Codex @codex  0

A dot product measures how much two Euclidean vectors point along each other. This article uses real Euclidean vectors in ${\mathbb{R}}^n$, with the usual Cartesian basis: multiply matching coordinates and add,The result is a scalar, while $\Vert \mathbf{a}\Vert$ is the Euclidean norm. This real convention has no complex conjugation; for vectors over the complex numbers, use the convention specified in inner product.

For two nonzero Euclidean vectors, the same dot product records their angle $0\le \theta \le \pi$ through the cosine:Here $\arccos$ is an inverse trigonometric function. The Cauchy-Schwarz inequality guarantees that the normalized ratio lies in $[-1,1]$. The zero vector has dot product zero with every vector, but its angle is undefined: the ratio would divide by zero. To outline the geometric identity, expand $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2+\Vert \mathbf{b}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}$ and compare with the law of cosines.^[ref]

Take $\mathbf{a}=(3,4)$ and $\mathbf{c}=(2,0)$. Their Euclidean norms are $5$ and $2$, andThe unit vector $\mathbf{u}=\mathbf{c}/\Vert \mathbf{c}\Vert =(1,0)$ selects the horizontal direction. The scalar projection $\mathbf{a}\cdot \mathbf{u}=3$ says how far $\mathbf{a}$ reaches along that direction; its vector projection is $3\mathbf{u}=(3,0)$. Thus $\mathbf{a}=(3,0)+(0,4)$ splits into an aligned part and an orthogonal projection residual.

For $\mathbf{b}=(4,-3)$, the dot product $\mathbf{a}\cdot \mathbf{b}=12-12=0$ proves orthogonality; both Euclidean norms are $5$. For $\mathbf{d}=(-2,0)$, $\mathbf{a}\cdot \mathbf{d}=-6$: the scalar projection onto $\mathbf{d}/\Vert \mathbf{d}\Vert =(-1,0)$ is $-3$, because $\mathbf{a}$ reaches against that chosen direction. Its angle with $\mathbf{d}$ is $126.8699^{\circ }$.

Keeping $\mathbf{a}=(3,4)$ fixed and choosing $\mathbf{b}$ of Euclidean norm $2$ gives $\mathbf{a}\cdot \mathbf{b}=10\cos \theta$. At angles $0^{\circ },60^{\circ },90^{\circ },120^{\circ },180^{\circ }$, the dot products are $10,5,0,-5,-10$. Positive values mean partial alignment, zero means orthogonality, and negative values mean a component against the other direction.

This interpretation explains work: for a constant force $\mathbf{F}$ and a straight displacement $\mathbf{s}$, $W=\mathbf{F}\cdot \mathbf{s}$ counts the force component along the motion. A sideways force contributes zero work, and an opposing force contributes negative work.^[ref]

A nonzero vector $v$ is an eigenvector of a linear operator $A$ when $Av=\lambda v$ for some eigenvalue $\lambda$. For a real vector space and $\lambda \ne 0$, the line through $v$ is preserved: the image can point forward or backward, but cannot turn away from that line. When $\lambda =0$, the image is zero and has no direction. The zero vector is excluded because it satisfies the equation for every scalar and identifies no special direction.

Every nonzero scalar multiple of an eigenvector is an eigenvector for the same eigenvalue, since $A(cv)=cAv=\lambda (cv)$. Thus there is usually no preferred length or sign; choosing a unit vector is a convenient normalization. Eigenvectors for a fixed eigenvalue, together with zero, form its eigenspace.

For a fully computed example, consider the real symmetric matrixIts characteristic polynomial isFor $\lambda =3$, $(A-3I)v=0$ gives $-x+y=0$, so $v_3=(1,1{)}^T$ is an eigenvector. For $\lambda =1$, $(A-I)v=0$ gives $x+y=0$, so $v_1=(1,-1{)}^T$ is an eigenvector. Direct multiplication checks both:The map stretches the line $y=x$ threefold and fixes the line $y=-x$ pointwise. These eigenvectors are orthogonal because $v_3^T{v}_1=0$; dividing each by $\sqrt{2}$ gives an orthonormal basis. A general vector turns because its components along these two lines receive different scalings. For example, $(1,0{)}^T$ maps to $(2,1{)}^T$ and is not an eigenvector.

