# spanning sets

• October 14th 2008, 02:41 AM
calculusgeek
spanning sets
Take http://www.mathhelpforum.com/math-he...6f4e228a-1.gif and give an example to show that it is possible that Span(S) = Span (S') even though http://www.mathhelpforum.com/math-he...79957b2d-1.gif and http://www.mathhelpforum.com/math-he...8a943da1-1.gif.

i having problems with this question as i know that i am looking for a set of vectors say S which are also in S' but in S' there has to be more vectors than in S in order for S' to be satisfied and not equal to S.Finding vectors is what im finding hard to grasp

if anyone could help it wolud be much apreciated trying to do it for months
• October 14th 2008, 04:14 AM
hatsoff
This is the case whenever $S'=S\cup T$, where every vector of $T$ is a linear combination of the vectors of $S$.

For example, let:

$S=\left\{\begin{array}{ccc}\left[\begin{array}{c}1\\0\\0\end{array}\right]&\left[\begin{array}{c}0\\1\\0\end{array}\right]&\left[\begin{array}{c}0\\0\\1\end{array}\right]\end{array}\right\}$

such that $span(S)=\mathbb{R}$; then let:

$T=\left\{\begin{array}{cc}\left[\begin{array}{c}1\\1\\0\end{array}\right]&\left[\begin{array}{c}0\\1\\1\end{array}\right]\end{array}\right\}$

which means $T\cup S=S'$, and:

$S'=\left\{\begin{array}{ccccc}\left[\begin{array}{c}1\\0\\0\end{array}\right]&\left[\begin{array}{c}0\\1\\0\end{array}\right]&\left[\begin{array}{c}0\\0\\1\end{array}\right]&\left[\begin{array}{c}1\\1\\0\end{array}\right]&\left[\begin{array}{c}0\\1\\1\end{array}\right]\end{array}\right\}$

which makes $span(S')=\mathbb{R}=span(S)$.