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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 basis is a linearly independent spanning set of vectors in a vector space. Every vector is a unique finite linear combination of basis vectors. Independence supplies uniqueness, and spanning supplies existence. An ordered basis turns the coefficients into a coordinate vector.

For $\mathcal{B}=(b_1,b_2)=((2,1),(1,2))$, the matrix with these vectors as columns isHence the coordinate vector of $v=(5,4)$ is $[v{]}_{\mathcal{B}}=B^{-1}v=(2,1)$, and $B[v{]}_{\mathcal{B}}=v$ recovers the usual coordinates. The determinant $3$ is nonzero, so this recovery works for every vector in ${\mathbb{R}}^2$. A basis need not have perpendicular or unit-length vectors; an orthonormal basis adds those geometric conveniences when an inner product is available.

The point of changing basis is to simplify the operation being studied. Diagonalization uses an eigenbasis to turn a coupled linear map into independent scalings. In a space of polynomials, the basis $1,x,x^2$ makes differentiation easy to encode, while orthogonal polynomials can make approximation errors easier to measure. Gilbert Strang's MIT lecture on independence, basis and dimension develops this connection between attainable columns, redundant columns and coordinates.^[ref]

A matrix whose columns are the coordinate vectors of an ordered basis in a fixed reference basis. It is an invertible matrix, and multiplication by it converts coordinates in the column basis to the reference coordinates: $v=B[v{]}_B$. For a real inner product space with orthonormal reference basis, its Gram matrix is $B^{T}B$, and its reciprocal basis has basis matrix $B^{-T}$.

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]

Fix the convention: the columns of the change-of-basis matrix $P$ are the new basis vectors written in the old basis. Thus $P$ converts new coordinate vectors into old ones; its matrix inverse converts old coordinate vectors into new ones. For the preceding basis,Its determinant is $-2$, so $P$ is an invertible matrix. Checking which direction $P$ converts is essential: different books may call either $P$ or $P^{-1}$ the change-of-basis matrix.

For a linear operator, convert the input into the old basis, apply the old matrix, and convert the output back. Reading that sequence from right to left givesFor example, let the old diagonal matrix stretch horizontal components by $2$ and vertical components by $3$. ThenThe output represents $2u+2v=(4,0)$, just as $A_{\mathrm{old}}(2,0{)}^T=(4,0{)}^T$. The matrices look different because their bases differ, but they encode the same linear operator. This relation is called matrix similarity; it preserves the underlying action.^[ref]

Apply $R_1$ to a column coordinate vector, then apply $R_2$ in the same fixed basis. The function composition isThus the rightmost matrix acts first. The product is again a rotation matrix, since $(R_2{R}_1{)}^T(R_2{R}_1)=I$ and $\det (R_2{R}_1)=1$. For planar rotations about the same origin, the sine addition formula and cosine addition formula giveFor a concrete composition of rotation matrices, first rotate $(2,0{)}^T$ by $\pi /3$ radians, reaching $(1,\sqrt{3}{)}^T$, then by $\pi /6$ radians:The total angle is $\pi /2$. Reversing these two planar rotations gives the same answer, but noncommuting rotations about fixed axes show why one cannot assume that property for rotations in three dimensions.^[ref]

A coordinate of a vector in a chosen basis is one entry of its coordinate vector. The corresponding coordinate functional in the dual basis extracts it. For an orthonormal basis, it can instead be read as an inner product with the basis vector, as in orthonormal eigenbasis. For a nonorthogonal basis, ordinary dot products need not give the coordinates.

The unique scalar coefficients in the finite linear combination expressing a vector in a chosen basis. Spanning gives existence, while linear independence makes the difference of two proposed representations have all zero coefficients. Ordering a finite basis turns these coefficients into a coordinate vector.

For an ordered basis $\mathcal{B}=(b_1,\dots ,b_n)$, the coordinate vector $[v{]}_{\mathcal{B}}=(a_1,\dots ,a_n{)}^T$ records the unique coefficients satisfying $v={\sum }_i{a}_i{b}_i$. Change of basis changes this representation while preserving the underlying vector.

