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Absolute homogeneity is the norm axiom $\Vert av\Vert =|a|\Vert v\Vert$ for every scalar $a$ and vector $v$. It distinguishes a norm from a merely homogeneous expression of a different degree; for example the fourth root in the box norm converts a degree-four expression into a degree-one norm.

Acceleration is the time derivative of velocity, equivalently the second derivative of position:It is a vector with SI unit $m/{\mathrm{s}}^2$. A change in direction produces acceleration even if speed stays constant. For example, a car rounding a bend at steady speed needs an inward resultant force by Newton's second law. Thus acceleration does not mean only speeding up. When $v>0$, the dot product gives $dv/dt=\mathbf{v}\cdot \mathbf{a}/v$: a perpendicular acceleration changes direction without changing speed at that instant.

Take counterclockwise circular motion about the origin with radius $R=2\mathrm{m}$ and constant angular speed $\omega =0.5rad/s$. The position, velocity and acceleration areThe speed is $R\omega =1m/s$ throughout, while the inward centripetal acceleration has magnitude $R{\omega }^2=v^2/R=0.5m/{\mathrm{s}}^2$. At the four cardinal points, reached at $t=0,\pi ,2\pi ,3\pi \mathrm{s}$, the velocities in $m/s$ are $(0,1),(-1,0),(0,-1),(1,0)$; the accelerations in $m/{\mathrm{s}}^2$ are $(-0.5,0),(0,-0.5),(0.5,0),(0,0.5)$. The vectors turn together, remaining perpendicular. ^[ref]

The proof outline is just two applications of the chain rule to $\mathbf{r}(t)$; ${\sin }^2(\omega t)+{\cos }^2(\omega t)=1$ then fixes both magnitudes. The usefulness is geometric: turning requires a force even when a speed reading remains steady, and doubling speed at fixed radius quadruples the required centripetal acceleration. The same mechanism appears in a circular orbit.

Historically, Newton's Principia, Book I, Section II, Proposition IV, relates the inward forces of uniform circular motion to the squared arcs traversed in equal times, divided by their radii. In modern notation, its dependence on speed and radius is $v^2/R$; Newton presents a geometric argument rather than the coordinate derivatives above. ^[ref]

For column coordinate vectors in the fixed Cartesian basis, an active planar rotation sends $(x,y{)}^T$ to $(x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta {)}^T$. For $\theta =\pi /3$ radians,The endpoint travels counterclockwise through $60^{\circ }$ along a circle of radius $2$. Its Euclidean norm stays $\sqrt{1^2+(\sqrt{3}{)}^2}=2$. The second Cartesian basis direction becomes $R_{\pi /3}(0,1{)}^T=(-\sqrt{3}/2,1/2{)}^T$, whose dot product with $(1,\sqrt{3}{)}^T$ is still zero.

A quick sign check is to look at the first column: $R_{\theta }(1,0{)}^T=(\cos \theta ,\sin \theta {)}^T$ must point above the horizontal axis for $0<\theta <\pi$. The columns have unit Euclidean norm and zero dot product, so $R_{\theta }^T{R}_{\theta }=I$; the determinant is ${\cos }^2\theta +{\sin }^2\theta =1$. Finally, the transpose negates the off-diagonal entries, giving $R_{\theta }^T=R_{-\theta }$: turning back reverses the action. This is the same matrix used for passive rotation of coordinates, though that interpretation keeps the vector fixed.

For a diagonalizable matrix, the active spectrum consists of eigenvalues whose eigenspaces have nonzero components of the starting vector in its spectral decomposition. These components are orthogonal projections for a self-adjoint operator, but need not be orthogonal projections for a general diagonalizable matrix. With a unique largest modulus in this active spectrum, the power method selects its eigendirection; an exactly absent dominant component is never created in exact arithmetic.

