§ 4. Linear Combinations and Systems of Linear Equations

Linear Combination and Span

Definition 4.1 : Linear Combination

Let \(V\) be a vector space and \(S\) a nonempty subset of \(V\). A vector \(v \in V\) is called a linear combination of vectors of \(S\) if there exist a finite number of vectors \(u_{1}, u_{2}, \ldots, u_{n}\) in \(S\) and scalars \(a_{1}, a_{2}, \ldots, a_{n}\) in \(F\) such that \(v=a_{1} u_{1}+a_{2} u_{2}+\cdots+a_{n} u_{n}\). In this case we also say that \(v\) is a linear combination of \(u_{1}, u_{2}, \ldots, u_{n}\) and call \(a_{1}, a_{2}, \ldots, a_{n}\) the coefficients of the linear combination.

Observe that in any vector space \(V\), \(0 v=0\) for each \(v \in V\). Thus the zero vector is a linear combination of any nonempty subset of \(V\).

Definition 4.2 : Span of \(S\)

Let \(S\) be a nonempty subset of a vector space \(V\). The span of \(S\), denoted \(\operatorname{span}(S)\), is the set consisting of all linear combinations of the vectors in \(S\). For convenience, we define span \((\varnothing)=\{0\}\).

Theorem 4.3 : \(\operatorname{span}(S)\) is a subspace, and \(\operatorname{span}(S)\) is a minimal subspace containing \(S\).

The span of any subset \(S\) of a vector space \(V\) is a subspace of \(V\). Moreover, any subspace of \(V\) that contains \(S\) must also contain the span of \(S\).

Proof

This result is immediate if \(S=\varnothing\) because span \((\varnothing)=\{0\}\), which is a subspace that is contained in any subspace of \(V\).

If \(S \neq \varnothing\), then \(S\) contains a vector \(z\). So \(0 z=0\) is in \(\operatorname{span}(S)\). Let \(x, y \in \operatorname{span}(S)\). Then there exist vectors \(u_{1}, u_{2}, \ldots, u_{m}, v_{1}, v_{2}, \ldots, v_{n}\) in \(S\) and scalars \(a_{1}, a_{2}, \ldots, a_{m}, b_{1}, b_{2}, \ldots, b_{n}\) such that

\[ x=a_{1} u_{1}+a_{2} u_{2}+\cdots+a_{m} u_{m} \quad \text { and } \quad y=b_{1} v_{1}+b_{2} v_{2}+\cdots+b_{n} v_{n} \]

Then

\[ x+y=a_{1} u_{1}+a_{2} u_{2}+\cdots+a_{m} u_{m}+b_{1} v_{1}+b_{2} v_{2}+\cdots+b_{n} v_{n} \]

and, for any scalar \(c\),

\[ c x=\left(c a_{1}\right) u_{1}+\left(c a_{2}\right) u_{2}+\cdots+\left(c a_{m}\right) u_{m} \]

are clearly linear combinations of the vectors in \(S\); so \(x+y\) and \(c x\) are in \(\operatorname{span}(S)\). Thus \(\operatorname{span}(S)\) is a subspace of \(V\).

Now let \(W\) denote any subspace of \(V\) that contains \(S\). If \(w \in \operatorname{span}(S)\), then \(w\) has the form \(w=c_{1} w_{1}+c_{2} w_{2}+\cdots+c_{k} w_{k}\) for some vectors \(w_{1}, w_{2}, \ldots w_{k}\) in \(S\) and some scalars \(c_{1}, c_{2}, \ldots, c_{k}\). Since \(S \subseteq W\), we have \(w_{1}, w_{2}, \ldots, w_{k} \in W\). Therefore \(w=c_{1} w_{1}+c_{2} w_{2}+\cdots+c_{k} w_{k}\) is in \(W\) by Exercise 3.20. Because \(w\), an arbitrary vector in \(\operatorname{span}(S)\), belongs to \(W\), it follows that \(\operatorname{span}(S) \subseteq W\).

