state has the displayed expansion, where $a_j$ are probability amplitudes. A nonzero linear combination can always be normalized; an arbitrary sum of normalized vectors need not already have norm one. Having more than one nonzero component depends on the chosen basis: the Hadamard basis vector $|+⟩$ is a quantum superposition of the two computational-basis states, but is a single vector of the Hadamard basis. ^[ref].

The distinction from classical ignorance is quantitative. Consider the normalized qubit quantum stateIts density operator, written in the computational basis, isPreparing $|0⟩$ with probability $3/4$ and $|1⟩$ with probability $1/4$, while discarding the preparation label, instead gives the mixed stateThe diagonal entries agree, so both preparations give the same computational-basis measurement probabilities. The off-diagonal entries of ${\rho }_{\varphi }$ retain quantum coherence and the relative quantum phase. In the Hadamard basis, the Born rule givesThe density matrix representation retains this distinction between a pure state and a mixed state. ^[ref].

For example, at $\varphi =0$, the quantum superposition yields the plus outcome with probability about $0.9330$, whereas the mixed state yields it with probability $0.5$. At $\varphi =\pi /2$ these particular probabilities agree, although the density matrices still differ: equality for one measurement does not establish equality of quantum states. The probability amplitude figure shows the comparison throughout a full cycle of relative quantum phase.

This is why a quantum superposition is useful: quantum gates act linearly on its probability amplitudes, and unitary operators preserve the full pure state information, including relative quantum phases. A Hadamard gate can turn otherwise hidden quantum coherence into different outcome probabilities. The beautiful feature is that the same basis weights can support different predictions when amplitudes recombine. The expansion specifies a preparation and its measurement predictions; it does not assign every unmeasured outcome a simultaneous classical value. See projective measurement for the separate question of the conditional state after a measurement.

The two-dimensional quantum torus is the algebra with invertible generators $X,Y$ and relation $YX=qXY$, for $q\in k^{\times }$. Its basis consists of $X^i{Y}^j$ with $i,j\in \mathbb{Z}$, with the same multiplication rule as the quantum plane. It is a right Noetherian ring: successive adjunction of $X^{-1}$, $Y$, and $Y^{-1}$ satisfies the hypotheses of the noncommutative Hilbert basis theorem.

Suppose an array $T_{ij}$ is specified in each right-handed orthonormal basis. If $w_i=T_{ij}{v}_j$ transforms as a vector for every vector $v$, then $T$ is a second-order Cartesian tensor under those basis changes. Indeed $v^{\prime }=Rv$, $w^{\prime }=Rw$ imply $T^{\prime }Rv=RTv$ for every $v$, hence $T^{\prime }=RT{R}^T$. The condition on every test vector is essential; one contraction is insufficient.

The rank-nullity theorem accounts for the input dimensions of a linear map: the number of independent invisible input directions plus the number of independent reachable output directions equals the input dimension. For $T:V\to W$ over a field, with finite-dimensional domain of a function $V$,Here $\dim \ker T$ is the nullity, and $\dim \mathrm{im}T$ is the rank of a linear map. The kernel lives in the input vector space, while the image of a linear map lives in the output vector space. For an $m$-by-$n$ matrix of matrix rank $r$, the formula is $n=(n-r)+r$: the number of columns counts input coordinates. ^[ref]

For a computed example over the real numbers, takeThe kernel is $\mathrm{span}\{(-2,1)\}$ and the image of a linear map is $\mathrm{span}\{(1,0)\}$. Each has dimension one, so the two input dimensions split as $2=1+1$. The vector $(3,1)$ can be writtenThe second summand is undetectable: $T(3,1)=T(1,2)$. The picture uses the standard dot product to choose a visible direction perpendicular to the kernel; this is a convenient geometric choice, not an extra hypothesis of the rank-nullity theorem.

This explains a practical tradeoff in a linear system. If $T(x_0)=b$, every other input with output $b$ is $x_0+k$ for $k\in \ker T$. The nullity therefore counts independent freedom in an existing solution; the rank counts independent reachable output directions. In the example, $T(x,y)=(3,0)$ has all solutions $(x,y)=(3-2t,t)$, while $T(x,y)=(3,1)$ has none. The cokernel records this failure of reachability.