For vector subspaces $U,W$ of a common vector space, this is a surjective linear map. Its kernel of a linear map is $\{(x,-x):x\in U\cap W\}$, which has a linear isomorphism to $U\cap W$ by $x\mapsto (x,-x)$. The rank-nullity theorem and the direct sum basis give the dimension formula for a sum of subspaces. The paired domain keeps the two contributions separate even when their underlying vectors overlap.

A value has cooperative-game additivity when adding two characteristic functions on the same player set adds their allocation vectors. Here $(v+w)(S)=v(S)+w(S)$ for every coalition. The sum of two simple cooperative games need not be simple; the axiom is naturally stated on transferable utility games.

For a bounded linear operator $T:H\to K$ between real or complex Hilbert spaces, its adjoint is the unique bounded linear operator $T^{\ast }:K\to H$ satisfyingThe inner product here is linear in its first argument. The adjoint operator transfers an output measurement back to the input: measuring $Tx$ against the vector $y$ gives exactly the same scalar as measuring $x$ against $T^{\ast }y$. It depends on both inner products.

In this real example,Both inner products equal $3$. The dashed lines collect all vectors giving that same measurement: the output line $u_1=3$ becomes the input line $2x_1+x_2=3$. The adjoint operator therefore moves the direction that describes the measurement. It generally differs from the inverse matrix: here $A^{-1}y=(1/2,0{)}^{\mathrm{T}}$.

For an explicitly complex example, use the standard inner product $⟨u,v⟩={\sum }_j{u}_j\overline{v_j}$ on ${\mathbb{C}}^2$. In the standard orthonormal basis, the adjoint operator is represented by the conjugate transpose:For $x=(1+i,2{)}^{\mathrm{T}}$ and $y=(1,i{)}^{\mathrm{T}}$, direct matrix multiplication givesand henceThe complex conjugate is essential to this agreement.

Existence follows from the Riesz representation theorem: for fixed $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ has a unique representing vector $T^{\ast }y$. The resulting adjoint operator has the same operator norm as $T$, andMIT's lecture 18 develops the representation argument and the conjugate transpose example.^[ref]

This transfer explains the normal equation for a linear inverse problem: a least-squares solution has residual $Tx-b$ orthogonal to every attainable direction $Th$, so $⟨h,T^{\ast }(Tx-b)⟩=0$ for every $h$, equivalently $T^{\ast }Tx=T^{\ast }b$. An output residual becomes an input condition through the same adjoint operator. All identities in this section concern bounded operators defined on the whole source Hilbert space; for a densely defined operator that is not bounded, use the adjoint of a densely defined operator and check its operator domain.

In the crystallographic convention, a Coxeter graph is admissible when it is realized by linearly independent unit vectors $v_i$ in a Euclidean space, with nonpositive mutual inner products and $4(v_i,v_j{)}^2=n_{ij}\in \{0,1,2,3\}$ for distinct vertices. The number $n_{ij}$ records a single, double or triple connection; equivalently, the displayed Gram matrix is positive-definite. In edge-label notation the nonzero connections have labels $3,4,6$. Forgetting multiplicity, the underlying graph is a forest: for any cycle of $k\ge 3$ distinct vertices, the sum of the corresponding unit vectors would have squared norm at most $k-k=0$, contradicting their linear independence. Multiple lines on one connection are labels and are not length-two cycles. This is the convention of Wehler's Lie algebra notes, section 6.3.

An affine dependence of points is a nonzero coefficient family satisfying the displayed identities. Equivalently, the augmented vectors $(x_i,1)$ are linearly dependent. Both identities are independent of the chosen origin. The absence of an affine dependence is affine independence.

Points $p_0,\dots ,p_k$ in an affine space are affinely independent if the only affine dependence displayed above has all coefficients zero, using any choice of origin to read the vector equation. Since the coefficients sum to zero, replacing the origin leaves that equation unchanged. Substituting ${\lambda }_0=-{\sum }_{i=1}^k{\lambda }_i$ shows that this condition is equivalent to linear independence of $p_1-p_0,\dots ,p_k-p_0$. It also says that an affine combination of the points has unique coefficients. In ${\mathbb{R}}^n$, at most $n+1$ points can be affinely independent; $n+1$ such points span a full-dimensional simplex.