Definition 4.4 : \(S\) generates (or spans) \(V\)

A subset \(S\) of a vector space \(V\) generates (or spans) \(V\) if \(\operatorname{span}(S)=V\). In this case, we also say that the vectors of \(S\) generate (or span) V.

Solving a System of Linear Equations

Example 4.5 : Solving a system of linear equations

We illustrate how to solve a system of linear equations by showing how to determine if the vector \((2,6,8)\) can be expressed as a linear combination of

\[ u_{1}=(1,2,1), \quad u_{2}=(-2,-4,-2), \quad u_{3}=(0,2,3), \quad u_{4}=(2,0,-3), \quad \text{and} \quad u_{5}=(-3,8,16). \]

Thus we must determine whether there are scalars \(a_{1},a_{2},a_{3},a_{4},a_{5}\) such that

\[ \begin{aligned} (2,6,8) &= a_{1}u_{1}+a_{2}u_{2}+a_{3}u_{3}+a_{4}u_{4}+a_{5}u_{5} \\ &= a_{1}(1,2,1)+a_{2}(-2,-4,-2)+a_{3}(0,2,3) +a_{4}(2,0,-3)+a_{5}(-3,8,16) \\ &= \bigl(a_{1}-2a_{2}+2a_{4}-3a_{5},\; 2a_{1}-4a_{2}+2a_{3}+8a_{5},\; a_{1}-2a_{2}+3a_{3}-3a_{4}+16a_{5}\bigr). \end{aligned} \]

Hence \((2,6,8)\) can be expressed as a linear combination of \(u_{1},u_{2},u_{3},u_{4},u_{5}\) if and only if there is a \(5\)-tuple \((a_{1},a_{2},a_{3},a_{4},a_{5})\) satisfying the system of linear equations

\[ \begin{alignedat}{6} a_{1} &-2a_{2} &\phantom{+2a_{3}} &+2a_{4} &-3a_{5} &= 2,\\ 2a_{1} &-4a_{2} &+2a_{3} &\phantom{+2a_{4}} &+8a_{5} &= 6,\\ a_{1} &-2a_{2} &+3a_{3} &-3a_{4} &+16a_{5} &= 8, \end{alignedat} \]

which is obtained by equating corresponding coordinates.

To solve this system, we replace it by another system with the same solutions, but which is easier to solve. The procedure expresses some unknowns in terms of others by eliminating certain unknowns from all but one equation.

To begin, we eliminate \(a_{1}\) from every equation except the first by adding \(-2\) times the first equation to the second, and \(-1\) times the first equation to the third. The result is the following new system:

\[ \begin{alignedat}{6} a_{1} &-2a_{2} &\phantom{+2a_{3}} &+2a_{4} &-3a_{5} &= 2,\\ \phantom{a_{1}-2a_{2}} &\phantom{+2a_{3}} &2a_{3} &-4a_{4} &+14a_{5} &= 2,\\ \phantom{a_{1}-2a_{2}} &\phantom{+2a_{3}} &3a_{3} &-5a_{4} &+19a_{5} &= 6. \end{alignedat} \]

In this case, it happened that while eliminating \(a_{1}\) from every equation except the first, we also eliminated \(a_{2}\) from every equation except the first. This need not happen in general. We now want to make the coefficient of \(a_{3}\) in the second equation equal to \(1\), and then eliminate \(a_{3}\) from the third equation. To do this, we first multiply the second equation by \(\frac12\), which produces

\[ \begin{alignedat}{6} a_{1} &-2a_{2} &\phantom{+2a_{3}} &+2a_{4} &-3a_{5} &= 2,\\ \phantom{a_{1}-2a_{2}} &\phantom{+2a_{3}} &a_{3} &-2a_{4} &+7a_{5} &= 1,\\ \phantom{a_{1}-2a_{2}} &\phantom{+2a_{3}} &3a_{3} &-5a_{4} &+19a_{5} &= 6. \end{alignedat} \]

