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For a basis $e_1,\dots ,e_n$ with cyclic indexing, the vectors $e_j+e_{j+1}$ have coefficient matrix $I+S$, where $S$ cyclically shifts coordinates. Its determinant is $1-(-1{)}^n$. They therefore form a basis exactly when $n$ is odd over a field of characteristic different from two. For even $n$, alternating coefficients produce a nontrivial linear dependence.

The two-dimensional Lie algebra of infinitesimal dilations and translations of a line has a basis $h,e$ with $[h,e]=e$. Its derived algebra is the abelian line spanned by $e$, but every nontrivial term of its lower central series of a Lie algebra is that same line. It is solvable and not nilpotent.

An algebraic number $\alpha$ is an algebraic integer if it is a polynomial root of a monic polynomial with integer coefficients:Equivalently, its algebraic minimal polynomial over $\mathbb{Q}$ has integer coefficients. The requirement that the leading coefficient be one matters: $1/2$ satisfies $2X-1=0$ and is an algebraic number, but is not an algebraic integer. In fact the rational numbers that are algebraic integers are exactly the integers.^[ref]

Some useful computed examples are:

Algebraic integers are closed under addition, subtraction, and multiplication. Thus they provide an arithmetic ring, while division is allowed only when the result is again an algebraic integer. The algebraic integer module criterion explains this closure without having to find a new polynomial for every sum and product.

The relation $[x,y]=-[y,x]$ for a bilinear Lie bracket. In characteristic zero it implies $[x,x]=0$. In characteristic two, the alternating identity must be required separately when defining a Lie algebra.

A basis is a linearly independent spanning family of vectors in a vector space.

Every linearly independent subset of a finite-dimensional vector space extends to a basis. Starting with its vectors, add a vector outside their current span until the span is the whole space. Each addition preserves linear independence, and the process stops because the dimension is finite. This also lets a linear functional on a vector subspace extend to the ambient space by assigning arbitrary values on the additional basis vectors.

A basis of a topology is a family of open sets such that every open set is a union of members of that family. Equivalently, every point of every open set lies in a basis member contained in it. The rectangles $U\times V$ form a basis for the product topology of two spaces. This topological meaning differs from a linear-algebra basis.

Give both position and differentiation generators of a Weyl algebra degree one. The ordered monomial basis gives $\dim F_j{A}_n=\left(\genfrac{}{}{0ex}{}{j+2n}{2n}\right)$ and ${\mathrm{gr}}_F{A}_n$ equal to a polynomial ring in $2n$ variables. Commuting a filtered element with any generator lowers its degree by at least one.

For an algebraic dual space $X^{\ast }={\mathrm{Hom}}_F(X,F)$, the linear map $J:X\to X^{\ast \ast }$ sends $x$ to evaluation at $x$. It is injective because linear functionals separate nonzero vectors: extend any nonzero vector to a basis and use its coordinate functional. If $X$ has a finite basis $e_i$ with dual basis $e_i^{\ast }$, every $B\in X^{\ast \ast }$ equals $J({\sum }_{i}B(e_i^{\ast })e_i)$, so $J$ is an isomorphism. For infinite-dimensional spaces, surjection does not follow from this finite sum argument.

Replace each positive power $x_i^d$ by $x_i$ in a polynomial. This gives a multilinear polynomial of no larger polynomial degree with the same values on $\{0,1{\}}^n$. In the polynomial method in combinatorics, these functions lie in a vector space with a basis consisting of the square-free monomials of the required polynomial degrees.

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]