Basis and dimension

Span

For a subset $S$ (possibly infinite) in a vector space $V$ over $F$ , the span of $S$ is the set of all finite linear combinations of vectors in $S$ :

\operatorname{span}(S) = \left\{ \lambda_1v_1+\cdots+\lambda_kv_k: k\geq 0,\ v_1,\ldots,v_k\in S,\ \lambda_1,\ldots,\lambda_k\in F \right\}.

For a finite set $S=\{v_1,\ldots,v_n\}$ , this is the same as allowing one coefficient for each listed vector. When $k=0$ , the sum is the empty sum $0$ .

[1]:Linear combination
[2]:Basis
[3]:A basis for a generated subspace
[4]:A basis of $P_1$
[5]:Spanning sets count against independent sets
17 more

Linear independence

Let $S$ be a subset of a vector space over $F$ . The set $S$ is linearly independent if every finite list of distinct vectors $u_1,\ldots,u_n\in S$ has the property that, for scalars $\lambda_1,\ldots,\lambda_n\in F$ ,

\lambda_1u_1+\cdots+\lambda_nu_n=0

implies

\lambda_1=\cdots=\lambda_n=0.

[1]:Linear combination
[2]:Span
[3]:Eigenvectors for distinct eigenvalues
[4]:When diagonalisation is possible
[5]:Column independence and nullity
[6]:Basis
14 more

Basis

A set $B$ in a vector space $V$ is a basis for $V$ if it is linearly independent and spans $V$ .

Spanning gives a representation for every vector in $V$ and independence makes it unique, as shown in linear independence. A basis is therefore a generating set with no redundant direction: removing any basis vector destroys spanning.

For an infinite set, span still uses finite linear combinations and independence means that every finite subset is independent. The calculations below use finite bases.

[1]:Span
[2]:Linear independence
[3]:Eigenbasis
[4]:Diagonalisation
[5]:A system with zero initial derivative
[6]:Values on a basis determine a linear map
23 more

Representation in a basis

If $B=\{u_1,\ldots,u_n\}$ is a basis for $V$ , then every vector $v\in V$ has a unique representation

v=\lambda_1u_1+\cdots+\lambda_nu_n, \qquad \lambda_1,\ldots,\lambda_n\in F.

Because $B$ spans $V$ , at least one representation of $v$ exists. If

v=\lambda_1u_1+\cdots+\lambda_nu_n =\mu_1u_1+\cdots+\mu_nu_n,

then subtracting the two expressions gives

(\lambda_1-\mu_1)u_1+\cdots+(\lambda_n-\mu_n)u_n=0.

Independence of $B$ forces every $\lambda_j-\mu_j$ to be zero, so the representation is unique.

[1]:Representation in a basis

A basis of $\mathbb R^3$

Consider

B=\left\{ \begin{pmatrix}1\\0\\2\end{pmatrix}, \begin{pmatrix}0\\-1\\1\end{pmatrix}, \begin{pmatrix}-2\\1\\0\end{pmatrix} \right\}.

The matrix with these vectors as columns row reduces to $I_3$ , so its columns are linearly independent and span $\mathbb R^3$ . Thus $B$ is a basis for $\mathbb R^3$ .

[routine]

[1]:Basis
[2]:The homogeneous-system test

A basis for a generated subspace

Let

S=\left\{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}-2\\1\end{pmatrix}, \begin{pmatrix}1\\1\end{pmatrix}, \begin{pmatrix}-1\\-2\end{pmatrix} \right\}.

The first two columns are leading columns when the matrix formed from $S$ is row reduced. They are independent and the remaining vectors are redundant, so

\left\{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}-2\\1\end{pmatrix} \right\}

is a basis for $\operatorname{span}(S)$ .

[routine]

A basis of $P_1$

For $p=a_0+a_1x$ , write

p=\lambda_1(1+x)+\lambda_2(x+2).

Equating coefficients gives

\lambda_1+2\lambda_2=a_0, \qquad \lambda_1+\lambda_2=a_1.

The coefficient matrix has determinant $-1$ , so this system has a unique solution for every $a_0,a_1$ . Therefore $\{1+x,x+2\}$ spans $P_1$ and is linearly independent, so it is a basis.

[routine]

Why dimension is well defined

A spanning set can never be smaller than an independent set in the same space. This single comparison shows that all bases have the same size.

Spanning sets count against independent sets

Let $S$ and $L$ be finite subsets of a vector space $V$ . If $\operatorname{span}(S)=V$ and $L$ is linearly independent, then

|S|\geq |L|.