An affine map combines a linear map with a translation. After choosing coordinate origins, it has the form $T:{\mathbb{R}}^p\to {\mathbb{R}}^q$, $T(x)=Mx+b$, where the $q\times p$ matrix $M$ is its linear part of an affine map and $b$ is a fixed vector. Equivalently, it preserves every affine combination. ^[ref]

An original planar example isIt stretches horizontally, applies a shear mapping, and then translates. The vertices $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$ become $A^{\prime }=(1,-2)$, $B^{\prime }=(5,-2)$ and $C^{\prime }=(3,0)$. A square grid becomes a grid of parallelograms; straight edges and their weighted points still fit together.

This map is not a linear map with respect to the displayed origins, because $T(0,0)=(1,-2)$ rather than zero. It also is not a Euclidean similarity: $AB$ has length $2$ and $A^{\prime }{B}^{\prime }$ has length $4$, while the right angle at $A$ becomes $45^{\circ }$. Its determinant is $2$, so it doubles planar area. These metric changes explain why affine geometry studies straightness, incidence and weighted points rather than treating every transformation as an isometry.

The linear part is an invertible matrix, giving the concise inverse $T^{-1}(u,v)=((u-v-3)/2,v+2)$. More generally an affine map is bijective exactly when its linear part of an affine map is bijective. Between equal finite-dimensional spaces, this means an invertible matrix; the translation cannot repair a loss of rank of a linear map.

Algebraic irreducibility excludes every proper nonzero invariant vector subspace, including those that are not closed. It is stronger than topological irreducibility, which excludes only closed invariant subspaces. For a nonzero irreducible action, every nonzero vector is cyclic in the algebraic sense: $A\xi =X$.

For a real conductivity tensor invariant under every rotation about a unit vector $n$, the electric current density is $J=a{E}_{\perp }+d{E}_{\parallel }+hn\times E_{\perp }$. Its collinearity with a nonzero real electric field $E$ has three cases. Axial fields are always mapped to an axial current. Purely transverse fields give a collinear current exactly when $h=0$, equivalently when this tensor is symmetric. Fields with nonzero axial and transverse parts require both $h=0$ and $a=d$, making the tensor isotropic. The zero vector is included as a scalar multiple; requiring a nonzero current also requires the relevant scalar conductivity to be nonzero.

To prove necessity, take the dot product of $J=\lambda E$ with $n\times E_{\perp }$. If $E_{\perp }\ne 0$ this gives $h=0$; the transverse and axial equations then respectively give $\lambda =a$ and, when $E_{\parallel }\ne 0$, $\lambda =d$. Reversing these steps proves sufficiency. Rotations about the axis alone permit $h\ne 0$; symmetry under a transverse half-turn would eliminate this antisymmetric term.

A bilinear form is alternating if $B(v,v)=0$ for every vector $v$. Expanding $B(v+w,v+w)=0$ gives $B(v,w)+B(w,v)=0$, hence $B(v,w)=-B(w,v)$. If the characteristic is not two, the converse follows from $B(v,v)=-B(v,v)$ by dividing by two. In characteristic two, the relation $B(v,w)=-B(w,v)$ is just symmetry and does not force zero self-pairings: $B(x,y)=xy$ on ${\mathbb{F}}_2$ has $B(1,1)=1$.^[ref]

The matrixgives an alternating bilinear form over every field. Over the real numbers, its value is the signed area of the parallelogram with sides $v,w$: $B((2,1),(1,3))=6-1=5$, and swapping the sides gives $-5$. Every self-pairing is zero, yet $\det J=1$, so this bilinear form is nondegenerate. Directly, pairing $(x,y)$ with $(1,0)$ gives $-y$, and pairing it with $(0,1)$ gives $x$; both vanish only when $(x,y)=0$.