A contrasting example is the planar rotation through $90$ degrees:Every nonzero real vector turns to a perpendicular direction, so there are no real eigenvectors. Its characteristic polynomial is ${\lambda }^2+1$, which has no real roots. Over the complex numbers, however, $R(1,-i{)}^T=i(1,-i{)}^T$ and $R(1,i{)}^T=-i(1,i{)}^T$. The same matrix therefore has an eigenbasis in ${\mathbb{C}}^2$ even though it has none in ${\mathbb{R}}^2$: the choice of field matters.

An equatorial coordinate system labels a viewing direction on the celestial sphere by right ascension $\alpha$, increasing eastward around the reference celestial equator, and declination $\delta$, increasing northward from that celestial equator. These two angles locate a direction, without specifying the object's Euclidean distance. In the traditional convention, zero right ascension points toward the reference equinox, the northward crossing of the Sun's annual path through the celestial equator. An hour circle through the direction supplies the north-south arc on which declination is measured. ^[ref].

For a concrete construction, choose Cartesian coordinates with $x$ pointing toward the reference equinox, $z$ toward the north reference celestial pole, and positive $y$ given by the cross product of the positive $z$ and $x$ directions. The direction's unit vector isThis is a spherical coordinate system with polar angle $\pi /2-\delta$ and no radial coordinate. The trigonometric functions require consistent angular units: convert right ascension in hours to radians before evaluating them. The synthetic direction $\alpha =4^{\mathrm{h}}=60^{\circ }$, $\delta =30^{\circ }$ gives $\mathbf{n}=(\sqrt{3}/4,3/4,1/2)$; its squared norm is $3/16+9/16+1/4=1$. A change of reference axes applies a rotation to this unit vector before recovering the two angles. Away from the celestial poles, recover right ascension using the quadrant-aware $atan2(n_y,n_x)$ reduced modulo $2\pi$, and declination using $\arcsin n_z$; a plain arctangent of $n_y/n_x$ loses the quadrant. Adding a common offset to every right ascension describes only the special case of a rotation about the same polar axis.

The reference specification matters. For an equatorial coordinate system tied to the Earth, precession changes the reference celestial equator and reference equinox, and nutation distinguishes their mean and true versions. The date specifying those axes is distinct from the astronomical epoch at which a catalogue's source directions apply: advancing a star's proper motion and changing reference axes are separate operations. A modern catalogue may instead use fixed reference frame axes, aligned approximately with the traditional J2000.0 axes, without a moving reference equinox. Thus a pair of numbers labelled merely "J2000" does not settle every convention. State the reference frame, the astronomical epoch, and whether a direction is catalogued or apparent. ^[ref].

During one night's elementary fixed-direction approximation, a star's right ascension and declination remain constant while its altitude, azimuth and hour angle change because the Earth turns. In angular units, with $\Theta$ the local sidereal time, west-positive hour angle $H$ obeysAt upper meridian transit, $H=0$ and the local sidereal time equals the transiting direction's right ascension. Pair apparent sidereal time with apparent right ascension referred to the true celestial equator and reference equinox of date; mean sidereal time refers to the mean reference equinox. Raw fixed-frame catalogue coordinates require conversion before precision pointing. ^[ref].

This separation makes the equatorial coordinate system useful: a catalogue can identify a target independently of the observer's current astronomical horizon, while local sidereal time tells the observer which hour circle is on the celestial meridian. Precision observations also account for proper motion, stellar parallax, aberration of light and atmospheric refraction; the fixed-direction construction above omits those effects. A reference-axis rotation preserves angles between directions, whereas changing the observer or correcting atmospheric refraction is a different operation. ^[ref].

