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A pivot column of a matrix is a column whose position contains a leading nonzero matrix entry in its row echelon form. The corresponding columns of the original matrix form a basis of its column space. The column positions survive row operations, although the column vectors themselves generally change.^[ref]

Differentiate the planar rotation matrix at angle zero:The even and odd powers in the matrix exponential therefore giveThis exact formula gives the whole one-parameter subgroup. Taking $X=J$ means unit angular velocity: at $t=\pi /2$, $\exp (tJ)=J$ sends $(1,0{)}^T$ to $(0,1{)}^T$, a counterclockwise quarter-turn. At $t=\pi$, it equals $-I$; at $t=2\pi$, it returns to $I$. Choosing $X=2J$ reaches the quarter-turn at $t=\pi /4$. Changing the generator changes the speed, while preserving the circular path.

For a moving vector $v(t)=R_{t}v(0)$, the ordinary differential equation $v^{\prime }(t)=Jv(t)$ says its velocity is perpendicular to its position. The beautiful compression is that one constant matrix specifies every angle of this endless family of rotations.

A plane is a two-dimensional affine subspace: a point $\mathbf{a}$ together with every linear combination of two linearly independent direction vectors $\mathbf{u},\mathbf{v}$,In three-dimensional Euclidean space, a Cartesian equation of a plane fixes one linear equation $\mathbf{n}\cdot \mathbf{x}=d$, with a nonzero normal vector $\mathbf{n}$. The two free directions are orthogonal to $\mathbf{n}$. A plane extends indefinitely; a rectangular patch in a drawing marks only part of it, and does not define an edge of the plane.

For example, the plane $2x-y+2z=3$ contains $\mathbf{a}=(1,1,1)$. The vectors $\mathbf{u}=(1,2,0)$ and $\mathbf{v}=(0,2,1)$ are linearly independent and satisfy $(2,-1,2)\cdot \mathbf{u}=(2,-1,2)\cdot \mathbf{v}=0$. Hence its points are exactly $(1+s,1+2s+2t,1+t)$: substitution gives $2x-y+2z=3$ for every real $s,t$, and every solution has $s=x-1$, $t=z-1$.

The description separates position from direction. A translation of the plane changes $\mathbf{a}$ and possibly $d$, while retaining its parallel directions. A plane through the origin is a linear subspace; the displayed example is not, since the zero vector fails its linear equation. This distinction matters when applying orthogonal projection: subtract an anchor point before projecting onto the direction linear subspace.

Historically, Euclid's Euclid's Elements by Euclid, Book I Definition 7, describes a plane surface through the straight lines lying on it. David E. Joyce's accompanying guide notes that this older use can include a bounded square or circular figure. That usage differs from the whole unbounded affine subspace used here.^[ref]

Viewed perpendicular to the magnetic field, the normal Zeeman effect has a $\pi$ component with linear polarization parallel to the magnetic field and two $\sigma$ components polarized perpendicular to it. Along the magnetic field, $\pi$ vanishes and the $\sigma$ pair has opposite circular polarizations. Pieter Zeeman described these geometry-dependent patterns in his lecture.^[ref]

The geometry can be derived from electric dipole radiation. Let the observation unit vector be $\mathbf{n}=(\sin \theta ,0,\cos \theta )$, with $\theta$ measured from the magnetic field. Radiation retains only the transverse part of the oscillating electric dipole moment $\mathbf{d}$:For $\pi$, take $\mathbf{d}\parallel \hat{\mathbf{z}}$; its angular light intensity factor is ${\sin }^2\theta$, so it vanishes on axis. For ${\sigma }^{\pm }$, use normalized complex dipoles $(\hat{\mathbf{x}}\pm i\hat{\mathbf{y}})/\sqrt{2}$. Their angular factors are $(1+{\cos }^2\theta )/2$. Projection gives circles on axis, lines at $\theta =\pi /2$, and generally elliptical polarization between. These factors assume equal dipole magnitudes; actual line light intensities also depend on initial quantum state populations, transition strengths and collection geometry. No absolute handedness name is needed: it depends on the viewing and phase conventions.

