[ALGEBRA] Spans

The span of a set is everything its vectors produce by linear combination. This page shows that the span is the smallest subspace containing the set and turns membership of the span into a linear system.

Combining chosen vectors

[DEFINITION]

Subspace

Let $V$ be a vector space over $F$ . A subset $S\subseteq V$ is a subspace of $V$ when it is a vector space over the same field $F$ and uses the addition and scalar multiplication inherited from $V$ . A proper subspace is a subspace that is not equal to $V$ .

[DEFINITION]

Linear combination

Let $v_1,\ldots,v_n$ be vectors in a vector space $V$ over $F$ . A linear combination of these vectors is an expression

\lambda_1v_1+\cdots+\lambda_nv_n, \qquad \lambda_1,\ldots,\lambda_n\in F.

The vectors are fixed while the scalars vary. For non-zero, non-parallel $u,v\in\mathbb R^3$ , the vectors

\lambda_1u+\lambda_2v,\qquad \lambda_1,\lambda_2\in\mathbb R,

form the plane through the origin determined by $u$ and $v$ , with each coefficient moving along one direction.

[THEOREM]

Closure under linear combinations

Every linear combination of a finite set of vectors in a vector space $V$ is an element of $V$ .

[PROOF]

For one vector, $\lambda_1v_1\in V$ by scalar closure. Suppose every combination of $k$ vectors lies in $V$ . If

w=\lambda_1v_1+\cdots+\lambda_kv_k\in V,

then scalar closure gives $\lambda_{k+1}v_{k+1}\in V$ and addition closure gives

w+\lambda_{k+1}v_{k+1}\in V.

Induction proves the claim for every finite number of vectors.

[1]:Closure under linear combinations

The smallest subspace containing the chosen vectors

[DEFINITION]

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 $0$ .

The span is the subspace generated by the chosen vectors. The empty-sum convention gives $\operatorname{span}(\varnothing)=\{0\}$ , which is needed for the zero vector space.

[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
18 more

[THEOREM]

The span is a subspace

If $S$ is a subset of a vector space $V$ , then $\operatorname{span}(S)$ is a subspace of $V$ . It is the smallest subspace of $V$ that contains $S$ .

[1]:Span
[2]:Subspace
[3]:Proof

[PROOF]

The empty combination gives $0\in\operatorname{span}(S)$ . If $u$ and $w$ are finite linear combinations of vectors in $S$ , concatenating their lists of terms expresses $u+w$ as another finite linear combination. For $\gamma\in F$ , multiplying every coefficient in a finite combination for $u$ by $\gamma$ expresses $\gamma u$ in the span. The subspace test therefore applies.

If $W$ is any subspace containing $S$ , then $W$ is closed under scalar multiplication and addition. It must therefore contain every linear combination of the vectors in $S$ , so $\operatorname{span}(S)\subseteq W$ . Hence the span is the smallest such subspace.

[1]:The span is a subspace

[EXAMPLE]

Lines and planes through the origin

For $v=(1,0,0)\in\mathbb R^3$ ,

\operatorname{span}\{v\} = \{(\lambda,0,0):\lambda\in\mathbb R\},

which is the line $y=z=0$ . The coefficient can be any real number, so the whole line is obtained; restricting it to an interval would give a bounded segment, which is not closed under scalar multiplication. Any plane through the origin containing this line is a larger subspace containing $v$ .

[routine]

[1]:Span

Spanning familiar spaces

The examples from here on are over $\mathbb R$ , so $P_n$ means $P_n(\mathbb R)$ and matrix entries are real.

[DEFINITION]

Spanning set

A subset $S$ in a vector space $V$ is a spanning set for $V$ when

\operatorname{span}(S)=V.

A spanning set may contain redundant vectors.

[1]:Span
[2]:span, independence and possible bases
[3]:Seen Questions
[4]:Standard spanning sets
[5]:Three vectors spanning $\mathbb R^3$
1 more

[EXAMPLE]

Standard spanning sets

The set

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

spans $\mathbb R^2$ , since every $(a,b)^T$ is

a\begin{pmatrix}1\\0\end{pmatrix} +b\begin{pmatrix}0\\1\end{pmatrix}.

The set

\{1,x,\ldots,x^n\}

spans $P_n$ , since each polynomial of degree at most $n$ is a linear combination of these powers. The four matrices

E_{11}=\begin{pmatrix}1&0\\0&0\end{pmatrix},\quad E_{12}=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad E_{21}=\begin{pmatrix}0&0\\1&0\end{pmatrix},\quad E_{22}=\begin{pmatrix}0&0\\0&1\end{pmatrix}

span $M_{2\times2}(\mathbb R)$ , because

\begin{pmatrix}a&b\\c&d\end{pmatrix} =aE_{11}+bE_{12}+cE_{21}+dE_{22}.

