# Polynomial Vector Spaces

1. Jun 17, 2008

### aredian

1. The problem statement, all variables and given/known data
Let S be the subspace P3 consisting of all polynomials P(x) such that p(0) = 0, and let T be the subspace of all polynomials q(x) such that q(1) = 0. Find a basis for S, T and S$$\cap$$T

2. Relevant equations

3. The attempt at a solution
I know that a basis is formed by linearly independant vectors which also generate the space thay belong to. And is true for polynomials that L.I. can be drawn for the coeff matrix that results from the vector grouping the terms by the powers of X.
What I am not sure is what p(0) = 0 and q(1) = 0 means. Is it 0, 0X, 0X^2 = 0 and 1, x, x^2 = 0?

I know it is supposed to be a simple question or at least that's how I see it, but I took my last math course about 7 years ago, and I can't find a resource to clarify that for the moment.

2. Jun 17, 2008

### Dick

3. Jun 17, 2008

### aredian

OK... Now Im a bit more confused. Does that means the vectors in p(0) = 0 grouped by the powers of X would be of the form (0 0 1)$$^{T}$$?

If so, then I take say 2 vectors, A and B, and need to verify they are L.I. and if they span S in order to declare them a basis of S.

How do I get a basis out of 2 vectors of the form (0 0 1)$$^{T}$$?

4. Jun 17, 2008

### Dick

If you are grouping powers in the order 1,x,x^2, then p(x)=a+bx+cx^2 becomes (a,b,c)^T. p(x)=0 tells you a=0. So the vectors in S look like (0,b,c)^T for any choice of b and c. Now tell me what a basis is.

5. Jun 17, 2008

### matt grime

You have chosen to identify the space you call P3 with column vectors with 3 elements. How have you done this? How did you associate the polynomials with constant coefficient 0 with a single vector (0,0,1)^t ?

6. Jun 17, 2008

### aredian

Since the vectors are of the type (a, b, 0) grouping by x^2, x, 1 then taking a linear combination of these, I can tell a(1,0,0)^T + b(0,1,0) = (x^2, x, 1) and use it to determine if they span S. They DO!. On the other hand the coeff matriz [(1,0,0)^T (0,1,0)^T] is non singular so these vectors are L.I. Thus a basis for S would be of the form {x^2, x}

Correct?

Thank you very much!!!

7. Jun 17, 2008

### Dick

Correct.

8. Jun 17, 2008

### aredian

Great!

Now for T... Are the vectors of T (p(1) = 0) of the form (a, b, -(a+b)) ? grouping well x^2, x, 1

Im not sure about the vectors in S $$\cap$$T. Are they of the form (a, b, c) for c=0 or only (a, b)?

Last edited: Jun 17, 2008
9. Jun 17, 2008

### Dick

They are of the form (a,b,c) where a+b+c=0, right? Can you use that relationship to eliminate a variable in the vector and find a basis? A basis vector for S intersect T has to be in both spaces.

10. Jun 17, 2008

### aredian

Ok... since c = -a -b I can use
ax^2 + bx + (-a -b)1. Correct?

This would yield the vectors of the form (a, b, -(a+b)) and their lineal combination would be of the form a(1,0,-1)^T + b(0,1,-1)^T = (x^2, x, 1). The coeff matrix is non singular so they are LI. However something is wrong with my algebra, because I can't seem to find the correct coeffs, such that they span T.

Where these the corect vectors at all?!?!?!?!

11. Jun 17, 2008

### Dick

Ok, a general vector in T is (a,b,-(a+b)) as you said. a*(1,0,-1)+b*(0,1,-1)=(a,b,-(a+b)), as you implied. So it looks me like they span.

12. Jun 18, 2008

### aredian

Yes indeed and the base is of the form {x^2-1, x-1}

Thanks!