SPAN

By John Gilbert and John Meth

I. BASICS OF SPAN

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

Definition: The Span of a set of vectors $S = \left\{ \ {\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k \ \right\}$ in a vector space $V$ is the set $$\textrm{Span}( \ S \ ) = \textrm{Span}\left\{\,{\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k\right\} = \left\{\, c_1 {\bf v}_1 + c_2 {\bf v_2} + \ldots + c_k {\bf v}_k\,\middle\vert \, c_j \ \in \ {\mathbb R}\,\right\} $$ of all linear combinations of the ${\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k$ and real numbers $\,c_j,\, 1 \le j \le k$.

When $$\textrm{Span}( \ S \ ) = \textrm{Span}\big\{\,{\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k\big\} \ = \ V$$ we say that $S = \left\{ \ {\bf v}_1,\, {\bf v}_2,\, \ldots, \, {\bf v}_k \ \right\}$ 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 ${\mathbb R}^3$.

Example 1: Determine $\textrm{Span}\{ {\bf u},\, {\bf v}\}$ in ${\mathbb R}^3$ when $${\bf u} \ = \ \left[\begin{array}{cc} 2 \\ -4 \\ -2 \end{array}\right], \qquad {\bf v} \ = \ \left[\begin{array}{cc} 0 \\ 1 \\ -3 \end{array}\right]\,. $$

Solution: A vector ${\bf x}$ lies in $\textrm{Span}\{ {\bf u},\, {\bf v}\}$ when the vector equation $$ s {\bf u} + t {\bf v} \ = \ {\bf x}$$ has a solution as an equation in $s,\, t$. Now the augmented matrix of this equation is

$$ \left[\begin{array}{cc} 2 & 0 & x \\ -4 & 1 & y \\ -2 & -3 & z \end{array} \right]\ \sim \ \left[\begin{array}{cc} 2 & 0 & x \\ 0 & 1 & 2x + y \\ -2 & -3 & z \end{array} \right] $$

$$\ \sim \ \left[\begin{array}{cc} 2 & 0 & x \\ 0 & 1 & 2x + y \\ 0 & -3 & x + z \end{array} \right]\qquad \qquad $$ $$\qquad \qquad \sim \ \left[\begin{array}{cc} 2 & 0 & x \\ 0 & 1 & 2x + y \\ 0 & 0 & 7x + 3y + z \end{array} \right].$$ But the original vector equation will not be consistent unless the right-most column is not a pivot column. Thus $x,\, y, \, z$ must satisfy the equation $$7x + 3y + z \ = \ 0\,,$$ and for each point in this plane the original vector equation will have a solution. So, geometrically, $\textrm{Span}\{ {\bf u},\, {\bf v}\}$ is the plane $7x + 3y + z = 0$ in ${\mathbb R}^3$.

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 $W.$

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.

II. EXAMPLES

When studying subspaces, we saw that every vector space $V$ is a subspace of itself, so in a trivial way we can express every vector space as a subspace. It is highly nontrivial that the same thing is true about spans.

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.

