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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.

A Cartesian basis is an ordered basis of unit vectors along the mutually perpendicular axes of a Cartesian coordinate system. It is an orthonormal basis of a real Euclidean space: $e_i\cdot e_j={\delta }_{ij}$, with ${\delta }_{ij}$ the Kronecker delta. In the plane, the familiar standard basis is $e_1=(1,0{)}^T$, $e_2=(0,1{)}^T$; the vector $p=(2,1{)}^T$ is $2e_1+e_2$. A rotated pair of perpendicular unit vectors is also a Cartesian basis, even though its basis vectors have different standard basis entries.

The advantage is that coordinates can be measured separately by dot products, and the squared norm is the sum of their squares:These are the real Euclidean space instances of the orthonormal basis expansion.^[ref]

An orthogonal basis only requires perpendicular basis vectors; their lengths need not be one. For example, $(2,0{)}^T,(0,3{)}^T$ are orthogonal, but $p=(2,1{)}^T$ has coordinates $(1,1/3{)}^T$, while its dot products with them are $4,3$. For an orthogonal basis $(a_i)$, the coordinate along $a_i$ is $(p\cdot a_i)/(a_i\cdot a_i)$. Dividing each nonzero basis vector by its norm makes it a unit vector without changing its direction. Unit length alone does not imply orthogonality: $(1,0{)}^T$ and $(1,1{)}^T/\sqrt{2}$ are both unit vectors but have dot product $1/\sqrt{2}$.

The examples below show why these distinctions matter: rotating a Cartesian basis changes the coordinate vector while keeping lengths easy to compute; using nonperpendicular basis vectors still gives unique coordinates, but requires the dual basis or a matrix inverse to extract them.

Rotate the standard basis through $30^{\circ }$ using the rotation matrix whose columns are the new basis vectors:Here $u\cdot v=0$ and $u\cdot u=v\cdot v=1$, so $Q$ is an orthogonal matrix with $Q^{T}Q=I$, where $T$ denotes transpose and $I$ is the identity matrix. For the same vector $p=(2,1{)}^T$, its coordinate vector in this Cartesian basis isThus $\alpha \approx 2.232$ and $\beta \approx -0.134$: travel forwards along $u$ and a little backwards along $v$ to reconstruct $p=\alpha u+\beta v$. The orthogonal projections onto the two axes give these signed coordinates because the measuring vectors are both perpendicular and of unit length.

The squared norm remainsThis is useful when choosing axes aligned with a geometric feature: the coordinates change but distances and dot products retain the same sum-of-products formula. The two directions of the change of basis must be kept straight: $Q$ reconstructs the vector in old coordinates, $p=Q[p{]}_{\mathcal{U}}$, and $Q^T$ extracts its new coordinates. Applying $Q$ to the old coordinate vector $p$ instead would actively rotate the vector.

A short proof explains both formulas. If $p=Qc$, multiplication by $Q^T$ gives $c=Q^{T}p$. Then $\Vert p{\Vert }^2=c^T{Q}^{T}Qc=c^{T}c$. Perpendicular unit-length columns make the matrix inverse equal the transpose; this is the computational convenience supplied by a Cartesian basis.

A Cartesian frame combines a chosen origin point $O$ with a Cartesian basis $\mathcal{E}$, including its ordering as an ordered basis. A point $P$ has coordinate vector $c$ when $\overrightarrow{OP}={\sum }_i{c}_i{e}_i$. The basis supplies measuring directions and unit lengths; the coordinate vector is the resulting tuple of numbers; the Cartesian frame additionally fixes where measurement starts. A vector already belongs to a vector space with a zero vector, so expressing it in a basis requires no additional choice of spatial origin.

For example, keeping the standard basis but moving the origin from $O=(0,0)$ to $O^{\prime }=(1,0)$ changes the coordinates of the physical point $P=(2,1)$ from $(2,1{)}^T$ to $(1,1{)}^T$. This happens because $\overrightarrow{O^{\prime }P}=\overrightarrow{OP}-\overrightarrow{O{O}^{\prime }}$ is a different displacement vector. By contrast, the two diagrams above keep $O$ fixed and change only the basis. In a fixed reference Cartesian frame, a new frame with origin column $o$ and basis-column matrix $Q$ describes points by $p=o+Qc$; the new coordinates are $c=Q^T(p-o)$ when $Q$ is an orthogonal matrix.