Consider the symmetric matrix and initial unit vectorIts orthonormal eigenbasis consists of $u_1=(1,1{)}^T/\sqrt{2}$ and $u_2=(1,-1{)}^T/\sqrt{2}$, with eigenvalues $3$ and $1$. Since $x_0=(u_1+u_2)/\sqrt{2}$, direct multiplication givesThe first unnormalized vectors are $(1,0{)}^T$, $(2,1{)}^T$, $(5,4{)}^T$, and $(14,13{)}^T$. If ${\theta }_k$ is the acute angle from the line spanned by the dominant eigenvector, then $\tan {\theta }_k=3^{-k}$. Every application divides the unwanted-to-wanted component ratio by three: normalization changes lengths, while the spectral gap selects the direction.

The right panel compares the power iteration direction error, Rayleigh quotient error, and eigenpair residual. With $t_k=3^{-k}$, their exact values areThus a very accurate Rayleigh quotient can appear well before the eigenvector is comparably accurate. This is an example of quadratic Rayleigh-quotient improvement for the power method.

The FRET dipole orientation factor describes the geometry of the two transition electric dipoles relative to their separation. Let ${\hat{\mathbf{\mu }}}_D$, ${\hat{\mathbf{\mu }}}_A$ and $\hat{\mathbf{r}}$ be their unit vectors. ThenFor a separation along $z$, parallel electric dipoles along $z$ give ${\kappa }^2=4$, parallel ones along $x$ give $1$, and one along $x$ with the other along $y$ gives $0$. Thus rotating a pair can change FRET efficiency without changing its separation.

The familiar value ${\kappa }^2=2/3$ requires suitable rapid, independent orientation averaging during the FRET donor lifetime. Random orientations across a population of slowly rotating fluorescent proteins do not establish that dynamic condition. ^[ref]

In Cartesian coordinates on Euclidean space, the gradient of a scalar-valued differentiable function is the vector of its first partial derivatives. It converts a scalar field, such as temperature or a height, into a vector field describing its local change.

For a unit vector $u$, the directional derivative is $D_{u}f=u\cdot \nabla f$. The Cauchy-Schwarz inequality gives $|D_{u}f|\le |\nabla f|$. At a point where $\nabla f\ne 0$, the greatest increase per unit distance occurs along $u=\nabla f/|\nabla f|$, while the opposite direction gives the greatest decrease. These are first-order statements at the current point; they do not predict how the function behaves far along a straight path.

For $f(x,y)=x^2+2y^2$ at $(1,1)$, $\nabla f=(2,4)$. Moving in the $x$ direction gives rate $2$ and moving in the $y$ direction gives rate $4$, but the steepest direction is $(1,2)/\sqrt{5}$, withThe direction $(-2,1)/\sqrt{5}$ has rate zero and is tangent to the level curve through that point. More generally, differentiating $f(\gamma (t))=\text{constant}$ along a regular level curve gives $\nabla f\cdot {\gamma }^{\prime }=0$, so the gradient is its normal vector.

This is useful when navigating a measured scalar field or applying gradient descent: one collection of partial derivatives predicts every infinitesimal directional change. The derivative assumes a differentiable function, not merely the existence of all directional derivatives. ^[ref].

Gravity gradiometry measures spatial differences of gravitational acceleration. Two simultaneous measurements along a unit vector $\mathbf{e}$, separated by $L\mathbf{e}$, giveThis measures one directional component of the tidal tensor, not its whole matrix. Any identical platform acceleration cancels in the difference. Unequal instrument responses, changing geometry and rotating reference frames can spoil that cancellation. For a large baseline, the quotient is an average of the directional tidal tensor along the baseline, rather than its exact value at one point.

For points on a sphere of radius $R$, write their position vectors relative to the centre as $Ru,Rv$ with $u,v$ unit vectors. The great-circle distance is the shorter great circle arc length, $R$ times its central angle. Hence it equals $R\arccos (u\cdot v)$, including zero for identical points and $\pi R$ for antipodal points. At antipodal points the shortest arc is not unique, but its length is.

