Definition

A linear combination of the vectors $u_1,...,u_k$ in $V$ is a vector of the form $$ \begin{align*} a_1u_1 + a_2u_2 + ... + a_ku_k, \quad \text{where $a_1, a_2, ... a_k \in \mathbf{R}$}. \end{align*} $$

Example 1

$(4,3) \in \mathbf{R}^2$. This vector can be written as

$$ \begin{align*} (4,3) = 4(1,0) + 3(0,1). \end{align*} $$

So $(4,3)$ is a linear combination of $(1,0)$ and $(0,1)$.

Example 2

$x^3 + 3x + \pi \in P_3$. in a linear combination of $x^3, x^2, x, 1$ with $a_1 = 1, a_2 = 0, a_3 = 3, a_4 = \pi$.

Span of of a Subset

Definition

Let $V$ be a vector space and let $S \in V$ be a subset. The span of $S$ is the set of all linear combinations of elements in $S$. $$ \begin{align*} Span(S) = \{a_1u_1 + a_2u_2 + ... + a_ku_k \quad|\quad a_1, a_2, ... a_k \in \mathbf{R}, u_1, u_2,...,u_k \in S\}. \end{align*} $$

Example 3

Let $S = \{\bar{0}\}$, then $Span(S) = \{\bar{0}\}.$

Example 4

Let $S = V$, then $Span(S) = V.$

Example 5

Let $S=\{(1,0),(0,1)\}$. $S \subset \mathbf{R}^2$. Then,

$$ \begin{align*} Span(S) &= \{a_1(1,0) + a_2(0,1) \ | \ a_1, a_2 \in \mathbf{R}\} \\ &= \{(a_1, a_2) \ | \ a_1, a_2 \in \mathbf{R}\} \\ &= \mathbf{R}^2. \end{align*} $$

Example 6

Let $S=\{(1,m)\}$. $S \subset \mathbf{R}^2$. Then,

$$ \begin{align*} Span(S) &= \{a_1(1,m) \ | \ a_1 \in \mathbf{R}\} \\ &= \{(a_1, ma_1) \ | \ a_1 \in \mathbf{R}\} \\ &= L_m. \end{align*} $$

This is the line through the origin with slope $m$.

Example 7

Let $S=\{(1,0,0),(0,0,1)\}$. $S \subset \mathbf{R}^3$. Then,

$$ \begin{align*} Span(S) &= \{a_1(1,0,0) + a_2(0,0,1) \ | \ a_1, a_2 \in \mathbf{R}\} \\ &= \{(a_1, 0, a_2) \ | \ a_1, a_2 \in \mathbf{R}\} \\ \end{align*} $$

This is the $xz$-plane in $\mathbf{R}^3$.

Theorem (part of 1.5 in the book)

For any nonempty subset $S \subset V$, $Span(S)$ is a subspace of $V$.

Proof: Let $S \subset V$ where $V$ is a vector space. By definition we know that $Span(S)$

$$ \begin{align*} \{a_1u_1 + a_2u_2 + ... + a_ku_k \ | \ a_1, a_2, ... a_k \in \mathbf{R}, u_1, u_2,...u_k \in S\} \end{align*} $$

We will show that $Span(S)$ is a subspace of (V) by verifying the three Subspace conditions.

We need to show that $\bar{0} \in Span(S)$. $S$ is nonempty so we can choose some vector $u \in S$. Since all linear combinations of $u$ are in $Span(S)$, then $0u \in Span(S)$. But $0u = \bar{0}$. Therefore, $\bar{0} \in Span(S)$ and we're done.
We need to show that $Span(S)$ is closed under addition. Suppose $u, v \in Span(S)$. We need to show that $u + v \in Span(S)$. We can write $u$ and $v$ as follows
$$ \begin{align*} u &= a_1u_1 + a_2u_2 + ... + a_ku_k \\ v &= b_1u_1 + b_2u_2 + ... + b_ku_k. \end{align*} $$
Then the sum can be written as,
$$ \begin{align*} u+v &= (a_1+b_1)u_1 + (a_2+b_2)u_2 + ... + (a_k+b_k)u_k. \end{align*} $$
So $u+v \in Span(S)$ as we wanted to show.
We need to show that $Span(S)$ is closed under scalar multiplication. Similar to the previous argument we can see that
$$ \begin{align*} cu &= (ca_1)u_1 + (ca_2)u_2 + ... + (ca_k)u_k. \end{align*} $$
So $cu \in Span(S)$ as required.

From (a), (b), (c), we can conclude that $Span(S)$ is a subspace of $V$.

Example 7

Is $x^3 - 3x + 5$ in $Span(\{x^3 + 2x^2 - x + 1, x^3 + 3x^2 - 1\})$?

For $x^3 - 3x + 5$ to be in the span, this means that $x^3 - 3x + 5$ can be written as a linear combinations of what’s inside the set. In other words, there exists $a_1, a_2 \in \mathbf{R}$ such that

$$ \begin{align*} x^3 - 3x + 5 &= a_1(x^3 + 2x^2 - x + 1) + a_2(x^3 + 3x^2 - 1) \\ &= (a_1+a_2)x^3 + (2a_1 - 3a_3)x^2 + (-a_1)x + (a_1-a_2) \end{align*} $$

These equations are the same when the coefficients are equal. So equivalently we can solve the following system of equations:

$$ \begin{align*} a_1 + a_2 = 1 \\ 2a_2 + 3a_2 = 0 \\ -a_1 = -3 \\ a_1 - a_2 = 5. \end{align*} $$

We can use Gaussian Elimination to solve this system by doing a forward pass followed by a backward pass.

$$ \begin{align*} \begin{pmatrix} 1 & 1 & 1 \\ 2 & 3 & 0 \\ -1 & 0 & -3 \\ 1 & -1 & 5 \end{pmatrix}. \end{align*} $$

We’ll put the matrix in Row Echelon Form. Starting with zeroing out the entries below the first leading entry in the first column.

$$ \begin{align*} \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & -2 \\ 0 & -2 & 4 \end{pmatrix}. \end{align*} $$

Continuing with the next leading entry to get.

$$ \begin{align*} \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. \end{align*} $$

This tells us that there is a solution so the answer is yes, we can write it $x^3 -3x + 5$ is in the span above.

Definition

$S \subset V$ generates $V$ if $Span(S)=V$.

Example 8

The set $\{(1,0),(0,1)\}$ generates $\mathbf{R}^2$.

Example 9

The following set

$$ \begin{align*} \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\} \end{align*} $$

generates the vector space of 2x2 symmetric matrices. How do we go about proving this?
Consider the matrix:

$$ \begin{align*} A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}, \end{align*} $$

If $A$ was symmetric, then we must have $a_{12} = a_{21}$ or $a_{ij} = a_{ji}$ if $i \neq j$. This means we can re-write $A$ as

$$ \begin{align*} a_{11} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + a_{12} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + a_{22} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \end{align*} $$

References:

Math416 by Ely Kerman