A transverse polarizer can separate $\pi$ from $\sigma$ even before the spectral line components are fully resolved. Along the magnetic field, a quarter-wave plate followed by a polarizer converts the two opposite circular polarizations into distinguishable linear settings. This explains why polarization analysis carried information beyond a small increase in line width.

Polarization of waves describes the direction and time-dependent geometry of a transverse wave's vector during its oscillation at a fixed position. In three spatial dimensions, a transverse wave can have two independent directions of oscillation, both perpendicular to its direction of propagation. For polarization of an electromagnetic wave, the relevant vector is the transverse electric field; for a mechanical transverse wave, it can be the transverse displacement. A purely longitudinal wave has no choice between these transverse directions, and the orientation of a static electric field is not polarization of waves.

Choose transverse axes $x,y$ in a Cartesian coordinate system. A monochromatic wave with a fixed phase difference has the real electric fieldHere $A_x,A_y$ are wave amplitudes, $\omega$ is the angular frequency, and $\delta$ is the phase difference of the $y$ oscillation relative to the $x$ oscillation. The vector's tip traces its path during the wave period $T=2\pi /\omega$. Both plotted coordinates are real electric field components; neither axis represents the imaginary part of a complex amplitude.

For nonzero wave amplitudes, $\delta =0$ or $\pi$ modulo $2\pi$ gives linear polarization: the vector moves back and forth along a fixed line. Equal nonzero wave amplitudes and $\delta =\pm \pi /2$ modulo $2\pi$ give circular polarization: the tip traces a circle of constant radius. Unequal nonzero wave amplitudes with $\delta =\pm \pi /2$ give a noncircular ellipse whose axes coincide with $x,y$. More generally, nonzero wave amplitudes and $\sin \delta \ne 0$ give elliptical polarization, with circular polarization as a special case. If exactly one wave amplitude vanishes, the nonzero wave has linear polarization for every $\delta$; if both vanish, there is no nonzero wave to assign a polarization of waves. ^[ref].

For example, take $\omega =2\pi {\mathrm{s}}^{-1}$, so the wave period is $T=1\mathrm{s}$. The choices $(A_x,A_y,\delta )=(2,1,0)$, $(1,1,\pi /2)$ and $(2,1,\pi /2)$, with electric field wave amplitudes in $V/m$, produce respectively the line segment $E_y=E_x/2$, the circle $E_x^2+E_y^2=1$ and the ellipse $E_x^2/4+E_y^2=1$, when coordinates are measured in $V/m$. At $t=T/4$, their tips are respectively $(0,0)$, $(0,-1)$ and $(0,-1)V/m$. The last two share a point without having the same path.

Resolving the electric field into two transverse components makes polarization of waves useful: the same frequency and wave amplitudes can describe different paths when their phase difference changes. An ideal linear polarizer selects the component along its pass axis, letting its orientation probe the direction of linear polarization through Malus's law. A fixed pair of wave amplitudes and a fixed phase difference describe one stable path; unpolarized light requires a statistical description rather than one fixed ellipse.

Consider a parent atomic nucleus at rest. Write its signed nuclear polarization along a fixed axis $\hat{\mathbf{z}}$ as $P=⟨J_z⟩/(I\hslash )$, where $I\hslash$ is its maximal spin component. Let $\theta$ be the angle of the emitted electron momentum to that axis and let $\beta =v/c$. Use $\mathcal{P}$ for the parity operator, so it is not confused with $P$.

For an axially symmetric leading-dipole model, retain a constant and the term linear in the polarization vector $\mathbf{P}=P\hat{\mathbf{z}}$. Rotational symmetry makes the directional term proportional to $\mathbf{P}\cdot \hat{\mathbf{p}}$. Define its dimensionless coefficient as $A\beta$. At fixed electron energy the relative intensity and normalized probability density function per solid angle are thereforeThe factor $\beta$ is the conventional separation of the speed dependence; rotational symmetry alone does not determine that dependence or $A$. The coefficient $A$ contains the decay dynamics. This leading model omits higher angular terms and detailed recoil and radiative effects. A partially oriented higher-spin ensemble can also have alignment that is not specified by its vector nuclear polarization alone.

