# Homework Help: Basis for a Subspace.

1. Aug 9, 2007

### tsMore

I am drawing some strange mental blank with one question in my final exam review.

1. The problem statement, all variables and given/known data

V is the set of all polynomials that are of the form p(x) = cx^2 + bx+a
U is a subset of V where all members satisfy the equation p(5) =0

Find a basis for U.

I am not sure why I am having so much of a problem with this one, it shouldn't be that hard. With 5 subed in you get polynomials of the form

25x^2+5x+a = 0

I guess I am having a hard time relating this a back to a basis. Isn't the basis for V just {x^2, x, 1)?

2. Aug 9, 2007

### Dick

Wrong. With 5 subbed in you get 25*c+5*b+a=0. That's a relation between the coefficients - not all polynomials satisfy it. The subspace has dimension 2 (why?), so can you find two linearly independent polynomials that do?

3. Aug 9, 2007

### malawi_glenn

Polynomials satisfying p(5) =0 on the form p(x) = cx^2 + bx+a can be expressed as factors: p(x) = A(x-5)(x+B) ; where A and B is a real number.

Edit: Dick was 5 seconds before me on this one too =P

4. Aug 9, 2007

### Dick

Ok, I'll close my eyes and count to 5 before I answer the next one.

5. Aug 9, 2007

### tsMore

Wow that was fast :) And you are probably hitting on exactly why I am confused as well. Guess I am am just phasing out with this.

I actually don't know why the subspace would need only two polynomials to span the space defined by p(5) =0. I was actually thinking it was three!
How did you determine that it was 2?

You are saying that 25*c+5*b+a=0 is the relation between the coefficients, but not all the polynomials for which p(5)=0 need to statisfy this. What am I missing there? I was trying to relate the coefficents together some how to come up with a definition of the space.

thanks again so much!

6. Aug 9, 2007

### Dick

All of the polynomials such that p(5)=0 DO satisfy 25*c+5*b+a=0!!! Name one that doesn't. V is spanned by your basis {1,x,x^2}, dimension 3. U is a subspace, it has smaller dimension. I can easily think of two polynomials that span it. I agree that you may be phasing out. Step out and get a breath of fresh air and take another look at the problem.

7. Aug 9, 2007

### HallsofIvy

The vector space P2, of all quadratic polynomials is 3 dimensional and is spanned by {x2, x, 1}. Any member of that space is of the form cx2+ bx+ a. The subspace of such polynomials for which f(5)= 0 must satisfy 25c+ 5b+ a= 0. All that has been said before.

From 25c+ 5b+ a= 0, you can get a= -25c-5b. In other words, all polynomials in that subspace are of the form (-25c- 5b)x2+ bx+ c.

One method I really like for finding a basis for such a subspace is to take the parameters (here b and c) equal to 1, 0 and 0, 1 successively. If b= 1 and c= 0, then (-25c- 5b)x2+ bx+ c is simply -5x1+ 5x and then, if b=0 and c= 1, -25x2+ 1. Those two polynomials form a basis for the 2 dimensional subspace of quadratic polynomials f(x), such that f(5)= 0.