# Thread: Is V in Span(S)

1. ## Is V in Span(S)

Hey,

I have a question that I'm not sure how to approach at all. The question is:

Let

S={[[-1],[2],[1]] , [[3],[1],[2]] , [[1],[5],[4]] , [[-6],[5],[1]]}

and

v=[[-5],[3],[0]]

Is v in span(S)?

So i know that a set of vectors ie. v, spans a subspace, S, if every vector in S can be written as a linear combination of vectors in v.

2. ## Re: Is V in Span(S)

Alright, let's us definition as you laid out. If v is a linear combination of vectors in S, then there exists constants $c_i$ such that

$c_1(-1,2,1) + c_2(3,1,2) + c_3(1,5,4) + c_4(-6,5,1) = (-5,3,0)$

by comparison we get the simultaneous system of equations

$-c_1 + 3c_2 + 1c_3 + -6c_4 = -5$
$2c_1 + c_2 + 5c_3 + 5c_4 = 3$
$1c_1 + 2c_2 + 4c_3 + 1c_4 = 0$

You can solve this system using a matrix. If such $c_i$ exist, then you have show v is in a span of S. If no such $c_i$ exist then v is not in the span of S.

3. ## Re: Is V in Span(S)

Alright, let's us definition as you laid out. If v is a linear combination of vectors in S, then there exists constants $c_i$ such that

$c_1(-1,2,1) + c_2(3,1,2) + c_3(1,5,4) + c_4(-6,5,1) = (-5,3,0)$

by comparison we get the simultaneous system of equations

$-c_1 + 3c_2 + 1c_3 + -6c_4 = -5$
$2c_1 + c_2 + 5c_3 + 5c_4 = 3$
$1c_1 + 2c_2 + 4c_3 + 1c_4 = 0$

You can solve this system using a matrix. If such $c_i$ exist, then you have show v is in a span of S. If no such $c_i$ exist then v is not in the span of S.
So when I row reduced that matrix, I ended up with [[1,0,2,3,2],[0,1,1,-1,-1],[0,0,0,0,0]]

Seeing as this means that there are infinitely many solutions does this mean that v is NOT in span(S)?

4. ## Re: Is V in Span(S)

No, the solution for $c_i$ need not be unique in order for v to be in the span (choose one linear combination and you're done). In fact, you are bound to have at least one free variable since you have four variables and three equations.

5. ## Re: Is V in Span(S)

Ok, so because there is a solution, even if it is infinitely many, we can say that v is in span s? Thank you very much for your time, you have been a great help by the way! So how could v be expressed as a linear combination of the vectors from s then? Am i correct in saying that it would be:

if we let c_3=t and c_4=s

(2-2t-3s)(-1,2,1)+(-1-t-s)(3,1,2)+t(1,5,4)+s(-6,5,1)=(-5,3,0)

6. ## Re: Is V in Span(S)

This looks correct. Well done.

7. ## Re: Is V in Span(S)

Thank you so much for your time!! I was dreading trying to figure out how to do this question, you have been such a great help

8. ## Re: Is V in Span(S)

No problem, I've been there.

I also forgot to mention the following; if you are comfortable in the rationale in setting up a matrix you can "short-cut" by inserting the spanning vectors into the matrix as columns from left to right (doesn't matter the order you insert the vectors in) with the vector that you want spanned at the end. (set up a couple of times and you'll see why)

9. ## Re: Is V in Span(S)

Thanks I'll be sure to check it out!