# Spanning

• Oct 4th 2011, 07:42 AM
dwsmith
Spanning
Let $S=\{v_1,v_1,...,v_n\}$ be a spanning set of V.

Prove that for every $w\in V$, the set $\{w\}\cup S$ is a lin. dep spanning set of V.

Seems trivially but not sure how to say it correctly.

If S isn't a lin ind set of vectors, some vectors in S can be removed to form a minimal spanning set say S' of basis vectors of v. If S is lin. ind., then S is a set of basis vectors for V.

By definition of span, any vector in V can be written as a linear combination of vectors in S or S' depending if S is lin ind. It seems to follow trivially. Is there a better way to say this?
• Oct 4th 2011, 01:00 PM
zoek
Re: Spanning
Quote:

Originally Posted by dwsmith
Let $S=\{v_1,v_1,...,v_n\}$ be a spanning set of V.

Prove that for every $w\in V$, the set $\{w\}\cup S$ is a lin. dep spanning set of V.

Seems trivially but not sure how to say it correctly.

If S isn't a lin ind set of vectors, some vectors in S can be removed to form a minimal spanning set say S' of basis vectors of v. If S is lin. ind., then S is a set of basis vectors for V.

By definition of span, any vector in V can be written as a linear combination of vectors in S or S' depending if S is lin ind. It seems to follow trivially. Is there a better way to say this?

Quote:

Originally Posted by dwsmith
By definition of span, any vector in V can be written as a linear combination of vectors in S

$w$ can be written as a linear combination of vectors in $S$
so, $\{w\}\cup S$ is lin. dep.
• Oct 4th 2011, 11:31 PM
Deveno
Re: Spanning
to amplify: we can write (because {v1,...,vn} spans V) w = a1v1 + a2v2 +...+ anvn.

therefore w - a1v1 - a2v2 - ... - anvn = 0. note that the coefficient of w is 1 ≠ 0, so we have found a linear combination

c1w + c2v1 + ...+ c(n+1)vn = 0 with not all the cj = 0. so {w,v1,...,vn} is linearly dependent.
• Oct 5th 2011, 03:06 PM
dwsmith
Re: Spanning
Quote:

Originally Posted by Deveno
to amplify: we can write (because {v1,...,vn} spans V) w = a1v1 + a2v2 +...+ anvn.

therefore w - a1v1 - a2v2 - ... - anvn = 0. note that the coefficient of w is 1 ≠ 0, so we have found a linear combination

c1w + c2v1 + ...+ c(n+1)vn = 0 with not all the cj = 0. so {w,v1,...,vn} is linearly dependent.

But other than that, what I have is ok?

I can add this along with what I have to make it more clear and stronger, correct?
• Oct 5th 2011, 03:46 PM
Deveno
Re: Spanning
you don't need to find the subset S'. all you need is the fact that S spans V, so you can express w as a linear combination of the vj.

what you did is take S, create a basis S' and then show {w}U S' was linearly dependent. you don't need the middle part, adding anything to a spanning set S

creates a linearly dependent set, whether or not S was linearly independent.