MATH 240

Fri. October 18th, 2019

Spanning Set Theorem

Consider the set of vectors $S=\{\begin{pmatrix} 0\\1 \end{pmatrix},\begin{pmatrix} 1\\0 \end{pmatrix},\begin{pmatrix} 2\\3 \end{pmatrix},\begin{pmatrix} 4\\7 \end{pmatrix}\}$ .

What is the span of $S$ ? $\text{Span}(S)=\R^2$ .

Is $S$ a basis? No, because it is a linearly dependent set – the vectors are in
$\R^2$ but there are four of them, and $4>2$ .

Is $\{\begin{pmatrix} 0\\1 \end{pmatrix},\begin{pmatrix} 2\\3 \end{pmatrix}\}$ ( $S$ with two vectors removed) a basis? Yes, because it is linearly independent and spans the entire space of $\R^2$ .

Theorem.

Let $S=\{v_1,...,v_p\}$ be a set of vectors in $V$ and let $H=\text{Span}(S)$ . If one of the vectors in $S$ , say $v_k$ , is a linear combination of the remaining vectors in $S$ , then the set formed by removing $v_k$ from $S$ also spans $H$ .
If $H\neq \{0\}$ , then some subset of $S$ is a basis for $H$ .

(Note that the basis produced by this method will be different depending on which vectors you remove.)

Bases for $\text{Nul}A, \text{Col}A$

For a matrix $A$ , a basis for $\text{Nul}A$ is formed from the solution set of $Ax=0$ , and a basis for $\text{Col}A$ is formed from pivot columns of $A$ .

Ex.
$\begin{aligned} A&=\begin{pmatrix} -3&6&-1&1&-7\\1&-2&2&3&-1\\2&-4&5&8&-4 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1&-2&0&-1&3\\0&0&1&2&-2\\0&0&0&0&0 \end{pmatrix}\\ \text{Basis for Col}A&=\{\begin{pmatrix} 1\\0\\0 \end{pmatrix},\begin{pmatrix} 0\\1\\0 \end{pmatrix}\}\\ \\ &x_2,\ x_4,\ x_5\ \text{are free}\\ x_1&=2x_2+x_4-3x_5\\ x_3&=-2x_4+2x_5\\ \begin{pmatrix} x_1\\x_2\\x_3\\x_4\\x_5 \end{pmatrix}&=x_2\begin{pmatrix} 2\\1\\0\\0\\0 \end{pmatrix}+x_4\begin{pmatrix} 1\\0\\-2\\1\\0 \end{pmatrix}+x_5\begin{pmatrix} -3\\0\\2\\0\\1 \end{pmatrix}\\ \text{Basis for Nul}A&=\{\begin{pmatrix} 2\\1\\0\\0\\0 \end{pmatrix},\begin{pmatrix} 1\\0\\-2\\1\\0 \end{pmatrix},\begin{pmatrix} -3\\0\\2\\0\\1 \end{pmatrix}\} \end{aligned}$

Coordinates

Why should we care about bases?

Theorem. Let $B=\{b_1,...,b_n\}$ be a basis for the vector space $V$ . Given any $x$ in $V$ , there are unique $c_1,...,c_n$ such that $x=c_1b_1+...+c_nb_n$ .

This means that once a basis is found for a vector space, we can uniquely identify every vector in that space by the list of scalars $c_1,...,c_n$ associated with that vector. We call these the coordinates of $x$ in $B$ , denoted by $[x]_B$ .