13.2: 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 \(\mathbb{P}_2\).