For $b_1=(1,0{)}^T$, $b_2=(1,1{)}^T$, the vector $v=(3,1{)}^T$ has coordinate vector $(2,1{)}^T$, since $2b_1+b_2=v$. These are the same coefficients extracted by the dual basis in its worked example. The vector is independent of the chosen basis; its list of coordinates is not. In an orthonormal eigenbasis, the coordinates additionally separate the action of a self-adjoint operator into independent scalar multiplications.

A covector is an element of the dual space, hence a linear functional on a vector space. In finite-dimensional coordinates, it acts as a row multiplied by a coordinate vector column. The row $\left(\begin{array}{cc}2& -1\end{array}\right)$ acts on $\left(\begin{array}{c}3\\ 1\end{array}\right)$ to give five. The terminology emphasizes its role as a scalar-valued measurement and its behavior under a change of basis.

For example, the total derivative of $f(x,y)=x^2+y^2$ at $(1,2)$ sends a small displacement $h$ to the linear change $df(h)=2h_x+4h_y$. This covector describes the first-order response. The gradient $\nabla f=(2,4)$ represents the same rule through the chosen Euclidean inner product. In differential geometry, the cotangent space collects these scalar-valued measurements of tangent vectors.

Let the columns of an invertible matrix $B$ be a new basis expressed in the old basis. The same vector has old coordinate vector $x$ and new coordinate vector $x^{\prime }=B^{-1}x$. If a covector has old row $\ell$, its new row is ${\ell }^{\prime }=\ell B$, sinceThus the scalar dual pairing stays unchanged. If covector coefficients are stored as a column $c={\ell }^T$, their new column is $c^{\prime }=B^{T}c$; this is the inverse transpose of the transformation $x\mapsto B^{-1}x$ on vector coordinates. These conventions express the covariance and contravariance of vectors. ^[ref].

For the basis $b_1=(1,0)$, $b_2=(1,1)$,The linear functional $\phi (x,y)=2x-y$ has rows $\ell =\left(\begin{array}{cc}2& -1\end{array}\right)$ and ${\ell }^{\prime }=\left(\begin{array}{cc}2& 1\end{array}\right)$. Both calculations give five: $2\cdot 3-1\cdot 1=2\cdot 2+1\cdot 1$. Changing the basis changes the descriptions, not the measured value.

Under $B=S^{-1}AS$, an eigenvector in old coordinates becomes $S^{-1}v$ in new coordinates. Indeed,Since $S^{-1}$ is an invertible matrix, the new coordinate vector is nonzero. The inverse conversion $S$ proves that the two eigenspaces have the same dimension, hence the same geometric multiplicity.

For $A=\mathrm{diag}(2,5)$, the eigenvector $e_2=(0,1{)}^T$ has eigenvalue $5$. In the oblique basis of the opening example, its coordinate vector isThe tuple changes from $(0,1)$ to $(-1,1)$, but reconstruction gives $-b_1+b_2=e_2$: the physical vertical direction is unchanged. Calling the eigenvectors unchanged is accurate only for the underlying vectors, not for their coordinate vectors.

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.

The left radical is $\{v\in V:B(v,w)=0\text{ for every }w\in V\}$. If $A$ is the matrix of a bilinear form, this condition is $v^{T}A=0$, so its coordinate vectors form $\ker A^T$. For $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, the bilinear form $B(v,w)=v_1{w}_2$ has left radical $\mathrm{span}\{(0,1)\}$.