The unit-integral condition follows without any model of a detector:Thus $f$ has unit angular average, while the probability density function in the polar angle alone contains the Jacobian determinant factor:In particular, $f$ itself is neither a normalized probability density function in $\theta$ nor a detector count. Positivity on $-1\le u\le 1$ gives $|a|\le 1$.

Under full spatial reflection, $\mathbf{p}\mapsto -\mathbf{p}$ but $\mathbf{J}\mapsto \mathbf{J}$. The latter follows already for orbital angular momentum from $\mathbf{r}\times \mathbf{p}\mapsto (-\mathbf{r})\times (-\mathbf{p})=\mathbf{r}\times \mathbf{p}$; spin has the same axial vector transformation. Hence $\mathbf{P}\cdot \hat{\mathbf{p}}$ is a pseudoscalar, and $\cos \theta \mapsto -\cos \theta$. If parity conservation held for this model, equal probabilities for the inverted process would force $a=0$. Conversely, $a=0$ in an unpolarized sample proves nothing about $A$.

Integrating over the two hemispheres gives a useful original check:Choose $A=-0.8$, $P=0.5$, and $\beta =0.75$, so $a=-0.3$. Then the forward and backward hemisphere probabilities are $0.425$ and $0.575$, and their counting asymmetry is $-0.15$. These are chosen teaching values, not measured values from the Wu experiment. At a single energy, the angular shape determines the product $AP\beta$; it cannot separately determine $A$ without a polarization calibration.

A polar vector transforms as ${\mathbf{v}}^{\prime }=Q\mathbf{v}$ under an orthogonal matrix $Q$, including an orientation-reversing transformation. Displacement, velocity and force are polar vectors. An axial vector instead transforms as ${\mathbf{a}}^{\prime }=(\det Q)Q\mathbf{a}$.

A position vector is the vector from a chosen origin to a point. Its time derivative is the velocity; differences of position vectors give displacement.

A real symmetric bilinear form is positive-definite when $B(v,v)>0$ for every nonzero vector. It defines an inner product and the associated norm $\sqrt{B(v,v)}$. Symmetry is part of this definition: positive diagonal values by themselves do not imply symmetry.^[ref]

The identity matrix $A=\mathrm{diag}(1,1)$ gives $B(v,w)=v_1{w}_1+v_2{w}_2$ and $B(v,v)=v_1^2+v_2^2$. For $v=(3,4)$ this is $25$, giving norm $5$. A weighted example, $A=\mathrm{diag}(4,1)$, gives norm $\sqrt{4x^2+y^2}$: the first coordinate costs twice as much in length. Such inner products let geometry assign different weights to different directions.

Every positive-definite bilinear form is nondegenerate: a vector in its radical of a bilinear form has $B(v,v)=0$, so positivity forces $v=0$. The converse fails, as $\mathrm{diag}(1,-1)$ shows. Even positivity of the associated quadratic form does not force a general bilinear form to be symmetric: $A=\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$ has $v^{T}Av=x^2+y^2$ but $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$.

The complex Killing form is not positive-definite on the entire complex Cartan subalgebra; its restriction to the Euclidean subspace of a Cartan subalgebra is a real positive-definite form.

A real symmetric matrix $A$ is positive definite whenfor every nonzero real vector $x$. It therefore defines the inner product $⟨x,y{⟩}_A=x^{T}Ay$ and is invertible.

The strict positivity of a quadratic form on every nonzero vector. A real symmetric matrix has positive definiteness exactly when its associated quadratic form does, making it a positive-definite matrix. A Hermitian form is positive definite when its value on $(v,v)$ is real and strictly positive for every nonzero vector.