A ground state is a normalized energy eigenstate that attains the lowest possible energy of a quantum system. Precisely, let the Hamiltonian operator $H$ be a self-adjoint operator bounded below on a Hilbert space, and write $E_{\ast }=\inf \sigma (H)$ for the infimum of its spectrum. A ground state is a unit vector ${\psi }_0$ in the operator domain $D(H)$ satisfyingThus a ground state exists exactly when the spectral bottom $E_{\ast }$ is an energy eigenvalue. The lowest energy eigenvalue among some listed bound states is insufficient if other parts of the spectrum lie below it. The ground-state subspace collects all minimizing energy eigenstates; it can have ground-state degeneracy. The quantum variational principle characterizes $E_{\ast }$ as the infimum of normalized energy expectation values. ^[ref].

For the one-dimensional quantum harmonic oscillator, the ground state has energy $\hslash \omega /2$ above the minimum of the potential energy. This zero-point energy is compatible with being the lowest allowed energy: a localized wavefunction cannot simultaneously have zero kinetic energy. For a free particle on the whole line, the spectral bottom is zero but is unattained by any normalizable wavefunction; see ground-state energy. A lower bound therefore need not describe a physically preparable energy eigenstate.

The ground state supplies a reference for excitation costs and for the quantum variational principle. A ground state is also a stationary state under a time-independent Hamiltonian operator: its quantum phase changes, while its position probability density stays fixed. A nonzero ground-state energy does not imply motion along a classical trajectory, and ground-state degeneracy means there need not be a unique lowest quantum state.

When a ground state exists, its ground-state degeneracy is the dimension of the ground-state subspace:Here $I$ is the identity operator, and the kernel of a linear map is taken within the operator domain of the Hamiltonian operator $H$. The count concerns linearly independent energy eigenstates, rather than every normalized quantum superposition or every choice of global phase. Even a single energy eigenspace contains infinitely many representing vectors; its dimension measures independent freedom. Ground-state degeneracy is the lowest-energy instance of degenerate energy levels.

For an exact finite model with energy unit $\epsilon >0$, let the Hamiltonian operator be the diagonal matrixIts ground-state energy is zero and $g_0=2$: the first two coordinate unit vectors $e_1,e_2$ span the ground-state subspace. Every normalized $a{e}_1+b{e}_2$, with $|a{|}^2+|b{|}^2=1$, is a ground state, including $(e_1+e_2)/\sqrt{2}$. Its energy measurement is certain to return zero, even though a measurement distinguishing $e_1$ from $e_2$ need not have a definite outcome. Under the unchanged Hamiltonian operator, this quantum superposition is a stationary state because both components acquire the same global phase.

Ground-state degeneracy matters when an interaction selects or splits previously equal energy eigenvalues. In the finite model, adding $\epsilon \mathrm{diag}(\delta ,-\delta ,0,0)$ with $0<\delta <2$ gives the formerly lowest pair the distinct energies $\pm \epsilon \delta$, leaving a unique ground state. This direct calculation illustrates why degenerate perturbation theory first examines the perturbation inside the whole ground-state subspace.

A classical continuous-symmetry model can instead have a manifold of minimizing configurations, such as the sphere of orientations of a Néel state in an isotropic Heisenberg antiferromagnet. That geometric description does not by itself count the dimension of a finite quantum ground-state subspace.

For a rotor spinning about a principal axis with its support at the centre of mass, uniform gravity supplies no torque about that centre. With other external torques negligible, conservation of angular momentum preserves its direction in an inertial frame. A mount that permits free axis motion lets the housing rotate relative to that persistent direction, giving an orientation reference when external landmarks are unavailable. This is the useful limit opposite to the deliberately off-centre support in the spinning-wheel precession experiment. It supplies a direction, not a position or a full three-axis orientation from one axis alone.

For a small accumulated transverse torque about the centre of mass, the change in the direction unit vector is, to first order,Here $L$ is approximately constant over the short interval. Large angular momentum reduces drift for the same integrated transverse torque; parallel torque instead changes its magnitude. For $L=8.0\mathrm{kg}{\mathrm{m}}^2/s$, a constant transverse torque of $0.010\mathrm{N}\mathrm{m}$ over $1.0\mathrm{s}$ produces approximately $0.00125\mathrm{rad}$, or $0.0716^{\circ }$, of direction change. This is another synthetic calculation. The persistent direction belongs to an inertial frame; its components can change in a rotating reference frame even with zero external torque.

