Thread: Given 5 polynomiums... How do you decide which is a basis for a vector space?

Okay, I think I get it now. :-)
So to use a new example:
If we have the subspace P(4)=0 in P_2(R), and there are the following polynomials:


In order to satisfy P(4)=0 we know that 1 root is (x-4) since x=4 and inserted it gives us 4-4=0. Thus if we look at the polynomials and factorize them with this root, we can sort out which polynomials that fit as a basis:

The first polynomial satisfies since: (x-4)(-0.5-0.5)=4-x
The second polynomial cannot be factorized with (x-4) so we'll sort that one out.
The third polynomial satisfies: (x-4)(1-3)=8-2x
The fourth polynomial satisfies: (x-4)(x-4)=16-8x+x^2
The fifth polynomial satisfies aswell: (x-4)(x-3)=12-7x+x^2

So 4 of our 5 polynomials seem to be a basis in our subspace which means that our subspace has 4 elements and thus 4 dimensions.

Was this a correct procedure?

no, you have to check for linear independence. P2(R) has only 3 dimensions to begin with, how could a 4 dimensional space be a subspace?

and the subspace has lots more than 4 elements, subspaces have an infinite number of elements. you're looking for a basis. a basis isn't just any collection of elements. they have to be linearly independent, and they have to span. this is important. you have to eliminate polynomials that can be written as linear combinations of other ones in your set.

Ah yes, sorry my bad.
So if we create an equation system with the 4 polynomials that we have and put them into a matrix and then apply reduced row echelon, we find that its rank is 2.
That means that our subspace has the dimension 2 since it's rank is equal to its dimension.
Therefore we need to find 2 elements (2 of our 4 polynomials) that together are linear independent and thus create a basis for our subspace.

Looking at the polynomials, (4-x) and (8-2x) are linear dependent since one is just multiplied with a factor 2.
But 16-8x+x^2 and 12-7x+x^2 are together ilnear independent and thus they form a basis for the subspace.
So these two polynomials are 1 out of many posibilities for a basis in the subspace.

well a basis has to be linearly independent...but to span the subspace we're looking at, they individual polynomials also have to BE in that space.