A change of basis rewrites a vector using a different ordered basis of the same vector space. The vector stays fixed; its coordinate vector changes because the measuring directions change. Both bases must describe every vector uniquely, so a change of basis is reversible.

Start with the Cartesian basis $e_1=(1,0)$, $e_2=(0,1)$, and choose the new basis vectors $u=(1,1)$ and $v=(1,-1)$. For the vector $z=(2,0)$,The second coordinate vector says to travel once along $u$ and once along $v$. Their vertical contributions cancel, leaving exactly the original horizontal arrow. The diagram shows two descriptions of that same arrow, with the same origin and endpoint.

Choosing a useful basis can reveal simple structure hidden by the old coordinate vectors. Here the new basis describes common and opposing contributions. For two equal measurements $(5,5)$, the new coordinate vector is $(5,0)$: only the common contribution is present. In larger examples, a suitable basis can concentrate important information into a few coefficients. Strang's lecture on change of basis develops this principle through applications: the reversible change of basis preserves information, while subsequently discarding small coefficients loses information.^[ref]

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]

For a nonzero real vector $x\in {\mathbb{R}}^n$, choose $\alpha$ with $|\alpha |=\Vert x{\Vert }_2$ and put $v=x-\alpha e_1$. When $v\ne 0$, the Householder reflectionsatisfies $Hx=\alpha e_1$: every component except the first becomes zero. Here $e_1$ is the first Cartesian basis vector. The calculation is short:If $v=0$, the vector is already at the selected target, so the identity matrix suffices.

In floating-point arithmetic, a customary choice isThen $|v_1|=|x_1|+\Vert x{\Vert }_2$, avoiding subtraction of nearly equal positive quantities in the first component. Either sign can be valid in exact arithmetic; the numerical choice protects the construction from catastrophic cancellation. ^[ref]

For $x=(3,4{)}^T$, this choice gives $\alpha =-5$ and $v=(8,4{)}^T$. Dividing $v$ by its Euclidean norm gives $u=(2,1{)}^T/\sqrt{5}$, exactly the normal in the geometric example.

Use a fixed right-handed Cartesian basis in three dimensions and active rotations in three dimensions through $\pi /2$ radians. The right-hand rule gives the following rotation matrices about the positive $x$ and $z$ axes:For the vector $v=(1,1,0{)}^T$, the two orders giveBoth outputs have Euclidean norm $\sqrt{2}$, but they point in different directions. Each arrow below is a vector from the origin; the curved paths track its endpoint during each rotation in three dimensions.

The axes in this example remain fixed throughout. A command to turn about an object's own current axis describes a different operation because that axis can move after the first turn. With an object's orthonormal basis represented by columns of $Q$, a fixed-axis rotation matrix $R$ acts by $RQ$, whereas the corresponding rotation expressed in the object's own basis acts by $QR$.^[ref]

Let $e_i^{\prime }(t)$ be a Cartesian basis fixed in the rotating frame. A vector $A=A_i^{\prime }{e}_i^{\prime }$ then satisfiesbecause a derivative of a body-fixed basis vector is ${\dot{e}}_i^{\prime }=\omega \times e_i^{\prime }$. Applying this transport formula twice to $r$ gives, for constant $\omega$,The second and third terms correspond to Coriolis acceleration and centrifugal acceleration in the rotating description.

For the bead, use cylindrical unit vectors $(e_r,e_{\varphi },e_z)$ with $e_z$ upward andProjection of $m\ddot{r}=-mg{e}_z+N$ along the wire tangent $\cos \theta e_r+\sin \theta e_z$ eliminates the smooth-wire normal force $N$ and gives

If $u=0$, any nonzero $w$, for instance the first Cartesian basis vector $e_1$, gives $H(w)u=0$ and $\alpha =0$. Suppose $u\ne 0$. Choosewhere ${\mathrm{sgn}}_{+}(r)=1$ for $r\ge 0$ and $-1$ otherwise. This choice ensures $w\ne 0$, including when $u$ is already on the first coordinate axis. Since ${\alpha }^2=u^{T}u$,Therefore the Householder reflection satisfiesThe orthogonality of $H(w)$, used here to check length preservation, gives $|\alpha |=\Vert H(w)u{\Vert }_2=\Vert u{\Vert }_2$. The chosen sign also avoids catastrophic cancellation in $w_1=u_1-\alpha$. This is a Householder reflection sending a vector to a coordinate axis with no zero-normal exception.

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]