# Linear Algebra: Linear Combinations

## Span

The span of a set of vectors $v_1, \ldots, v_n$ in a vector space $V$ over $F$ is the set of all possible linear combinations of $v_1, \ldots, v_n$. The span of a set of vectors does not have a fancy symbol to denote it, and is instead written as $\text{span}(v_1, \ldots, v_n)$. What a plain bagel! In set builder notation, we can write the span out like this:

$$\text{span}(v_1, \ldots, v_n) = \left\{\sum\limits_{i}^{n} a_i v_i : a_i \in F \right\}$$

The span of an empty set of vectors is defined to be $\{ 0 \}$, the set containing only the zero vector.

If the span of a set of vectors $v_1, \ldots, v_n$ is equal to $V$, then the vectors are said to span $V$.

A vector space $V$ is called a finite-dimensional vector space if it is spanned by a finite number of vectors. Otherwise it is called an infinite-dimensional vector space.

## Problems

1. Determine whether $\langle 3, 2 \rangle \in \text{span}\left( \langle 1, 1 \rangle, \langle 1, 2 \rangle \right)$.

Determining whether a vector is an element of the span of some vectors is the same as determining whether it's a linear combination of those vectors. So if you read the previous section on linear combinations and though, "Aw gee, that looks easy, I'll skip it," well guess what, double whammy!

$\langle 3, 2 \rangle = a_1 \langle 1, 1 \rangle + a_2 \langle 1, 2 \rangle \\ \langle 3, 2 \rangle = \langle a_1, a_1 \rangle + \langle a_2, 2a_2 \rangle \\ \langle 3, 2 \rangle = \langle a_1 + a_2, a_1 + 2a_2 \rangle \\$

There are two equations and two unknowns, so this vector has a shot of being in that span:

$3 = a_1 + a_2 \\ a_1 = 3 - a_2 \\$

$2 = a_1 + 2a_2 \\ 2 = 3 - a_2 + 2a_2 \\ -1 = a_2 \\$

$a_1 = 3 - (-1) \\ a_1 = 4 \\$

Looks like we've got a winner, folks. That vector is indeed part of the span:

$\langle 3, 2 \rangle = 4 \langle 1, 1 \rangle - \langle 1, 2 \rangle \\$

It must feel good to belong to something important.

2. Determine whether $\langle 2, 3 \rangle \in \text{span}\left(\langle 1, i \rangle, \langle 0, i \rangle \right)$.

$\langle 2, 3 \rangle = a_1 \langle 1, i \rangle + a_2 \langle 0, i \rangle \\ \langle 2, 3 \rangle = \langle a_1, ia_1 \rangle + \langle 0, ia_2 \rangle \\ \langle 2, 3 \rangle = \langle a_1, ia_1 + ia_2 \rangle \\$

There are two equations and two unknowns, so the vector might be in the span. Let's solve and find out:

$2 = a_1$

$3 = ia_1 + ia_2 \\ 3 = 2i + ia_2 \\ 3 - 2i = ia_2 \\ \dfrac{3}{i} - 2 = a_2 \\$

Well, it's kind of awkward and gangly, like most complex numbers, but it'll do:

$\langle 2, 3 \rangle = 2 \langle 1, i \rangle + \left(\dfrac{3}{i} -2\right) \langle 0, i \rangle \\$ $3. Show that the span of a set of vectors is a subspace. To show that any set is a subspace, they must fulfill the three requirements outlined in the Subspaces section. Let$v_1, \ldots, v_n$be vectors in a vector space$V$over$F$. Additive identity:$0$is a linear combination of$v_1, \ldots, v_n$, namely$0 = 0v_1 + \ldots 0v_n$. Therefore$0 \in \text{span}\left( v_1, \ldots, v_n \right)$. Closed under addition: Let$a$and$b$be in$\text{span}(v_1, \ldots, v_n)$such that$a = a_1v_1 + \ldots + a_nv_n$and$b = b_1v_1 + \ldots + b_nv_n$, where each$a_i$and$b_i$are in$F$. Then$ a + b = (a_1v_1 + \ldots + a_nv_n) + (b_1v_1 + \ldots + b_nv_n) \\ a + b = (a_1 + b_1)v_1 + \ldots + (a_n + b_n)v_n $Therefore$a + b$is also in$\text{span}(v_1, \ldots, v_n)$, so$\text{span}(v_1, \ldots, v_n)$is closed under addition. Closed under addition: Let$a$be in$\text{span}(v_1, \ldots, v_n)$such that$a = a_1v_1 + \ldots + a_nv_n$, and let$c \in F$. Then$ ca = c(a_1v_1 + \ldots + a_nv_n) \\ca = (ca_1)v_1 + \ldots + (ca_n)v_n$Therefore$ca$is also in$\text{span}(v_1, \ldots, v_n)$, so$\text{span}(v_1, \ldots, v_n)$is closed under scalar multiplication. 4. Show that the span of a set of vectors is the smallest subspace containing all the vectors in the set. Let$v_1, \ldots, v_n$be a set of vectors in a vector space$V$over a field$F$. Each vector$v_i$is in$\text{span}(v_1, \ldots, v_n)$, as$v_i = 1v_i + \displaystyle\sum\limits_{j \neq i} 0 v_j$. Conversely, as subspaces are closed under scalar addition and multiplication, each subspace containing all of$v_1, \ldots, v_n$contains all linear combinations of$v_1, \ldots, v_n$and therefore contains$\text{span}(v_1, \ldots, v_n)$. Therefore,$\text{span}(v_1, \ldots, v_n)$is the smallest subspace of$V$containing all of$v_1, \ldots, v_n$. 5. Let$F^n$be a vector space over$F$, where$F$is a field (such as$\mathbb{R}$or$\mathbb{C}$). Consider the vectors$e_1, \ldots, e_n$, where$e_1 = \langle 1, 0, \ldots, 0 \rangle, e_2 = \langle 0, 1, \ldots, 0 \rangle, \ldots, e_n = \langle 0, 0, \ldots, 1 \rangle$. Show that these vectors span$F^n$. By the definition of span, we see that$ \text{span}(e_1, e_2, \ldots, e_n) = \left\{ c_1e_1 + c_2e_2 + \ldots + c_ne_n : c_i \in F \right\} \\ $From here we just simplify and follow our noses:$ \text{span}(e_1, e_2, \ldots, e_n) = \left\{ \langle c_1, 0, \ldots, 0 \rangle + \langle 0, c_2, \ldots, 0 \rangle + \ldots + \langle 0, 0, \ldots, c_n \rangle : c_1, c_2, \ldots, c_n \in F \right\} \\ \text{span}(e_1, e_2, \ldots, e_n) = \left\{ \langle c_1, c_2, \ldots, c_n \rangle : c_1, c_2, \ldots, c_n \in F \right\} \\ \text{span}(e_1, e_2, \ldots, e_n) = F^n \\ \$