Think of a positive-definite quadratic form as an energy cost for a displacement: every direction costs something, and zero cost identifies the origin. For example, the real symmetric matrixis a positive-definite matrix because the two squares can vanish together only at $(0,0)$. The mixed term couples the coordinates, but cannot overwhelm their positive contributions: $q(1,1)=8$ and $q(1,-1)=4$. The graph is a bowl, and its positive level sets are ellipses with axes along the eigenvectors, rather than along the original coordinates.

The real and complex settings require different symmetry assumptions. For a real symmetric matrix use $q(x)=x^{T}Ax$ with $x\in {\mathbb{R}}^n$; for a complex Hermitian matrix use $q(z)=z^{\ast }Az$ with $z\in {\mathbb{C}}^n$, where $z^{\ast }$ is the conjugate transpose. The latter expression is real, so strict positivity is meaningful. A complex quadratic form such as $z_1^2+z_2^2$ has no such positivity: it vanishes at $(1,i)$ and takes negative real values at $(i,0)$. A Hermitian form uses complex conjugation, which changes this example to $|z_1{|}^2+|z_2{|}^2$.

For a real matrix $M$ without a symmetry hypothesis, its quadratic form depends only on its symmetric part of a matrix:Positive eigenvalues of $M$ alone do not establish positive definiteness of this quadratic form. For example, $M=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has both eigenvalues equal to $1$, yet $q(1,-1)=1-4+1=-2$. The relevant test is positivity of the symmetric part of a matrix, not the spectrum of an arbitrary matrix.^[ref]

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 $c^{\ast }Gc=\Vert {\sum }_i{c}_i{x}_i{\Vert }^2$.

Vary the coupling in the opening example. The eigenvectors $e_{+}=(1,1{)}^T/\sqrt{2}$ and $e_{-}=(1,-1{)}^T/\sqrt{2}$ form a fixed orthonormal eigenbasis, with eigenvalues $3+t$ and $3-t$. Writing a vector as $u{e}_{+}+v{e}_{-}$ givesFor $|t|<3$, both directions cost energy and $A_t$ is a positive-definite matrix. At $t=3$, the quadratic form is $6u^2$: the entire line spanned by $e_{-}$ has zero energy, and $A_t$ is a singular positive semidefinite matrix. At $t=-3$, the flat line is instead spanned by $e_{+}$. For $|t|>3$, the quadratic form is an indefinite quadratic form; one direction lowers the energy without bound as its displacement grows. At $t=4$, for instance, $q_t(e_{+})=7$ while $q_t(e_{-})=-1$.

The geometric transition is precise. For $|t|<3$, $q_t=1$ is an ellipse. At $t=3$, it consists of the two parallel lines $u=\pm 1/\sqrt{6}$, extending freely in the zero-energy direction. For $t>3$, it is a hyperbola. The boundary case retains nonnegative energy but loses both the bounded positive level sets and the unique zero-energy vector.

If a real matrix $B$ has a positive diagonal matrix $W$ such that $WB$ is a symmetric matrix, then $H=W^{1/2}B{W}^{-1/2}$ is a symmetric matrix related to $B$ by a similarity transformation: $B^{T}W=WB$ gives $H^T=H$. Thus $B$ is a diagonalizable matrix with real eigenvalues. If ${\mathbf{z}}^{T}WB\mathbf{z}\le 0$ for all real vectors $\mathbf{z}$, its eigenvalues are nonpositive. This converts symmetry in a weighted inner product into Euclidean symmetry.

A matrix exponential defines a positive matrix semigroup if $e^{tA}$ maps nonnegative vectors to nonnegative vectors for all $t\ge 0$. Here the inequality means entrywise nonnegativity, with zero entries allowed. In finite dimensions this holds exactly when the real matrix $A$ has nonnegative off-diagonal entries: if $A+\gamma I\ge 0$, expand $e^{tA}=e^{-\gamma t}{\sum }_{k\ge 0}{t}^k(A+\gamma I{)}^k/k!$; conversely, the off-diagonal entries of $e^{tA}=I+tA+O(t^2)$ force $a_{ij}\ge 0$ for $i\ne j$. Zero column sums additionally preserve $1^{T}c$ under $c^{\prime }=Ac$.