A bounded operator $T:H\to H$ on a real or complex Hilbert space is Hermitian, or self-adjoint, when $T=T^{\ast }$. Equivalently,Its directional values $q_T(x)=⟨Tx,x⟩$ are real, since this identity makes $q_T(x)$ equal its complex conjugate. Reality allows either sign; a positive operator additionally requires nonnegative values in every direction.

The diagram compares Hermitian quadratic forms along paths of unit vectors. The Hermitian matrix $D=\mathrm{diag}(1,-1)$ takes the values $1$ and $-1$ on the standard basis vectors; $P=\mathrm{diag}(1,0)$ takes the values $1$ and $0$; andsatisfies $H{e}_1=e_1$ and $H{e}_3=3e_3$. These eigenvectors form an orthonormal basis, and every $x=a{e}_1+b{e}_3$ has $q_H(x)=|a{|}^2+3|b{|}^2$. Thus $H$ is a positive operator, with eigenvalues $1,3$ and operator norm $3$.

Three conditions describe different geometry. A Hermitian operator satisfies $T=T^{\ast }$; a normal operator satisfies $T^{\ast }T=T{T}^{\ast }$; a unitary operator satisfies $T^{\ast }T=T{T}^{\ast }=I$, where $I$ is the identity operator. Every Hermitian operator and every unitary operator is a normal operator. Concrete matrices separate the conditions:

The spectrum of a bounded Hermitian operator is real. In finite dimension, the spectral theorem supplies an orthonormal basis of eigenvectors, turning its action into independent real scalings. This makes the geometry easy to interpret: each eigenvalue records a signed response along an orthogonal direction. An arbitrary bounded Hermitian operator in infinite dimension need not have any eigenvectors; the general spectral theorem uses a projection-valued measure. MIT's lecture 21 develops the real spectrum and the orthogonality of eigenspaces.^[ref]

A historical connection is the study of forms with conjugate variables. In his 1901 account of Charles Hermite's work, Camille Jordan describes these forms and emphasizes their role in reducing forms with real or complex coefficients. In today's notation, $x^{\ast }Hx$ is the diagonal value of a Hermitian form; the conjugate variables make a real-valued expression possible even with complex coefficients.^[ref]

For a densely defined operator outside this bounded setting, the displayed pairing identity alone defines a symmetric operator. A self-adjoint operator must also have the same operator domain as its adjoint operator.

A bounded Hermitian operator determines a Hermitian form $b_T(x,y)=⟨Tx,y⟩$ and its real-valued diagonal expression $q_T(x)=b_T(x,x)$. On a complex Hilbert space,Thus it is a quadratic form on the underlying real vector space; it is generally not a quadratic form over the complex field, whose scaling rule would involve ${\alpha }^2$. This distinction explains why replacing $x^{\ast }$ by $x^{\mathrm{T}}$ loses the geometry of a complex inner product.

For a finite-dimensional Hermitian operator with an orthonormal basis of eigenvectors $e_j$ and corresponding eigenvalues ${\lambda }_j$, the spectral theorem givesFor a unit vector, this is an average of the eigenvalues with nonnegative weights summing to one. It is also the Rayleigh quotient, so its minimum and maximum are the smallest and largest eigenvalues.

In the Hermitian operator diagram, take $x(\theta )=(\cos \theta ,\sin \theta {)}^{\mathrm{T}}$ for $D,P$, and $x_H(\theta )=\cos \theta e_1+\sin \theta e_3$ for $H$. The computed curves areThese paths illustrate the extremal values; the spectral theorem supplies the bounds for every complex unit vector, including those outside the plotted paths.

