Span


1. Basics of Span

A general way of creating subspaces of a vector space is in terms of linear combinations.

Let $V$ be a vector space and $S = \left\{ \ {\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k \ \right\} \subseteq V$ be a finite subset of $V.$ By a Linear Combination of the vectors in $S$ we mean any vector ${\bf v} \in V$ of the form $$ {\bf v} = \alpha_1 {\bf v}_1 + \alpha_2 {\bf v}_2 + \cdots + \alpha_k {\bf v}_k $$ where $\alpha_1 , \alpha_2 , \dots , \alpha_k \in \mathbb{R}.$

The set of all linear combinations of vectors in $S$ is called the Span of $S:$ $$\textrm{Span}( \ S \ ) = \textrm{Span}\left\{\,{\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k\right\} = \left\{\, c_1 {\bf v}_1 + c_2 {\bf v_2} + \ldots + c_k {\bf v}_k\,\middle\vert \, c_1 , \dots , c_k \ \in \ {\mathbb R}\,\right\}.$$

When $$\textrm{Span}( \ S \ ) = \textrm{Span}\big\{\,{\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k\big\} \ = \ V$$ we say that $S$ Spans $V,$ or that $S$ is a Spanning Set for $V.$


Theorem 1: The $\,\textrm{Span}\big\{\,{\bf v}_1,\, {\bf v}_2,\,\ldots, \, {\bf v}_k\big\}$ of vectors ${\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k$ in a vector space $V$ is a subspace of $V,$ hence a vector space.

Proof: We check that $$\,\textrm{Span}\big\{\,{\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k\big\}$$ has the three properties of a subspace. For ease of notation, suppose $k = 2$

Property Z: Take $c_1 = c_2 = 0$. Then $\textrm{Span}\{{\bf v}_1,\, {\bf v}_2\}$ contains $ 0 {\bf v}_1 + 0 {\bf v}_2 = {\bf 0}_V$.

Property AC: Fix arbitrary vectors in $\textrm{Span}\{{\bf v}_1,\, {\bf v}_2\}$, say $${\bf u} \,=\, c_1 {\bf v}_1 + c_2 {\bf v}_2, \quad {\bf v} \,=\, d_1 {\bf v}_1 + d_2 {\bf v}_2,$$

then $${\bf u} + {\bf v} \,=\, (c_1+d_1){\bf v}_1 + (c_2+d_2){\bf v}_2\,,$$ which shows that ${\bf u}+{\bf v}$ belongs to $\textrm{Span}\{{\bf v}_1,\, {\bf v}_2\}$.

Property SC: Now fix an arbitrary scalar $k$. Then $$k {\bf u} \,=\, k(c_1 {\bf v}_1 + c_2 {\bf v}_2) \,=\, (k c_1){\bf v}_1 + (k c_2){\bf v}_2\,,$$ which shows that $k {\bf u}$ belongs to $\textrm{Span}\{{\bf v}_1,\, {\bf v}_2\}$.

Thus $\textrm{Span}\{{\bf v}_1,\, {\bf v}_2\}$ is a subspace of $V$.

    For a single nonzero vector, ${\bf v},$ $\textrm{Span}({\bf v}) = \{ t \ {\bf v} \, | \, -\infty \, < t \, < \,\infty\,\}$ consists of all scalar multiples of ${\bf v}$. Geometrically, this is a Line in ${\mathbb R}^n$ passing through the origin at $t = 0$. By Fundamental Theorem 1, this line is a subspace of ${\mathbb R}^n$. Can you see why the line $\{ {\bf u} + t {\bf v} \ | \, -\infty < t < \infty \, \}$ through ${\bf u}$ in the direction of ${\bf v}$ is not a subspace? Similarly, the span of two vectors is usually a plane through the origin, hence a subspace of ${\mathbb R}^3$, but a plane not passing through the origin is not a subspace of ${\mathbb R}^3$.

Another way to describe $\textrm{Span}( \ S \ )$ is to say it is the smallest subspace of $V$ containing the set $S$.

Theorem 2: If $W$ is a subspace of a vector space $V,$ and $S = \left\{ \ {\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k \ \right\}$ is a subset of $W,$ then $\textrm{Span}( \ S \ )$ is a subspace of $W.$ $$ S \subseteq W \subseteq V \quad \Rightarrow \quad \textrm{Span}(S) \subseteq W \subseteq V.$$

We use this fact frequently in solving differential equations. For example, let $L[ \ y \ ] = y^{\, \prime \prime} + y = 0$. We know that $V_L$ is a subspace of $C(\mathbb{R})$, and we know that $S = \left\{ \cos t , \sin t \right\} \subseteq V_L$ since both $\cos t$ and $\sin t$ are solutions to the differential equation. Thus, by Theorem 2, we know that $\textrm{Span}( \ S \ )$ is a subspace of $V_L:$ $$\textrm{Span}( \ S \ ) = \left\{ \ c_1 \cos t + c_2 \sin t \ \middle\vert \ c_1 , c_2 \in \mathbb{R} \ \right\} \subseteq V_L.$$ The fact that they are equal, $\textrm{Span}( \ S \ ) = V_L,$ comes from a theorem in differential equations. See Differential Equations and Their Applications by Martin Braun, section 2.1, Theorem 2.

2. Examples

When studying subspaces, we saw that every vector space $V$ is a subspace of itself, so every vector space is a subspace. Similarly, every vector space is a span.

Theorem 3: Every finite dimensional vector space, $V \cong \mathbb{R}^n,$ can be expressed as the span of a finite set $S.$ That is, there exist vectors ${\bf v}_1 , \dots , {\bf v}_k \in V$ such that $$V = \textrm{Span}\left\{ {\bf v}_1, \dots , {\bf v}_k \right\}.$$

WARNING: The spanning set $S$ is NOT unique. For example, the $x$-axis in $\mathbb{R}^2$ can be expressed as $$\left\{ \left[ \begin{array}{c} x \\ 0 \end{array} \right] \ \middle\vert \ x \in \mathbb{R} \right\} = \textrm{Span}\left\{ \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] \right\} = \textrm{Span}\left\{ \left[ \begin{array}{c} -1 \\ 0 \end{array} \right] \right\} = \textrm{Span}\left\{ \left[ \begin{array}{c} \sqrt{2} \\ 0 \end{array} \right] \right\}.$$
Let us use this theorem to construct spanning sets for all the vector spaces (finite-dimensional only) that we have seen so far.