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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]

A circulant matrix applies the same rule at every position of a finite ring: moving both the input and output indices one step leaves its entries unchanged. Its first column determines the whole matrix. For a length-$N$ vector $c$, index rows and columns by $0,\dots ,N-1$ and defineThe indices use modular arithmetic. Thus each column is the previous column shifted down one place, with the last entry wrapping around to the top. Multiplication is exactly circular convolution:This makes a circulant matrix a natural model for a periodic finite difference rule or a filter acting on a periodic record. Strang's MIT lecture develops this connection between circulant matrices, circular convolution and the discrete Fourier transform.^[ref]

For example, the first column $c=(2,-1,0,-1{)}^T$ givesHere the indices wrap around modulo four. Each output compares a value with its two neighbors; this is the Graph Laplacian of a four-vertex cycle graph. A constant input gives zero, while an alternating input gives a large response. The Fourier diagonalization of a circulant matrix makes this distinction exact: its four eigenvalues are $0,2,4,2$.

The plotted discrete Fourier modes are complex vectors, shown through their real and imaginary parts. Each entry has modulus one; the Euclidean norm of each length-four vector is two. In particular, modes $k=1$ and $k=3$ have the same real part but opposite imaginary parts, so they are distinct eigenvectors with the same eigenvalue.

The useful surprise is that every length-$N$ circulant matrix shares the same basis of eigenvectors, whatever its first column. One change of basis therefore simplifies an entire family of linear maps, instead of requiring a separate eigenvalue computation for each one. Bamieh's tutorial develops the discrete Fourier transform from precisely this simultaneous diagonalization problem.^[ref]

Closed paths at vertex $i$ form a basis of $e_{i}kQ{e}_i$. For nonzero finite linear combinations $f,g$, take their largest path lengths $a,b$. In the length-$a+b$ part of $fg$, a concatenated path has a unique cut into lengths $a,b$. Thus a product of nonzero top-degree coefficients cannot cancel, and $fg\ne 0$. The corner is a noncommutative domain and has no idempotents except zero and its identity $e_i$. This argument works with oriented cycles and proves indecomposability of the vertex projective module of a path algebra without falsely assuming that its endomorphism ring is local.

A complete flag in an $n$-dimensional vector space is a nested sequence of vector subspaces with $\dim V_i=i$. An action preserves such a flag exactly when its matrices are upper triangular in a basis adapted to the flag. The Lie theorem supplies an invariant complete flag for every complex finite-dimensional Lie algebra representation of a solvable Lie algebra.

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]

For a local basis $e_a$ of the tangent bundle, an affine connection has components defined by ${\nabla }_{e_c}{e}_b={\Gamma }^a{}_{bc}{e}_a$. If $V=V^b{e}_b$, then $({\nabla }_{e_c}V{)}^a=e_c(V^a)+{\Gamma }^a{}_{bc}{V}^b$. The connection components depend on the chosen basis and have an inhomogeneous transformation law; they are not themselves a tensor field. In a coordinate basis, a torsion-free connection has symmetry in the two lower indices.

Local coordinates $x^{\mu }$ induce the coordinate-basis tangent vectors ${\partial }_{\mu }=\partial /\partial x^{\mu }$. Under a coordinate change they transform by the chain rule.

A coordinate functional extracts one coefficient in a specified finite basis. It is an element of that basis's dual basis, rather than a coordinate rule independent of the basis. For the standard basis of $F^n$, the $i$th coordinate functional sends $(a_1,\dots ,a_n)$ to $a_i$.

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.

A vector $v$ can be included in a diagonal basis for a real quadratic form precisely when $v\ne 0$ and either its quadratic value is nonzero or it lies in the radical of a bilinear form. A nonzero quadratic value gives the direct splitting $\mathbb{R}v\oplus v^{\perp }$ and induction. A nonzero radical vector can be included in a radical basis. Conversely, an isotropic vector in a diagonal basis is orthogonal to every basis vector, and must lie in the radical.