Vector ($v\in V$)

A vector is an element of a vector space. Its defining operations are vector addition and scalar multiplication. An arrow in a Euclidean space is one example; a polynomial or a function can also be a vector when the chosen vector space uses the corresponding addition and scalar multiplication.

For instance, the polynomials $p(x)=1+x$ and $q(x)=x-x^2$ belong to the real vector space of polynomials of degree at most two. Their vector addition gives $(p+q)(x)=1+2x-x^2$, and scalar multiplication gives $(2p)(x)=2+2x$. In the monomial basis $(1,x,x^2)$ these are the coordinate vectors $(1,1,0)$, $(0,1,-1)$, $(1,2,-1)$ and $(2,2,0)$ respectively. The vectors are the polynomials; these tuples are their representations in a chosen basis.

A general vector space does not itself prescribe lengths or angles. A norm supplies a size, while an inner product also supplies angle information and orthogonality. The familiar arrow geometry below belongs to Euclidean vectors, where this extra structure is available.

An axial vector transforms as ${\mathbf{a}}^{\prime }=(\det Q)Q\mathbf{a}$ under an orthogonal matrix $Q$. It agrees with a polar vector under a proper rotation matrix and has the additional determinant sign under an orientation-reversing transformation. Angular momentum and angular velocity are axial vectors.

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 Euclidean vector is a vector in a finite-dimensional real inner product space. Choosing an orthonormal basis identifies it with its real vector components in ${\mathbb{R}}^n$. A geometric arrow represents a Euclidean vector by a length and a direction; translating the arrow without changing either represents the same Euclidean vector. The zero vector has length zero and no distinguished direction.^[ref]

For points $P=(1,1)$ and $Q=(4,5)$ in a Cartesian coordinate system, the displacement is $Q-P=(3,4)$. The arrow from $(-2,0)$ to $(1,4)$ represents the same Euclidean vector, even though its endpoints differ. An endpoint is a position; their difference is a vector. In an orthonormal basis $({\mathbf{e}}_1,{\mathbf{e}}_2)$,The Euclidean norm comes from the inner product: $\Vert \mathbf{v}{\Vert }_2^2=⟨\mathbf{v},\mathbf{v}⟩$. The sum of squared vector components is valid in an orthonormal basis; an arbitrary basis requires the appropriate Gram matrix. This distinction lets the geometric vector stay the same while its coordinate vector changes.^[ref]

Vector addition is componentwise. With $\mathbf{u}=(2,1)$ and $\mathbf{w}=(1,3)$,Geometrically, translate the second arrow so its tail starts at the first arrow's head. The vector sum joins the starting point to the final endpoint. Reversing the order gives the other route around the same parallelogram, so vector addition is commutative. This is the useful bookkeeping behind adding successive displacements, velocities, or forces: direction information survives the calculation. For a force, the Euclidean vector alone does not specify its point of application, which also affects the torque.^[ref]

The magnitude of a vector sum is generally different from the sum of the magnitudes. Here $\Vert \mathbf{u}+\mathbf{w}{\Vert }_2=5$, whereas $\Vert \mathbf{u}{\Vert }_2+\Vert \mathbf{w}{\Vert }_2=\sqrt{5}+\sqrt{10}>5$. The triangle inequality explains why the direct displacement cannot exceed the length of this two-segment route. A return displacement $(-3,-4)$ makes the net vector sum zero, although the lengths of the three travelled segments sum to a positive number.

The dot product exposes direction quantitatively: for nonzero Euclidean vectors, $⟨\mathbf{u},\mathbf{w}⟩=\Vert \mathbf{u}{\Vert }_2\Vert \mathbf{w}{\Vert }_2\cos \theta$. For the displayed pair, the dot product is $2\cdot 1+1\cdot 3=5$, so $\cos \theta =1/\sqrt{2}$ and the angle is $45^{\circ }$. For orthogonal vectors, that dot product is zero. Dividing by magnitudes separates direction from size, which is the purpose of vector normalization.

A unit vector is a vector of norm one in a normed vector space. For a nonzero vector $v$, vector normalization produces one by dividing by $\Vert v\Vert$. In a real Euclidean space, this keeps the direction and removes the magnitude. The zero vector cannot be normalized because its norm is zero.