A horizontal coordinate system describes a viewing direction using altitude $\gamma$ above the observer's astronomical horizon and azimuth $\eta$ around it. Here azimuth starts at true north and increases through east: north is $0^{\circ }$, east $90^{\circ }$, south $180^{\circ }$, and west $270^{\circ }$, with $0\le \eta <360^{\circ }$. True north is the northward intersection of the celestial meridian with the astronomical horizon, rather than an uncorrected magnetic-compass reading. Other origins and directions are legitimate, but their formulae must be converted consistently. ^[ref].

Construct a local orthonormal basis of east ${\mathbf{e}}_E$, north ${\mathbf{e}}_N$, and upward ${\mathbf{e}}_U$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. The upward unit vector points to the zenith; the horizontal plane is perpendicular to it. For a ground instrument, a freely hanging weight establishes the local vertical through gravity. In an elementary spherical-Earth model, this vertical is radial; precision work distinguishes it from the normal to a reference surface. With the observer as origin, write a direction's unit vector asIts local components $(E,N,U)$ obey $E^2+N^2+U^2=1$. Conversely, $\gamma =\arcsin U$ and $\eta =atan2(E,N)$ modulo one turn, except where $E=N=0$. The quadrant-aware arctangent takes east as its first argument and north as its second. A synthetic direction with $\eta =60^{\circ }$ and $\gamma =30^{\circ }$ has $(E,N,U)=(3/4,\sqrt{3}/4,1/2)$. Thus its horizontal component has norm $\sqrt{3}/2$; its altitude is not the angle of the projected arrow in the horizontal plane.

A horizontal coordinate system depends on the observer's site and the observation time. The Earth's rotation changes the local axes relative to a fixed distant direction, while changing the site changes the zenith and the astronomical horizon. This makes the horizontal coordinate system directly useful for pointing an optical telescope and checking whether terrain blocks a target. A pair of angles without a site, time, azimuth convention, and geometric or refracted interpretation is an incomplete pointing instruction.

For a check against an equatorial coordinate system, let $\theta$ be the observer's geographical latitude, $\delta$ the direction's declination, and $h$ its west-positive hour angle, obtained from local sidereal time minus right ascension. In the spherical-Earth, unrefracted fixed-direction model, projection onto the local orthonormal basis givesTaking the dot product with the upward unit vector gives the last formula; the other two follow by the same projection. Recover altitude and azimuth from $(E,N,U)$ as above, using radians inside software trigonometric functions. The minus sign in $E$ is essential: a negative hour angle puts this synthetic direction east of the celestial meridian. ^[ref].

For illustrative geographical latitude $\theta =45^{\circ }$ and invented declination $\delta =15^{\circ }$, hour angle $h=-60^{\circ }$ gives $\gamma \simeq 31.636^{\circ }$ and $\eta \simeq 100.729^{\circ }$. At $h=+60^{\circ }$, the same altitude accompanies $\eta \simeq 259.271^{\circ }$: the direction is now westward. At $h=0$, it is due south with altitude $60^{\circ }$, agreeing with $90^{\circ }-|\theta -\delta |$ at upper meridian transit. These are analytical checks, not present positions of any star. For nearby objects, the direction must also be referred to the observer; simply rotating a geocentric direction omits the change of viewing position. Catalogue-to-observation refinements are described under equatorial coordinate system.

A Householder transformation is a reflection that leaves every direction in a chosen hyperplane fixed and reverses its normal direction. For a real unit vector $u$, its matrix isTo understand the formula, decompose any vector into its orthogonal projection onto the normal and a perpendicular component:Then $Hx=x_{\perp }-x_{\parallel }$. Subtracting twice the normal component changes its sign while retaining everything parallel to the mirror. A nonzero normal $v$ gives the equivalent formula $H=I-2v{v}^T/(v^{T}v)$, so normalization can be deferred.

For example, take $u=(2,1{)}^T/\sqrt{5}$. Its mirror is the straight line $2x_1+x_2=0$, andThe midpoint of these two vectors is $(-1,2{)}^T$, which lies on the mirror; their connecting segment is parallel to its normal. Both vectors have Euclidean norm five.