$n$-space, $\mathbb{R}^n$
- In $\mathbb{R}^n$, we have the following vectors: $$e_1 = \left[ \begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right], \quad e_2 = \left[ \begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \right], \quad \cdots, \quad e_n = \left[ \begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \end{array} \right].$$ Consider $\textrm{Span}\{e_1, e_2, \dots, e_n \} \subseteq \mathbb{R}^n:$ $$\textrm{Span}\left\{e_1, e_2, \dots, e_n \right\} = \left\{ \alpha_1 \cdot e_1 + \alpha_2 \cdot e_2 + \cdots + \alpha_n \cdot e_n \, \middle\vert \, \, \alpha_1, \alpha_2, \dots, \alpha_n \in \mathbb{R} \right\} =$$ $$= \left\{ \ \alpha_1 \cdot \left[ \begin{array}{c} 1 \\ 0 \\\vdots \\ 0 \end{array} \right] + \alpha_2 \cdot \left[ \begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \right] + \cdots + \alpha_n \cdot \left[ \begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \end{array} \right] \ \ \middle\vert \, \, \alpha_1, \alpha_2, \dots, \alpha_n \in \mathbb{R} \ \right\} =$$ $$= \left\{ \ \left[ \begin{array}{c} \alpha_1 \\ 0 \\ \vdots \\ 0 \end{array} \right] + \left[ \begin{array}{c} 0 \\ \alpha_2 \\ \vdots \\ 0 \end{array} \right] + \cdots + \left[ \begin{array}{c} 0 \\ 0 \\ \vdots \\ \alpha_n \end{array} \right] \ \ \middle\vert \, \, \alpha_1, \alpha_2, \dots, \alpha_n \in \mathbb{R} \ \right\} =$$ $$= \left\{ \ \left[ \begin{array}{c} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n \end{array} \right] \ \ \middle\vert \, \, \alpha_1, \alpha_2, \dots, \alpha_n \in \mathbb{R} \ \right\} = \mathbb{R}^n.$$ Thus, the most fundamental vector space is immediately defined as a span: $$\mathbb{R}^n = \textrm{Span}\{e_1, e_2, \dots, e_n \}.$$
- Every line through the origin in $\mathbb{R}^2$ can be realized as the span of a non-zero vector. In particular, the line with slope $m$ is given by $$\left\{ \left[ \begin{array}{c} x \\ y \\ \end{array} \right] \, \middle\vert \, \, y = m x \right\} = \left\{ \left[ \begin{array}{c} x \\ m x \\ \end{array} \right] \, \middle\vert \, \, x \in \mathbb{R} \right\} = \textrm{Span}\left\{ \left[ \begin{array}{c} 1 \\ m \\ \end{array} \right] \right\}$$ and the vertical line (with infinite slope) is given by $$\left\{ \left[ \begin{array}{c} 0 \\ y \\ \end{array} \right] \, \middle\vert \, \, y \in \mathbb{R} \right\} = \textrm{Span}\left\{ \left[ \begin{array}{c} 0 \\ 1 \\ \end{array} \right] \right\}.$$ Similarly, every plane in $\mathbb{R}^3$ can be expressed as the spane of two vectors. See Example 1 above.
- It is crucial for us to express the null space of a matrix $A$ as a span. This is the reason we have been so insistent on expressing our solutions in vector form. For example, in your last homework, LinAlg7HW.html , we expressed the solution to Beezer's Example CNS1 as $$\textrm{Nul}(A) = \left\{ \ r \left[\begin{array}{c} -2 \\ 3 \\ 1 \\ 0 \\ 0 \\ \end{array}\right] + s \, \left[\begin{array}{c} -1 \\ -4 \\ 0 \\ -2 \\ 1 \\ \end{array}\right] \ \middle\vert \ r , s \in \mathbb{R} \ \right\}.$$ Now we can easily see how to express this as a span: $$\textrm{Nul}(A) = \left\{ \ r \left[\begin{array}{c} -2 \\ 3 \\ 1 \\ 0 \\ 0 \\ \end{array}\right] + s \, \left[\begin{array}{c} -1 \\ -4 \\ 0 \\ -2 \\ 1 \\ \end{array}\right] \ \middle\vert \ r , s \in \mathbb{R} \ \right\} = \textrm{Span}\left\{ \ \left[\begin{array}{c} -2 \\ 3 \\ 1 \\ 0 \\ 0 \\ \end{array}\right] , \ \left[\begin{array}{c} -1 \\ -4 \\ 0 \\ -2 \\ 1 \\ \end{array}\right] \ \right\}.$$

Matrices:
Let's look at some examples of expressing different matrix spaces as spans.

$$M_{2 \times 3}(\mathbb{R}) = \textrm{Span}\left\{ \ \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] , \ \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] , \ \left[ \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} \right] , \ \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{array} \right] , \ \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right] , \ \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] \ \right\}.$$
$$D_3 = \textrm{Span}\left\{ \ \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] , \ \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] , \ \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] \ \right\}.$$
$$UT_2 = \textrm{Span}\left\{ \ \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right] , \ \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right] , \ \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} \right] \ \right\}.$$

Polynomials:

$$\mathcal{P}_n = \textrm{Span}\{1, t, t^2, \dots, t^n\} \textrm{.}$$
Similar to our last homework set, Exercise S.T30: $$W = \left\{ \ p \in \mathcal{P}_8 \ \middle\vert \ p \textrm{ has terms of only even degree } \right\} = \textrm{Span}\{1, t^2, t^4, t^6, t^8 \} \textrm{.}$$
$$V = \left\{ \ p \in \mathcal{P}_5 \ \middle\vert \ p(0) = 0 \ \right\} = \textrm{Span}\{ t, t^2, t^3, t^4, t^5 \} \textrm{.}$$

Complex Numbers, $\mathbb{C}:$

The complex numbers themselves are a span, $\mathbb{C} = \textrm{Span}\left\{ 1, i \right\}$.
The real numbers are a subspace of $\mathbb{C}:$ $\mathbb{R} = \textrm{Span}\left\{ 1 \right\} \subseteq \mathbb{C}.$
The imaginary numbers are also subspace of $\mathbb{C}:$ $V = \left\{ z \in \mathbb{C} \ \middle\vert \ z = \beta i \textrm{ for some } \beta \in \mathbb{R} \right\} = \textrm{Span}\left\{ i \right\} \subseteq \mathbb{C}.$

Differential Equations:

Consider the homogeneous equation $$L [ \ y \ ] = y^{\ \prime \prime} - y = 0.$$ If we solve this equation using the roots of the characteristic polynomial, $p(r) = r^2-1$, we get $V_L = \textrm{Span} \left\{ e^t , e^{-t} \right\}.$ On the other hand, if we solve this equation using series solutions, we get $V_L = \textrm{Span} \left\{ \cosh(t) , \sinh(t) \right\}.$ Let us explore this using the idea of span. The definitions of $\cosh(t)$ and $\sinh(t)$ are $$\cosh(t) = \frac{1}{2} e^t + \frac{1}{2} e^{-t} \quad \quad \textrm{ and } \quad \quad \sinh(t) = \frac{1}{2} e^t - \frac{1}{2} e^{-t}$$ which shows us that both $\cosh(t)$ and $\sinh(t)$ are linear combinations of $e^t$ and $e^{-t}$. Thus $\cosh(t), \sinh(t) \in \textrm{Span} \left\{ e^t , e^{-t} \right\},$ a vector space. So by Theorem 2, we have $\textrm{Span} \left\{ \cosh(t) , \sinh(t) \right\} \subseteq \textrm{Span} \left\{ e^t , e^{-t} \right\}.$
Again, using the definitions of $\cosh(t)$ and $\sinh(t),$ we have that $\cosh(t)+\sinh(t) = e^t$ and $\cosh(t)-\sinh(t) = e^{-t}.$ This means $e^t$ and $e^{-t}$ are linear combinations of $\cosh(t)$ and $\sinh(t),$ so $e^t, e^{-t} \in \textrm{Span} \left\{ \cosh(t) , \sinh(t) \right\},$ a vector space. So by Theorem 2, we have $\textrm{Span} \left\{ e^t , e^{-t} \right\} \subseteq \textrm{Span} \left\{ \cosh(t) , \sinh(t) \right\}.$ Together, $\textrm{Span} \left\{ \cosh(t) , \sinh(t) \right\} \subseteq \textrm{Span} \left\{ e^t , e^{-t} \right\}$ and $\textrm{Span} \left\{ e^t , e^{-t} \right\} \subseteq \textrm{Span} \left\{ \cosh(t) , \sinh(t) \right\}$ imply that the two sets are equal: $$V_L = \textrm{Span} \left\{ e^t , e^{-t} \right\} = \textrm{Span} \left\{ \cosh(t) , \sinh(t) \right\}.$$
Consider the homogeneous equation $$L [ \ y \ ] = y^{(3)} + 4y^{\ \prime} = 0.$$ If we solve this equation using the roots of the characteristic polynomial, $p(r) = r^3+4r = r(r^2+4)$, we get $V_L = \textrm{Span} \left\{ \cos(2t) , \sin(2t), 1 \right\}.$ Using trigonometric identities, we can see that $$V_L = \textrm{Span} \left\{ \cos(2t) , \sin(2t), 1 \right\} = \textrm{Span} \left\{ \cos^2(t) , \cos(t)\sin(t), \sin^2(t) \right\},$$ which are the degree 2 monomials in $\cos(t)$ and $\sin(t)$. We will see this example done in the homework.
Let $$A = \left[ \begin{array}{cc} 2 & 0 \\ 0 & -3 \\ \end{array} \right],$$ then $$V_A = \textrm{Span} \left\{ \left[ \begin{array}{c} e^{2t} \\ 0 \\ \end{array} \right] , \left[ \begin{array}{c} 0 \\ e^{-3t} \\ \end{array} \right] \right\}.$$ We can see by plugging in that these are solutions to the system $$\frac{d}{dt} \vec{x}(t) = \left[ \begin{array}{cc} 2 & 0 \\ 0 & -3 \\ \end{array} \right] \vec{x}(t),$$ and we know from our text that $V_A$ is a vector space, so by Theorem 2 we have $$\textrm{Span} \left\{ \left[ \begin{array}{c} e^{2t} \\ 0 \\ \end{array} \right] , \left[ \begin{array}{c} 0 \\ e^{-3t} \\ \end{array} \right] \right\} \subseteq V_A.$$ We will know that this is the entirety of $V_A$ after we have explored the idea of a basis in a later lesson.
Let $$A = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right].$$ Then $\frac{d}{dt} \vec{x}(t) = A \vec{x}(t)$ is the second order, homogeneous equation $L [ \ y \ ] = y^{\ \prime \prime} - y = 0$ (from the example above), converted into a linear system. Using the solutions to the O.D.E., and the fact that $x_2 = x_1^{\ \prime},$ we see that $$V_A = \textrm{Span} \left\{ \left[ \begin{array}{c} e^t \\ e^t \\ \end{array} \right] , \left[ \begin{array}{c} -e^{-t} \\ e^{-t} \\ \end{array} \right] \right\} = \textrm{Span} \left\{ \left[ \begin{array}{c} \cosh(t) \\ \sinh(t) \\ \end{array} \right] , \left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \\ \end{array} \right] \right\}.$$