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

A Hermitian form is positive semidefinite when $h(v,v)\ge 0$ for every vector. It need not define an inner product, because a nonzero vector may have zero squared norm. The Cauchy-Schwarz inequality still holds: applying nonnegativity to $v+zw$ and minimizing the quadratic expression in $z$ gives $|h(v,w){|}^2\le h(v,v)h(w,w)$ when $h(w,w)>0$; when $h(w,w)=0$, varying $z$ forces $h(v,w)=0$. Thus its zero-norm vectors are exactly the radical of a Hermitian form. Quotienting this radical of a Hermitian form gives a positive inner product; taking its Hilbert space completion then gives a Hilbert space. This is the final positivity step in the Gupta-Bleuler null-state quotient.

A real symmetric or complex Hermitian matrix $A$ is positive semidefinite when $x^{†}Ax\ge 0$ for every vector $x$, equivalently when all its eigenvalues are nonnegative.

Power iteration finds a dominant eigenvalue and its eigenvector by repeatedly applying a matrix and rescaling the resulting vector. The essential idea is selective amplification: components along different eigenvectors grow at different rates, so one direction can eventually dominate. It is the computational form of the power method and is useful when applying a large sparse matrix to a vector is cheap but computing its whole spectral decomposition is expensive.

For a square matrix over the real numbers or complex numbers $A$, choose a unit vector $x_0$ and repeatHere $\Vert \cdot {\Vert }_2$ is the Euclidean norm. If $y_{k+1}=0$, stop before division: the current vector is an eigenvector for the zero eigenvalue, and this iteration cannot continue. Otherwise, in exact arithmetic,Repeated normalization keeps the stored vectors at manageable size without changing their directions; it does not guarantee that an individual multiplication cannot exceed the range of floating-point arithmetic. The dominant eigenvalue means largest modulus, which need not mean largest signed real value.^[ref]

Suppose $A\in {\mathbb{C}}^{n\times n}$ is a diagonalizable matrix, $n\ge 2$, with an eigenbasis $v_1,\dots ,v_n$ and spectral decompositionAssume the dominant eigenvalue ${\lambda }_1$ is a simple eigenvalue andThen the normalized power iteration approaches the one-dimensional eigenspace of ${\lambda }_1$. Its asymptotic direction-error bound has factor $\gamma =|{\lambda }_2/{\lambda }_1|<1$. This spectral gap is a gap in modulus: if $g=|{\lambda }_1|-|{\lambda }_2|$, then $\gamma =1-g/|{\lambda }_1|$. A small relative gap gives slow convergence.^[ref]

The short proof is the exact eigenbasis identityThe remaining sum has norm bounded by a constant times ${\gamma }^k$. This proves convergence of direction, allowing a changing sign or complex factor. A tiny $c_1$ or a poorly conditioned, nonorthogonal eigenbasis can delay visible convergence: the condition number of $V$ affects constants, while $\gamma$ controls the eventual rate. The bound need not be an equality; if the initial coefficient of the second-largest eigenvalue vanishes, the active spectrum of the power method gives a faster rate. When $\gamma =0$, all unwanted components disappear after one multiplication.^[ref]

The hypothesis is $c_1=(V^{-1}{x}_0{)}_1\ne 0$, not an ordinary nonzero inner product with the right eigenvector. Let $w_1^{\ast }$ be the first row of $V^{-1}$, a left eigenvector normalized so $w_1^{\ast }{v}_j$ is $1$ for $j=1$ and $0$ otherwise; $\ast$ denotes conjugate transpose. Then $c_1=w_1^{\ast }{x}_0$. For a Hermitian matrix, choose an orthonormal eigenbasis and the criterion becomes $v_1^{\ast }{x}_0\ne 0$: the starting vector must be nonorthogonal to the dominant eigenvector. For a general diagonalizable matrix, the nonorthogonal eigenbasis makes these two tests different, as illustrated under power iteration failure cases.