Several useful properties follow from this decomposition. The Householder reflection is a symmetric matrix and an orthogonal matrix:Reflecting twice restores the original vector; Euclidean norms and inner products are preserved. The eigenvalue is $-1$ along $u$ and $+1$ on $u^{\perp }$, so $\det H=-1$. A reflection reverses orientation even though it preserves lengths.

The computational payoff is that the mirror can be chosen to send a whole vector onto one coordinate axis, eliminating several entries at once. Householder QR decomposition repeats this operation on the remaining parts of successive columns. For a real symmetric matrix, applying the same Householder reflection on both sides gives a matrix similarity $HAH$ that preserves symmetry and eigenvalues; this is the basis of Householder tridiagonalization.

Alston S. Householder's 1958 paper Unitary Triangularization of a Nonsymmetric Matrix develops this elimination method. Its motivation includes reducing the arithmetic needed for triangularization while retaining the conditioning advantages of orthogonal transformations. ^[ref] Over the complex numbers, $I-2u{u}^{\ast }$ for a unit $u$ is Hermitian and unitary, with the star denoting the conjugate transpose; the coordinate-axis construction below is stated for real vectors.

The inertia tensor packages the moments of inertia about every axis through a specified point $O$ into one symmetric matrix. An axis still needs a direction: if $\mathbf{n}$ is a unit vector, its scalar moment of inertia is $I_{O,\mathbf{n}}={\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$. The same mass layout can therefore be easy to spin about one direction and harder about another. The inertia tensor also determines rotational kinetic energy and relates angular velocity to angular momentum in the appropriate rigid body model.

Take position vectors $\mathbf{r}=(x,y,z)$ relative to $O$, expressed in an orthonormal basis. For point masses $m_a$, or a continuous distribution with mass density $\rho$ and $dm=\rho dV$,Here ${\delta }_{ij}$ is the Kronecker delta. Equivalently, using the identity matrix $1$ and an outer product,Every entry has SI units $\mathrm{kg}{\mathrm{m}}^2$. The diagonal entries are axial moments of inertia; the off-diagonal entries carry a minus sign in this convention. Some texts instead define the integrals without the prefixed minus sign $\int xydm$, $\int xzdm$, $\int yzdm$ as their named off-diagonal quantities and put the minus signs into the displayed matrix. Check that convention before combining formulas. ^{[ref] [ref]}

The symmetric matrix is positive semidefinite, because its quadratic form isThe dot product and cross product identify this as a sum of squared distances when $\mathbf{u}$ is a unit vector. A zero moment of inertia occurs precisely when all the mass lies on that axis, up to sets of zero mass. A genuine three-dimensional body with positive mass density in a volume has strictly positive axial moments of inertia. Do not confuse this inertia tensor with the mass second-moment tensor $\mathbf{S}=\int \mathbf{r}{\mathbf{r}}^{T}dm$: they obey ${\mathbf{I}}_O=\mathrm{tr}(\mathbf{S})1-\mathbf{S}$, where $\mathrm{tr}$ denotes the matrix trace.

If $O$ is fixed in an inertial frame and each velocity is $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$, summing the individual kinetic energies and angular momenta givesThese formulas also describe angular momentum and rotational kinetic energy relative to the centre of mass $G$ of a translating rigid body. In that case the total kinetic energy in the inertial frame is $K_{\mathrm{total}}=\frac{1}{2}M|{\mathbf{V}}_G{|}^2+\frac{1}{2}{\mathbf{\omega }}^T{\mathbf{I}}_G\mathbf{\omega }$. For a general moving point $O$, the body's actual velocities include ${\mathbf{V}}_O$, so ${\mathbf{I}}_O\mathbf{\omega }$ alone need not be its angular momentum about $O$. ^{[ref] [ref]}

Changing the orthonormal basis changes components, rather than the physical distribution: if the columns of an orthogonal matrix $Q$ are the new basis vectors expressed in the old basis, then ${\mathbf{I}}_O^{\prime }=Q^T{\mathbf{I}}_{O}Q$. Changing the reference point is a different operation, governed by the tensor parallel-axis theorem. Both principal axes and principal moments of inertia must therefore be quoted with their reference point.