In characteristic different from two, a skew-symmetric matrix of odd integer size has zero determinant, since $\det A=\det (-A^T)=(-1{)}^n\det A$. Thus an alternating bilinear form on a finite odd-dimensional vector space is a degenerate bilinear form. The stronger statement that the rank of an alternating bilinear form is even holds in all characteristics: splitting off a two-dimensional block with matrix $J$ repeatedly proves it. A nondegenerate alternating bilinear form defines a symplectic vector space, the linear setting for signed area rather than squared length.

For an isotropic vector order parameter, an analytic scalar Landau free energy near the origin contains powers of $\mathbf{m}\cdot \mathbf{m}$. Odd powers of its magnitude are rotationally invariant but nonanalytic as functions of the Cartesian components at zero; excluding them requires analyticity, not merely writing the magnitude as a nonnegative scalar. With $d>0$, changing $b$ from positive to negative continuously creates minima at $|\mathbf{m}|=\sqrt{-b/(2d)}$, with an arbitrary direction.^[ref]

For nonzero real vectors in a Euclidean space, their angle $\theta \in [0,\pi ]$ is determined by the displayed dot product formula. The Cauchy-Schwarz inequality puts the quotient in $[-1,1]$, so the inverse cosine is defined. Perpendicular vectors have angle $\pi /2$, parallel vectors pointing the same way have angle zero, and those pointing in opposite directions have angle $\pi$. A zero vector has no uniquely defined direction and hence no angle by this definition.

For a solenoidal vector field $J$ with compact support and suitable integrability, apply the divergence theorem to $x_i{x}_{j}J$. It gives zero boundary term and hence the displayed antisymmetry. Equivalently, for any constant vector $C$, the vector triple product identity yields $\int (C\cdot r)JdV=\frac{1}{2}C\times \int (J\times r)dV$. A piecewise smooth field must have no normal jump at its support boundary; otherwise its weak divergence contains a surface contribution and the argument fails.

Areal velocity measures how fast a position vector from a chosen origin sweeps signed swept area. It connects a geometric picture of motion to angular momentum: a long position vector with slow transverse motion can sweep the same area per unit time as a short one with fast transverse motion. Constancy requires a dynamical condition, given by the torque criterion for constant areal velocity.

Choose a fixed origin $O$ in an inertial frame and a fixed plane through $O$, with constant unit vector $\mathbf{n}$ normal to it. For a differentiable trajectory in that plane, let $\mathbf{r}(t)$ be position relative to $O$ and $\mathbf{v}=\dot{\mathbf{r}}$ its velocity. Positive swept area follows the right-hand rule about $\mathbf{n}$. In positively oriented Cartesian coordinates in the plane, and in plane polar coordinates with angle $\theta$ in radians,Here the dot product selects the normal component of the cross product. Motion away from or towards $O$ alone contributes no area. The signed specific angular momentum component $h_n=\mathbf{n}\cdot (\mathbf{r}\times \mathbf{v})$ obeysfor a constant mass $m$. If $\mathbf{n}$ is chosen along a nonzero conserved specific angular momentum vector, then $h_n=h=|\mathbf{r}\times \mathbf{v}|$ and the familiar positive formula is $\dot{A}=h/2$. Reversing the orientation changes the sign; the instantaneous unsigned swept-area rate is $|\dot{A}|$.

A distribution of quantum state populations over atomic magnetic sublevels with a nonzero expectation value of the angular momentum operator vector.

Place equal masses at the six vertices $\pm {\mathbf{e}}_i$ of a regular octahedron and identical unstressed axial springs on its twelve edges. For the invariant axial displacement family ${\mathbf{u}}_{\pm i}=\pm q_i{\mathbf{e}}_i$, the quadratic potential and kinetic energies areHence $m\ddot{\mathbf{q}}=-k(I+11^{\mathsf{T}})\mathbf{q}$. The breathing vector $1=(1,1,1)$ has squared frequency $4k/m$, while every traceless axial vector with $q_1+q_2+q_3=0$ has squared frequency $k/m$. Orthogonal projection onto these subspaces solves any initial condition in this family without an eighteen-dimensional characteristic polynomial.