Let $V$ be a vector space over a specified field $F$, and let $v_1,\dots ,v_n\in V$. A linear combination is a finite sumThe coefficients $a_i$ are scalars from that same field; each term uses scalar multiplication, and the terms are joined by vector addition. For coefficients in the real numbers, negative values reverse arrows and fractional values scale them by fractional amounts. The collection of attainable vectors is the linear span. The column picture in Strang's MIT notes interprets solving a linear system as choosing coefficients to reach a target vector.^[ref]

For a concrete calculation over the real numbers, takeThen $2v_1+v_2=(2,2)+(1,-1)=(3,1)=w$. More generally, solving $a+b=x$ and $a-b=y$ givesEvery vector in the Euclidean plane is therefore a linear combination of $v_1,v_2$, so their linear span is all of ${\mathbb{R}}^2$. The target $(3,1)$ has coefficients $(2,1)$; the target $(0,2)$ has coefficients $(1,-1)$, since $(1,1)-(1,-1)=(0,2)$. The displayed formula also makes the coefficients unique. Consequently $(v_1,v_2)$ is an ordered basis, with coordinate vector $[(x,y){]}_{(v_1,v_2)}=((x+y)/2,(x-y)/2{)}^T$.

The payoff is a reusable language for synthesis: linear combinations build vectors from chosen ingredients, linear span decides which targets are reachable, and linear independence decides whether the coefficients are unique. The ingredients can also be polynomials or functions, as long as they belong to a vector space over the specified field.

A finite family of vectors in a vector space over a specified field $F$ is linearly dependent if it has a nontrivial linear relation: ${\sum }_i{a}_i{v}_i=0$ with $a_i\in F$ and at least one coefficient nonzero. One can then solve for a vector whose coefficient is nonzero as a linear combination of the others; removing that entry preserves the linear span. Repeated vectors in a list give linear dependence: for example $(v,v)$ has the linear relation $1v+(-1)v=0$. A list keeps both entries even though the set $\{v,v\}$ has only one element. Strang's notes illustrate redundant columns by expressing one as the sum of two others.^[ref]

Keep $v_1=(1,1)$ and $v_2=(1,-1)$, and add $v_3=(2,0)=v_1+v_2$. The three vectors still have linear span ${\mathbb{R}}^2$, but nowThe two coefficient triples $(2,1,0)$ and $(1,0,1)$ differ by the nontrivial linear relation $(-1,-1,1)$. In fact all linear combination representations of this target areTo check both completeness and correctness, use $c_1{v}_1+c_2{v}_2+c_3{v}_3=(c_1+c_3)v_1+(c_2+c_3)v_2$. Linear independence of $v_1,v_2$ forces $c_1+c_3=2$ and $c_2+c_3=1$, giving exactly the displayed family. Redundancy therefore preserves attainable targets while allowing their coefficients to move. For solving a linear system, these alternatives matter; for assigning unique coordinate vectors, one instead chooses a basis.

The following example shows why linear independence depends on the scalar field.

Choose an ordered basis $(e_1,\dots ,e_n)$ of a finite-dimensional vector space. The matrix of a bilinear form records all pairings of basis vectors, with $A_{ij}=B(e_i,e_j)$. Writing $v={\sum }_i{v}_i{e}_i$ and $w={\sum }_j{w}_j{e}_j$, separate linearity giveswhere the last expression uses the coordinate vectors of $v,w$. Conversely, every matrix $A$ defines a bilinear form by this formula.^[ref]

For a concrete bilinear form on ${\mathbb{R}}^2$, takeThen $B((1,0),(0,1))=2$ but $B((0,1),(1,0))=0$: a bilinear form need not be a symmetric bilinear form. For $v=(2,1{)}^T$ and $w=(-1,3{)}^T$, first compute $Aw=(5,3{)}^T$, then $v^{T}Aw=2\cdot 5+1\cdot 3=13$. Reversing the inputs gives $Av=(4,1{)}^T$ and $w^{T}Av=-4+3=-1$.

Under change of basis, let the columns of an invertible matrix $P$ be the new basis vectors in old coordinates. Thus $v=P{v}^{\prime }$ and $w=P{w}^{\prime }$, soThis is matrix congruence, rather than the matrix similarity $P^{-1}AP$ for an endomorphism. Both inputs have changed coordinates, which explains the two factors of $P$. For the displayed $A$ and $P=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, direct multiplication gives $A^{\prime }=\left(\begin{array}{cc}1& 3\\ 1& 4\end{array}\right)$; its $(1,2)$ entry is $B(e_1,e_1+e_2)=3$.^[ref]