For an $m$-by-$n$ matrix, injectivity requires matrix rank $n$, and surjectivity requires matrix rank $m$. When $m\ne n$, these conditions differ. For example, $B:{\mathbb{R}}^3\to {\mathbb{R}}^2$ with $B(x,y,z)=(x+2y,0)$ has rank one and nullity two, because $\ker B=\mathrm{span}\{(-2,1,0),(0,0,1)\}$. The input accounting is $3=2+1$, although only one of two output directions is reached.

Proof outline: extend a basis $k_1,\dots ,k_q$ of the kernel to a basis $k_1,\dots ,k_q,v_1,\dots ,v_r$ of $V$. Every output is a linear combination of $T(v_1),\dots ,T(v_r)$ because the $k_i$ vanish under $T$. If a linear combination of the $T(v_j)$ is zero, the corresponding combination of the $v_j$ lies in the kernel; linear independence of the full basis forces all its coefficients to vanish. Thus the added basis vectors give a basis of the image of a linear map. Counting them proves the formula.

Consider the linear map $A:{\mathbb{R}}^3\to {\mathbb{R}}^2$ represented by the rectangular matrixTo compute its kernel, solve the homogeneous linear system $x+2y=0$, $y+z=0$. Taking $y=t$ gives $x=-2t$, $z=-t$, soThus $v$ represents a null direction of a linear map: decreasing the first input by two, increasing the second by one, and decreasing the third by one changes neither output. Solving the homogeneous linear system is the concrete computation behind a kernel; the column space instead decides which outputs are attainable.^[ref]

Now prescribe the nonzero output $b=(3,2{)}^T$. The linear system $Au=b$ becomes $x+2y=3$, $y+z=2$. Again take $y=t$:Direct substitution gives $A{u}_0=(3,2{)}^T$, and $A(u_0+tv)=A{u}_0+tAv=b$. Conversely, the two linear equations force exactly this form, so no solutions have been omitted. This fiber is an affine subspace of dimension one in the three-dimensional input vector space. It is parallel to the kernel, but does not pass through the origin. For instance, $u_0=(3,0,2{)}^T$ and $u_1=(1,1,1{)}^T$ both give $b$, while their difference $u_1-u_0=v$ gives zero. Adding $u_0$ and $u_1$ gives output $2b$, which is another way to see that this nonzero-output fiber is not a vector subspace.

The first and third columns of $A$ are $(1,0{)}^T$ and $(0,1{)}^T$, a basis of the output vector space. Hence $A$ has matrix rank two and reaches every output. The rank-nullity theorem predicts nullity $3-2=1$ before we solve any linear system. It provides a useful completeness check: one nonzero kernel vector already forms a basis here, so a second independent null direction of a linear map would signal a calculation error. In general, matrix rank $r$ for an $m\times n$ matrix leaves $n-r$ independent kernel directions, rather than $m-r$.^[ref]

A right inverse of a function $f:X\to Y$ is a function $g:Y\to X$ satisfying $f\circ g={\mathrm{id}}_Y$. It forces $f$ to be a surjective function. A surjective linear map onto a finite-dimensional vector space has a linear right inverse: lift a basis of the target and extend by linearity.

A rotation matrix is an orthogonal matrix with determinant $1$.

A rotation matrix turns vectors while preserving their shape relationships. For column coordinate vectors in an orthonormal basis, its defining conditions arewhere $R^T$ is the transpose and $I$ the identity matrix. In the usual Cartesian basis of the plane, an active counterclockwise planar rotation through an angle $\theta$ usesHere the vector moves and the basis stays fixed: $R_{\pi /2}(1,0{)}^T=(0,1{)}^T$. Positive angles turn from the positive horizontal axis towards the positive vertical axis. The two columns are precisely the rotated basis vectors; the sine and cosine entries encode their new directions. Turning the basis instead uses passive rotation of coordinates, which has the opposite sign.

This distinction makes rotation matrices useful for keeping a geometric action separate from its coordinate description. For a rigid body, a rotation matrix can encode the directions of an attached orthonormal basis or convert between that basis and a fixed orthonormal basis, with different meanings for its input and output.^[ref]

The orthogonal matrix condition preserves every dot product:Consequently Euclidean norms and angles between nonzero vectors are preserved, as is the Euclidean norm of the difference between any two endpoints. The additional determinant condition preserves orientation of a vector space and oriented volume: for any square matrix $V$ whose columns are the input vectors, $\det (RV)=\det V$. An orthogonal matrix of determinant $-1$, such as a reflection matrix, preserves dot products but reverses orientation of a vector space. The matrix inverse is especially simple: $R^{-1}=R^T$.^[ref]