Find basis for the subsp of P2 spanned by the vectors....

**v **= 1+x^{2}

**q **= x^{2}

**r **= -2+2x^{2}

**s **= -3x

dim P_{2} = 3 (c1, x, x^{2})

c1 c2 2c3 0 0

0 0 0 -3c4 0

c1 0 -2c3 0 0

I know a basis in linearly independent c1 = c2 = c3 = c4 = 0

I know that to span means the minimum amount of vectors required to "encompass" a subspace without a vector being a linear combination of the others, and therefore if one of the vectors is removed the vectors wouldn't account for all the possible combinations in the space.

I think I'm trying to show that I can remove one of these vectors and still account for all the combinations....or maybe I have no idea what I'm doing. Do I just start removing vectors and see if they are linearly independent and spanning? Just looking seems like I could remove **q **and **r**. Please help me get started with the tactics of this problem. Thanks

Anthony

Re: Find basis for the subsp of P2 spanned by the vectors....

, so a basis for would consist of 3 linearly independent vectors, so you would just have to remove one vector from your list and then show that the others are independent and span an arbitrary polynomial.

What you had:

If we want to see if the vectors span any arbitrary polynomal , then we have the matrix

What might be simpler:

Try constructing r as a linear combination of p and q, then remove from the list. then apply the same logic as above with the linear combination of p,q,s to see what the constants have to be (dependent on a,b,d in the polynomial)

Re: Find basis for the subsp of P2 spanned by the vectors....

Ok.

So I proved that r was a linear combination where c1 = -2 and c2 = 4

then I checked that the matrix

1 1 0 0

0 0 -3 0

1 0 0 0

gave me c1 = c2 = c3 = 0 and it did. This means they're linearly independent. Now I'm left to make sure that I span

1 0 1 ax^{2}

0 -3 0 bx

1 0 0 c

where column one is c1, two is c2 and three is c3. C2 = -b/3. c3 = a-c but what is c1? I've confused myself

Re: Find basis for the subsp of P2 spanned by the vectors....