MATH 240

Fri. October 18th, 2019


Spanning Set Theorem

Consider the set of vectors S={(01),(10),(23),(47)}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 SS? Span(S)=R2\text{Span}(S)=\R^2.

Is SS a basis? No, because it is a linearly dependent set – the vectors are in
R2\R^2 but there are four of them, and 4>24>2.

Is {(01),(23)}\{\begin{pmatrix} 0\\1 \end{pmatrix},\begin{pmatrix} 2\\3 \end{pmatrix}\} (SS with two vectors removed) a basis? Yes, because it is linearly independent and spans the entire space of R2\R^2.

Theorem.

  1. Let S={v1,...,vp}S=\{v_1,...,v_p\} be a set of vectors in VV and let H=Span(S)H=\text{Span}(S). If one of the vectors in SS, say vkv_k, is a linear combination of the remaining vectors in SS, then the set formed by removing vkv_k from SS also spans HH.
  2. If H{0}H\neq \{0\}, then some subset of SS is a basis for HH.

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


Bases for NulA,ColA\text{Nul}A, \text{Col}A

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

Ex.
A=(361171223124584)(120130012200000)Basis for ColA={(100),(010)}x2, x4, x5 are freex1=2x2+x43x5x3=2x4+2x5(x1x2x3x4x5)=x2(21000)+x4(10210)+x5(30201)Basis for NulA={(21000),(10210),(30201)}\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={b1,...,bn}B=\{b_1,...,b_n\} be a basis for the vector space VV. Given any xx in VV, there are unique c1,...,cnc_1,...,c_n such that x=c1b1+...+cnbnx=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 c1,...,cnc_1,...,c_n associated with that vector. We call these the coordinates of xx in BB, denoted by [x]B[x]_B.

Ex. Let B={(13),(40)}B=\{\begin{pmatrix} 1\\3 \end{pmatrix},\begin{pmatrix} 4\\0 \end{pmatrix}\}, and let [x]B=(711)[x]_B=\begin{pmatrix} 7\\11 \end{pmatrix}.
x=7(13)+11(40)=(5121)x=7\begin{pmatrix} 1\\3 \end{pmatrix}+11\begin{pmatrix} 4\\0 \end{pmatrix}=\begin{pmatrix} 51\\21 \end{pmatrix}