1. ## Sapnning Sets

So to determine if a set is a spanning set, you find if the determinant is a non-zero, yes? What do you do for non-square matricies?

2. Originally Posted by kaylakutie
So to determine if a set is a spanning set, you find if the determinant is a non-zero, yes? What do you do for non-square matricies?

Your question makes no sense: think carefully what you want to ask and then ask it within the proper context. This is college/university level stuff, we must try to be a little more accurate.

Tonio

3. Oh, sorry. Did you know what I meant though? I turned the set into a matrix by multiplying by scalars c1, c2..cn. By looking at them as like linear combinations.

Ex:
S={2,x,x+3,3x^2}

(0,0,0) = c1(0,0,2)+c2(0,1,0)+c3(0,1,3)+c4(3,0,0).
And then it becomes a matrix from this system:
0c1+0c2+0c3+3c4=0
0c1+1c2+1c3+0c4=0
2c1+0c1+3c3+0c4=0

and it reduced to
[[1, 0, 3/2, 0], [0,1,1,0], [0,0,0,1]]

So I said non-trivial and therefore dependent. And now I want to see if it's a spanning set so that i can tell if it forms a basis for p2. I guess it wouldn't because it's dependent anyways.. but I'd still like to know how to find if something if it spans p2 / if it's a spanning set. Thanks you!!

4. Originally Posted by kaylakutie
Oh, sorry. Did you know what I meant though? I turned the set into a matrix by multiplying by scalars c1, c2..cn. By looking at them as like linear combinations.

Ex:
S={2,x,x+3,3x^2}

(0,0,0) = c1(0,0,2)+c2(0,1,0)+c3(0,1,3)+c4(3,0,0).
And then it becomes a matrix from this system:
0c1+0c2+0c3+3c4=0
0c1+1c2+1c3+0c4=0
2c1+0c1+3c3+0c4=0

and it reduced to
[[1, 0, 3/2, 0], [0,1,1,0], [0,0,0,1]]

So I said non-trivial and therefore dependent. And now I want to see if it's a spanning set so that i can tell if it forms a basis for p2. I guess it wouldn't because it's dependent anyways.. but I'd still like to know how to find if something if it spans p2 / if it's a spanning set. Thanks you!!

Oh, now we're talking college/university level...almost! So you have a set of polynomials and you want to check whether this set spans some vector space of polynomials...right? What space I've no idea, but your writing the polynomials as coordinate vectors is a very good idea to find out whether the set is linearly independent or not. Sinceyou wrote the polynomials as 3-D coordinate vectors I deduce we're talking here of the v.s. of all (real...? complex...?) polynomials of degree less than or equal 2, and since this v.s.'s dimension is 3 we already know this set must be lin. dependent, so its only chance to be a spanning set is if we can dig out of it a set with 3 lin. ind. vectors. What you did is correct, and now it isn't hard to check that $\displaystyle \{2,x,3x^2\}$ is lin. ind. and thus a basis and, thus again, a spanning set of the vector space (if I guessed correctly and the v.s. is what I said it is).

Tonio

5. Thanks. I think I get it. The x+3 is redundant, thus linearly dependent. So while it does span p2, it does not form a basis for S in p2. However, the set, {2,x,3x^2} spans p2 AND is linearly independent, so that one is thus a basis. Not the original set though.

Yes?

6. Another one from our text book.
15. Find a basis for the subspace of R^3 spanned by S.
{(4,4,8),(1,1,2),(1,1,1)}

It would seem to me that (4,4,8) can be written as a linear combination of the other two (4 times (1,1,2)). So I would think that it would be dependent. And (1,1,2) and (1,1,1) by themselves can't span R^3. I would think there would be no basis for this one. But the book lists this as an answer:

{(1,1,0),(0,0,1)}

Looks like they got that by row reducing. But is the book wrong or am I super confused still? I would say there is no basis.

7. Or do they mean a subspace within R^3.. not R^3 as a whole? ??????

8. Originally Posted by kaylakutie
Thanks. I think I get it. The x+3 is redundant, thus linearly dependent. So while it does span p2, it does not form a basis for S in p2. However, the set, {2,x,3x^2} spans p2 AND is linearly independent, so that one is thus a basis. Not the original set though.

Yes?

Indeed.

Tonio