MATH 240

Wed. September 4th, 2019


1.3: Vectors

Adding vectors

Scaling vectors


Span

Span{v1,v2}={c1v1+c2v2 (c1,c2 are scalars)}\text{Span}\{v_1, v_2\}=\{c_1v_1+c_2v_2 \text{ ($c_1, c_2$ are scalars)}\}


Properties of  Rn\ \R^n

For all u,v,wu,v,w in Rn\R^n, and c,dc,d in R\R:
u+v=v+u(u+v)+w=u+(v+w)u+0=0+u=uu+(u)=0c(u+v)=cu+cv(c+d)u=cu+du(cd)u=c(du)1u=u\begin{aligned} u+v&=v+u\\ (u+v)+w&=u+(v+w)\\ u+0=0+u&=u\\ u+(-u)&=0\\ c(u+v)&=cu+cv\\ (c+d)u&=cu+du\\ (cd)u&=c(du)\\ 1u&=u \end{aligned}

ux1+vx2=w has a solution iff w is in Span{u,v}.ux_1+vx_2=w\text{ has a solution iff } w \text{ is in Span}\{u,v\}.


Example.

Is (13)\begin{pmatrix} 1\\3 \end{pmatrix} in the span of (11)\begin{pmatrix} 1\\1 \end{pmatrix} and (22)\begin{pmatrix} 2\\2 \end{pmatrix}?
(11)x1+(22)x2=(13)(121123)(121002)\begin{gathered} \begin{pmatrix} 1\\1 \end{pmatrix}x_1+\begin{pmatrix} 2\\2 \end{pmatrix}x_2=\begin{pmatrix} 1\\3 \end{pmatrix}\\ \begin{pmatrix} 1&2&1\\1&2&3 \end{pmatrix} \rightarrow \begin{pmatrix} 1&2&1\\0&0&2 \end{pmatrix} \end{gathered}
The system has no solution; therefore (13)\begin{pmatrix} 1\\3 \end{pmatrix} is not in the span.