9.2: Spanning Sets

Outcomes

1. Determine if a vector is within a given span.

In this section we will examine the concept of spanning introduced earlier in terms of $\mathbb{R}^n$. Here, we will discuss these concepts in terms of abstract vector spaces.

Consider the following definition.

Definition $\PageIndex{1}$: Subset

Let $X$ and $Y$ be two sets. If all elements of $X$ are also elements of $Y$ then we say that $X$ is a subset of $Y$ and we write $$X \subseteq Y\nonumber$$

In particular, we often speak of subsets of a vector space, such as $X \subseteq V$. By this we mean that every element in the set $X$ is contained in the vector space $V$.

Definition $\PageIndex{2}$: Linear Combination

Let $V$ be a vector space and let $\vec{v}_1, \vec{v}_2, \cdots, \vec{v}_n \subseteq V$. A vector $\vec{v} \in V$ is called a linear combination of the $\vec{v}_i$ if there exist scalars $c_i \in \mathbb{R}$ such that $$\vec{v} = c_1 \vec{v}_1 + c_2 \vec{v}_2 + \cdots + c_n \vec{v}_n\nonumber$$

This definition leads to our next concept of span.

Definition $\PageIndex{3}$: Span of Vectors

Let $\{\vec{v}_{1},\cdots ,\vec{v}_{n}\} \subseteq V$. Then $$\mathrm{span}\left\{ \vec{v}_{1},\cdots ,\vec{v}_{n}\right\} = \left\{ \sum_{i=1}^{n}c_{i}\vec{v}_{i}: c_{i}\in \mathbb{R} \right\}\nonumber$$

When we say that a vector $\vec{w}$ is in $\mathrm{span}\left\{ \vec{v}_{1},\cdots ,\vec{v}_{n}\right\}$ we mean that $\vec{w}$ can be written as a linear combination of the $\vec{v}_1$. We say that a collection of vectors $\{\vec{v}_{1},\cdots ,\vec{v}_{n}\}$ is a spanning set for $V$ if $V = \mathrm{span} \{\vec{v}_{1},\cdots ,\vec{v}_{n}\}$.

Consider the following example.

Example $\PageIndex{1}$: Matrix Span

Let $A = \left [ \begin{array}{rr} 1 & 0 \\ 0 & 2 \end{array}\right ]$, $B = \left [ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array}\right ]$. Determine if $A$ and $B$ are in $$\mathrm{span}\left\{ M_1, M_2 \right\} = \mathrm{span} \left\{ \left [ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right ], \left [ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array}\right ] \right\}\nonumber$$

Solution

First consider $A$. We want to see if scalars $s,t$ can be found such that $A = s M_1 + t M_2$. $$\left [ \begin{array}{rr} 1 & 0 \\ 0 & 2 \end{array}\right ] = s \left [ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right ] + t \left [ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array}\right ]\nonumber$$ The solution to this equation is given by $$\begin{aligned} 1 &= s \\ 2 &= t\end{aligned}$$ and it follows that $A$ is in $\mathrm{span} \left\{ M_1, M_2 \right\}$.

Now consider $B$. Again we write $B = sM_1 + t M_2$ and see if a solution can be found for $s, t$. $$\left [ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array}\right ] = s \left [ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right ] + t \left [ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array}\right ]\nonumber$$ Clearly no values of $s$ and $t$ can be found such that this equation holds. Therefore $B$ is not in $\mathrm{span} \left\{ M_1, M_2 \right\}$.

Consider another example.

Example $\PageIndex{2}$: Polynomial Span

Show that $p(x) = 7x^2 + 4x - 3$ is in $\mathrm{span}\left\{ 4x^2 + x, x^2 -2x + 3 \right\}$.

Solution

To show that $p(x)$ is in the given span, we need to show that it can be written as a linear combination of polynomials in the span. Suppose scalars $a, b$ existed such that $$7x^2 +4x - 3= a(4x^2+x) + b (x^2-2x+3)\nonumber$$ If this linear combination were to hold, the following would be true: $$\begin{aligned} 4a + b &= 7 \\ a - 2b &= 4 \\ 3b &= -3 \end{aligned}$$

You can verify that $a = 2, b = -1$ satisfies this system of equations. This means that we can write $p(x)$ as follows: $$7x^2 +4x-3= 2(4x^2+x) - (x^2-2x+3)\nonumber$$

Hence $p(x)$ is in the given span.

Consider the following example.

Example $\PageIndex{3}$: Spanning Set

Let $S = \left\{ x^2 + 1, x-2, 2x^2 - x \right\}$. Show that $S$ is a spanning set for $\mathbb{P}_2$, the set of all polynomials of degree at most $2$.

Solution

Let $p(x)= ax^2 + bx + c$ be an arbitrary polynomial in $\mathbb{P}_2$. To show that $S$ is a spanning set, it suffices to show that $p(x)$ can be written as a linear combination of the elements of $S$. In other words, can we find $r,s,t$ such that: $$p(x) = ax^2 +bx + c = r(x^2 + 1) + s(x -2) + t(2x^2 - x)\nonumber$$

If a solution $r,s,t$ can be found, then this shows that for any such polynomial $p(x)$, it can be written as a linear combination of the above polynomials and $S$ is a spanning set.

$$\begin{aligned} ax^2 +bx + c &= r(x^2 + 1) + s(x -2) + t(2x^2 - x) \\ &= rx^2 + r + sx - 2s + 2tx^2 - tx \\ &= (r+2t)x^2 + (s-t)x + (r-2s) \end{aligned}$$

For this to be true, the following must hold: $$\begin{aligned} a &= r+2t \\ b &= s-t \\ c &= r-2s\end{aligned}$$

To check that a solution exists, set up the augmented matrix and row reduce: $$\left [ \begin{array}{rrr|r} 1 & 0 & 2 & a \\ 0 & 1 & -1 & b \\ 1 & -2 & 0 & c \end{array} \right ] \rightarrow \cdots \rightarrow \left [ \begin{array}{rrr|c} 1 & 0 & 0 & \frac{1}{2} a + 2b + \frac{1}{2}c\\ 0 & 1 & 0 & \frac{1}{4}a - \frac{1}{4}c \\ 0 & 0 & 1 & \frac{1}{4}a - b - \frac{1}{4}c \end{array} \right ]\nonumber$$

Clearly a solution exists for any choice of $a,b,c$. Hence $S$ is a spanning set for $\mathbb{P}_2$.