[routine]

[1]:Spanning set

[DEFINITION]

Column space

If $S=\{v_1,\ldots,v_n\}$ is a set of vectors in $\mathbb R^m$ and $A=[\,v_1\ \cdots\ v_n\,]$ , then the span of $S$ is the column space of $A$ :

\operatorname{col}(A)=\operatorname{span}(S) =\{Ac:c\in\mathbb R^n\},

where $c$ records the coefficients of the columns. Thus multiplying $A$ by $c$ is exactly the linear combination represented by those coefficients.

Membership as a linear system

Because $Ac$ is a linear combination of the columns of $A$ , asking whether a vector lies in a span is asking whether a linear system has a solution.

[METHOD]

Test membership in a span

To decide whether $u\in\mathbb R^m$ lies in $\operatorname{span}\{v_1,\ldots,v_n\}$ , form $A=[\,v_1\ \cdots\ v_n\,]$ and solve

Ac=u.

A solution for the scalar coefficients exists if and only if $u$ belongs to the span, equivalently to $\operatorname{col}(A)$ .

Use row reduction on the augmented matrix $[A\mid u]$ . A contradictory row means that $u$ is not in the span.

[1]:Span
[2]:Column space
[3]:A vector outside a span
[4]:Three vectors spanning $\mathbb R^3$
[5]:Membership in a polynomial span
[6]:A set that does not span $P_2$

[EXAMPLE]

A vector outside a span

Let

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

To test whether $(1,2,3)^T$ belongs to $\operatorname{span}(S)$ , row-reduce

\left[ \begin{array}{cc|c} 1&2&1\\ 0&1&2\\ 2&-1&3 \end{array} \right] \sim \left[ \begin{array}{cc|c} 1&0&-3\\ 0&1&2\\ 0&0&11 \end{array} \right].

The last row is contradictory, so the system has no solution and

\begin{pmatrix}1\\2\\3\end{pmatrix}\notin\operatorname{span}(S).

[routine]

[1]:Test membership in a span

[EXAMPLE]

Three vectors spanning $\mathbb R^3$

For

A= \begin{bmatrix} 1&1&2\\ 2&0&1\\ 3&2&-1 \end{bmatrix},

row reduction gives

A\sim \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}.

Thus $A c=b$ has a solution for every $b\in\mathbb R^3$ . The columns

\begin{pmatrix}1\\2\\3\end{pmatrix},\qquad \begin{pmatrix}1\\0\\2\end{pmatrix},\qquad \begin{pmatrix}2\\1\\-1\end{pmatrix}

span $\mathbb R^3$ .

[routine]

[1]:Test membership in a span
[2]:Spanning set

[EXAMPLE]

Membership in a polynomial span

Let

q_1=x^3+2x^2,\qquad q_2=x-2,

and write

p=a_0+a_1x+a_2x^2+a_3x^3.

Then

\alpha q_1+\beta q_2 =-2\beta+\beta x+2\alpha x^2+\alpha x^3.

The constant and linear coefficients require $a_0=-2\beta$ and $a_1=\beta$ ; the quadratic and cubic coefficients require $a_2=2\alpha$ and $a_3=\alpha$ . Thus $p\in\operatorname{span}\{q_1,q_2\}$ exactly when

a_0=-2a_1 \qquad\text{and}\qquad a_2=2a_3.

[routine]

[1]:Test membership in a span

[EXAMPLE]

A set that does not span $P_2$

Consider

q_1=x+2x^2,\qquad q_2=x-2,\qquad q_3=x^2+1.

For

p=a_0+a_1x+a_2x^2,

the equation

\alpha q_1+\beta q_2+\gamma q_3=p

gives

-2\beta+\gamma=a_0,\qquad \alpha+\beta=a_1,\qquad 2\alpha+\gamma=a_2.

The first two equations give $\gamma=a_0+2\beta$ and $\alpha=a_1-\beta$ . Substituting into the third gives the condition

a_2=a_0+2a_1.

Conversely, when this condition holds, those choices of $\alpha$ and $\gamma$ solve all three equations for any $\beta$ . The polynomial $p=x^2$ fails the condition, so these three polynomials do not span $P_2$ .

[routine]

[1]:Test membership in a span
[2]:Spanning set

Linear independence asks when the coefficients in a span are unique.

Spans

Subspace

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