Basis (linear algebra) · aiwiki.page

In linear algebra, a basis of a vector space is a set of vectors that both spans the space and is linearly independent. Every vector in the space can therefore be expressed uniquely as a finite linear combination of basis vectors. A basis provides coordinates for abstract vectors, connecting vector spaces with calculations using numbers and matrices. A space generally has many different bases, but their common size defines its dimension. (ocw.mit.edu)

Definition and uniqueness

Let $V$ be a vector space over a field $F$ . A subset $B\subseteq V$ is a basis if it satisfies two conditions:

Spanning: its linear span is $V$ , meaning every vector of $V$ is a finite linear combination of elements of $B$ .
Independence: it has linear independence, meaning a finite linear combination of distinct elements of $B$ equals zero only when every coefficient is zero.

For a finite basis $B=\{b_1,\ldots,b_n\}$ , these conditions imply that each $v\in V$ has exactly one expression

v=a_1b_1+\cdots+a_nb_n,\qquad a_i\in F.

Spanning guarantees existence; independence guarantees uniqueness, since subtracting two representations produces a combination equal to zero. (ocw.mit.edu)

Equivalently, a basis is a minimal spanning set or a maximal linearly independent set, where minimality and maximality refer to inclusion. No basis contains the zero vector. The zero vector space has the empty set as its basis. (math.ucdavis.edu)

Examples and the scalar field

The standard basis of $F^n$ consists of the vectors $e_1,\ldots,e_n$ , where $e_i$ has a $1$ in position $i$ and zeros elsewhere. In $\mathbb R^2$ , the vectors $(1,1)$ and $(1,-1)$ also form a basis, because

(x,y)=\frac{x+y}{2}(1,1)+\frac{x-y}{2}(1,-1).

Basis vectors need not have unit length or be perpendicular. (web.mit.edu)

Vectors need not be geometric arrows. The space of polynomials of degree at most $m$ , including the zero polynomial, has basis

1,x,x^2,\ldots,x^m.

The space of $r\times s$ matrices has a basis consisting of the $rs$ matrices with exactly one nonzero entry, equal to $1$ . (math.ucdavis.edu)

The scalar field matters. The complex numbers form a one-dimensional vector space over themselves, with basis $\{1\}$ , but a two-dimensional vector space over the real numbers, with basis $\{1,i\}$ . Thus “a basis of a space” presupposes its scalar field. (math.ucdavis.edu)

Dimension and constructing bases

All bases of a finite-dimensional space contain the same number of vectors. This number is its dimension. Consequently, in an $n$ -dimensional space, any $n$ independent vectors form a basis, and any spanning set containing exactly $n$ vectors is a basis. (ocw.mit.edu)

An independent set can be extended to a basis by adding suitable vectors; a finite spanning set can be reduced to a basis by removing redundant vectors. A basis of a linear subspace can likewise be extended to a basis of the containing finite-dimensional space. (math.ucdavis.edu)

For computational purposes, place candidate vectors in the columns of a matrix and apply Gaussian elimination. The pivot-column indices identify a basis for the column space, using the corresponding columns of the original matrix. Their number is the matrix rank. For $n$ vectors in $F^n$ , the resulting square matrix gives a basis exactly when it is invertible, equivalently when its determinant is nonzero. (ocw.mit.edu)

Coordinates and change of basis

An ordered basis specifies the order of its vectors. Relative to $B=(b_1,\ldots,b_n)$ , the coefficient column

[v]_B=(a_1,\ldots,a_n)^{\mathsf T}

is the coordinate vector of $v$ . The mapping $v\mapsto[v]_B$ is a bijective linear map from $V$ to $F^n$ . Coordinates depend on the chosen basis, whereas the vector itself does not. (web.mit.edu)

For another ordered basis $C=(c_1,\ldots,c_n)$ , let $P$ have columns $[c_1]_B,\ldots,[c_n]_B$ . Then

[v]_B=P[v]_C,\qquad [v]_C=P^{-1}[v]_B.

The inverse matrix exists because both lists are bases. If a linear operator has matrix $A_B$ in basis $B$ , its matrix in basis $C$ is

A_C=P^{-1}A_BP.

When a basis of eigenvectors exists, this change of basis produces diagonalization. (web.mit.edu)

Orthonormal and infinite-dimensional bases

In a finite-dimensional space equipped with an inner product, an orthonormal basis consists of mutually orthogonal unit vectors. The Gram–Schmidt process converts any basis into an orthonormal one. With the convention that the inner product is linear in its first argument, coordinates simplify to

v=\sum_{j=1}^{n}\langle v,e_j\rangle e_j.

These coefficients also describe orthogonal projections onto the individual basis directions. (math.ucdavis.edu)

An algebraic basis in an infinite-dimensional space is often called a Hamel basis. Every vector still uses only finitely many basis elements. For example, $1,x,x^2,\ldots$ is a Hamel basis of the space of all polynomials. Assuming the axiom of choice, every vector space has an algebraic basis, although this existence statement need not provide an explicit construction. (math.ucdavis.edu)

In an infinite-dimensional Hilbert space, “orthonormal basis” usually means an orthonormal set whose finite linear combinations are dense. Vectors may require convergent infinite expansions, rather than finite sums. Such a basis is therefore not generally a Hamel basis: it incorporates the space’s topology and convergence, not just its algebraic operations. (www2.math.upenn.edu)