For $\mathbf{v}=(3,4)$ with Euclidean norm $5$, the corresponding unit vector isThe Euclidean vector can therefore be reconstructed as $\mathbf{v}=5\hat{\mathbf{v}}$: size and direction are distinct pieces of information. The standard basis vectors are familiar unit vectors, but a unit vector need not point along a coordinate axis.

Vector normalization converts a nonzero vector in a normed vector space into a unit vector. Absolute homogeneity of a norm givesThe direction is unchanged by multiplying by a positive real scalar. For $c>0$, $\Vert cv\Vert =c\Vert v\Vert$ and $\widehat{cv}=\hat{v}$; for a negative real scalar, $\widehat{cv}=-\hat{v}$. Thus $(3,4)$ and $(6,8)$ normalize to the same unit vector, while $(-3,-4)$ normalizes to its negative. This decomposition is useful for specifying a direction independently of a force, velocity or displacement magnitude.

In a complex normed vector space, a nonzero complex scalar $c$ instead gives $\widehat{cv}=(c/|c|)\hat{v}$. The remaining complex argument is not removed by normalization. For a pure quantum state, vectors differing only by this overall global phase represent the same physical state; vector normalization alone does not identify them.

The zero vector cannot be normalized this way. The construction also applies in a function space: an eigenfunction of a Sturm-Liouville problem can be scaled to unit weighted norm byThe positive weight $w$ specifies the weighted inner product. Multiplication by this nonzero constant preserves the eigenvalue equation and its homogeneous boundary conditions.

Vector normalization is not a linear map. With ${\mathbf{e}}_1=(1,0)$ and ${\mathbf{e}}_2=(0,1)$,Consequently, adding individually normalized directions does not generally give the direction of the original vector sum. Compute the intended vector sum first, then normalize it if it is nonzero. Cancellation is a further boundary case: $v+(-v)=0$ has no normalized direction even though both nonzero summands do. A numerical algorithm must check that case before dividing by the norm.

The cross product $a\times b$ of two three-dimensional vectors is perpendicular to both, has magnitude $|a||b|\sin \theta$, and is oriented by the right-hand rule.

For a unit vector $\mathbf{n}$, real $\alpha \ne 0$, and real $\beta$, the displayed linear equation has the unique vector solutionTo prove it, split vectors into components parallel and orthogonal to $\mathbf{n}$. On the parallel line, the linear map is multiplication by $\alpha$. On the perpendicular plane, the operator $N\mathbf{v}=\mathbf{n}\times \mathbf{v}$ obeys $N^2=-I$, by the vector triple product. Therefore $(\alpha I-\beta N)(\alpha I+\beta N)=({\alpha }^2+{\beta }^2)I$ on that plane. Combining its inverse with the parallel solution gives the formula. This avoids component calculations in equations with an axial cross product term.

For any real three-dimensional vectors $a,b$, the displayed linear equation has the unique solutionTaking a dot product with $b$ first yields $(1+|b{|}^2)(b\cdot x)=a\cdot b$. Taking a cross product with $b$ and using the vector triple product identity givesSubstitution into the original equation cancels the longitudinal terms and proves the formula. Conversely direct substitution verifies it for all $b$, including $b=0$, since the denominator is positive. If $C_{b}x=b\times x$, the underlying identity for linear maps is $(I-C_b)(I+C_b+b{b}^T)=(1+|b{|}^2)I$, using $C_b^2=b{b}^T-|b{|}^{2}I$.

For a fixed unit vector $g\in {\mathbb{R}}^3$, the linear map $r\mapsto r\times g$ has kernel $\mathbb{R}g$ and image the perpendicular plane $g^{\perp }$. Thus $r\times g=f$ is solvable exactly when $f\cdot g=0$, and then its full solution set isFor the proof, a cross product with $g$ must be perpendicular to $g$. Conversely the vector triple product identity gives $g\times (r\times g)=r-(r\cdot g)g$, which determines the perpendicular orthogonal projection of $r$. The remaining parallel component is arbitrary. Finally $(g\times f)\times g=f$ when $f\cdot g=0$, verifying sufficiency.

Geometrically, the solution set is a straight line with direction $g$. Its nearest point to the origin is $g\times f$ and its distance from the origin is $|f|$. For a general nonzero fixed vector $a$, the corresponding formula is $r=(a\times f)/|a{|}^2+ta$, again subject to $f\cdot a=0$.