Next we add \(-3\) times the second equation to the third, obtaining

\[ \begin{alignedat}{5} a_{1} &-2a_{2} &\phantom{{}+2a_{3}} &+2a_{4} &-3a_{5} &= 2,\\ \phantom{a_{1}} &\phantom{-2a_{2}} &a_{3} &-2a_{4} &+7a_{5} &= 1,\\ \phantom{a_{1}} &\phantom{-2a_{2}} &\phantom{a_{3}} &\phantom{+1}a_{4} &-2a_{5} &= 3. \end{alignedat} \]

We continue by eliminating \(a_{4}\) from every equation except the third. This yields

\[ \begin{alignedat}{5} a_{1} &-2a_{2} &\phantom{{}+a_{3}} &\phantom{{}+2a_{4}} &+a_{5} &= -4,\\ \phantom{a_{1}} &\phantom{-2a_{2}} &a_{3} &\phantom{{}-2a_{4}} &+3a_{5} &= 7,\\ \phantom{a_{1}} &\phantom{-2a_{2}} &\phantom{{}+a_{3}} &\phantom{+1}a_{4} &-2a_{5} &= 3. \end{alignedat} \]

System (4) is a system of the desired form: it is easy to solve for the first unknown present in each equation (\(a_{1},a_{3},a_{4}\)) in terms of the other unknowns (\(a_{2},a_{5}\)). Rewriting, we find that

\[ \begin{aligned} a_{1} &= 2a_{2}-a_{5}-4,\\ a_{3} &= -3a_{5}+7,\\ a_{4} &= 2a_{5}+3. \end{aligned} \]

Thus for any choice of scalars \(a_{2}\) and \(a_{5}\), a vector of the form

\[ (a_{1},a_{2},a_{3},a_{4},a_{5}) = (2a_{2}-a_{5}-4,\; a_{2},\; -3a_{5}+7,\; 2a_{5}+3,\; a_{5}) \]

is a solution to the original system. In particular, setting \(a_{2}=0\) and \(a_{5}=0\) gives the solution \((-4,0,7,3,0)\). Therefore

\[ (2,6,8)=-4u_{1}+0u_{2}+7u_{3}+3u_{4}+0u_{5}, \]

so \((2,6,8)\) is a linear combination of \(u_{1},u_{2},u_{3},u_{4},u_{5}\).

Concept 4.6 : Simplifying a system of linear equations

The procedure illustrated in Example 4.5 uses three types of operations to simplify the original system:

1. interchanging the order of any two equations in the system;
2. multiplying any equation in the system by a nonzero constant;
3. adding a constant multiple of any equation to another equation in the system.

In Section 18, we prove that these operations do not change the set of solutions to the original system. Note that we employed these operations to obtain a system of equations that had the following properties:

1. The first nonzero coefficient in each equation is one.
2. If an unknown is the first unknown with a nonzero coefficient in some equation, then that unknown occurs with a zero coefficient in each of the other equations.
3. The first unknown with a nonzero coefficient in any equation has a larger subscript than the first unknown with a nonzero coefficient in any preceding equation.

Exercise

Exercise 4.11

Prove that \(\operatorname{span}(\{x\})=\{a x: a \in F\}\) for any vector \(x\) in a vector space. Interpret this result geometrically in \(\mathbb{R}^{3}\).

Exercise 4.13

Show that if \(S_{1}\) and \(S_{2}\) are subsets of a vector space \(V\) such that \(S_{1} \subseteq S_{2}\), then \(\operatorname{span}\left(S_{1}\right) \subseteq \operatorname{span}\left(S_{2}\right)\). In particular, if \(S_{1} \subseteq S_{2}\) and \(\operatorname{span}\left(S_{1}\right)=V\), deduce that \(\operatorname{span}\left(S_{2}\right)=V\).