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
Determine whether $[3, 2] \in \text{span}\left( [1, 1], [1, 2] \right)$.
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!
$[3, 2] = a_1 [1, 1] + a_2 [1, 2] \\ [3, 2] = [a_1, a_1] + [a_2, 2a_2] \\ [3, 2] = [a_1 + a_2, a_1 + 2a_2] \\ $
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:
$ [3, 2] = 4[1, 1] - [1, 2] \\$
It must feel good to belong to something important.
Determine whether $[2, 3] \in \text{span}\left([1, i], [0, i] \right)$.
$[2, 3] = a_1 [1, i] + a_2 [0, i] \\ [2, 3] = [a_1, ia_1] + [0, ia_2] \\ [2, 3] = [a_1, ia_1 + ia_2] \\ $
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:
$ [2, 3] = 2[1, i] + \left(\dfrac{3}{i} -2\right) [0, i] \\$
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.
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 $v_1, \ldots, v_n$.
Let $F^n$ be a vector space over $F$, where $F$ is a field. Consider the vectors $e_1, \ldots, e_n$, where $e_1 = [1, 0, \ldots, 0], e_2 = [0, 1, \ldots, 0], \ldots, e_n = [0, 0, \ldots, 1]$. 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\{ [c_1, 0, \ldots, 0] + [0, c_2, \ldots, 0] + \ldots + [0, 0, \ldots, c_n] : c_1, c_2, \ldots, c_n \in F \right\} \\ \text{span}(e_1, e_2, \ldots, e_n) = \left\{ [c_1, c_2, \ldots, c_n] : c_1, c_2, \ldots, c_n \in F \right\} \\ \text{span}(e_1, e_2, \ldots, e_n) = F^n \\ $