Because $S$ spans $V$ , each vector of $L$ is a linear combination $\ell_j=\sum_i a_{ij}s_i$ of the vectors in $S$ . Suppose $|L|>|S|$ . The coefficient matrix $A=(a_{ij})$ then has more columns than rows, so $Ac=0$ has a non-zero solution $c=(c_\ell)_{\ell\in L}$ . This relation lifts to $\sum_j c_j\ell_j=\sum_i(\sum_j a_{ij}c_j)s_i=0$ , that is,

\sum_{\ell\in L}c_\ell\ell=0,

which contradicts the linear independence of $L$ . Hence $|S|\geq|L|$ .

[1]:Spanning sets count against independent sets

All bases have the same size

Any two finite bases of a vector space contain the same number of vectors.

[1]:Basis
[2]:Spanning sets count against independent sets
[3]:Proof
[4]:Dimension

If $B_1$ and $B_2$ are bases, the counting theorem gives $|B_1|\geq|B_2|$ by taking $B_1$ as the spanning set and $B_2$ as the independent set. Reversing their roles gives $|B_2|\geq|B_1|$ , so the cardinalities are equal.

[1]:All bases have the same size

This result covers every basis of a space with a finite basis $B_0$ , because no basis of such a space can be infinite. An infinite basis would contain a finite subset $L$ with more than $|B_0|$ vectors. That subset is independent, while $B_0$ spans $V$ , which contradicts the counting theorem.

Dimension

If a vector space $V$ has a finite basis, then $V$ is finite-dimensional and its dimension, denoted $\dim(V)$ , is the common number of vectors in all bases of $V$ .

The zero vector space has the empty basis, so $\dim(\{0\})=0$ . If $V$ has no finite basis, it is called infinite-dimensional. For example, $P(\mathbb R)$ is infinite-dimensional: no finite list of polynomials can span polynomials of arbitrarily high degree.

[1]:Basis
[2]:All bases have the same size
[3]:A system with zero initial derivative
[4]:Nullity
[5]:Rank
[6]:Rank-nullity
8 more

Using dimension to recognise a basis

Once the dimension is known, a set with the right number of vectors needs only one of the two basis checks.

Counting with dimension

Let $V$ be finite-dimensional.

If a set $S$ of $n$ vectors spans $V$ , then $n\geq\dim(V)$ . If $n=\dim(V)$ , then $S$ is a basis.
If a set $S$ of $m$ vectors is linearly independent in $V$ , then $m\leq\dim(V)$ . If $m=\dim(V)$ , then $S$ is a basis.

[1]:Basis
[2]:Dimension
[3]:Span
[4]:Linear independence
[5]:Proof
[6]:Two properties and the dimension are enough
[7]:Counting prevents spanning
[8]:An orthonormal basis of $\mathbb R^3$
7 more

Compare $S$ with a basis $B$ of $V$ in the preceding counting theorem. When $S$ spans, taking $S$ as the spanning set and $B$ as the independent set gives $|S|\ge|B|=\dim(V)$ . If equality holds but $S$ is dependent, one vector can be removed without changing its span. That would give a spanning set with fewer than $\dim(V)$ vectors, a contradiction.

When $S$ is independent, take $B$ as the spanning set and $S$ as the independent set to get $|S|\le\dim(V)$ . If equality holds but $S$ does not span, adjoining a vector outside its span gives an independent set with more than $\dim(V)$ vectors, again a contradiction.

[1]:Counting with dimension

[COROLLARY]

Two properties and the dimension are enough

For a set $S$ of $n$ vectors in a finite-dimensional vector space $V$ , any two of the following imply that $S$ is a basis for $V$ :

$S$ is linearly independent;
$S$ spans $V$ ;
$n=\dim(V)$ .

[1]:Basis
[2]:Counting with dimension

Standard bases and dimensions

The standard basis of $\mathbb R^2$ is

\left\{ \begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix} \right\}.

The standard polynomial basis is $\{1,x,\ldots,x^n\}$ for $P_n(\mathbb R)$ . For $M_{2\times2}(\mathbb R)$ , the standard matrix basis is

\left\{ \begin{pmatrix}1&0\\0&0\end{pmatrix}, \begin{pmatrix}0&1\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\1&0\end{pmatrix}, \begin{pmatrix}0&0\\0&1\end{pmatrix} \right\}.

More generally, the standard bases give

\dim(\mathbb R^n)=n, \qquad \dim(P_n(\mathbb R))=n+1, \qquad \dim(M_{m\times n}(\mathbb R))=mn.

Each count is the number of free coefficients: a polynomial in $P_n(\mathbb R)$ has $n+1$ coefficients and an $m\times n$ matrix has one for each of its $mn$ entries.

[routine]

[1]:Dimension