# 9.2: Spanning Sets

- Page ID
- 14553

- 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.

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 \(V\).

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.

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.

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.

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.

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 \(\mathbb{P}_2\).