Basis for Polynomial Subspace

  • #1

Homework Statement



Let [tex]P_4(\mathbb{R})[/tex] be the vector space of real polynomials of degree less than or equal to 4.

Show that {[tex]{f \in P_4(\mathbb{R}):f(0)=f(1)=0}[/tex]}

defines a subspace of V, and find a basis for this subspace.

The Attempt at a Solution



Since [tex]P_4(\mathbb{R})[/tex] is spanned by (1,z,...,z4), I think that this subspace will be spanned by a list/set of 4 elements - since f(0)=0 means there is no constant term. And obviously, by the definition of basis, these elements will have to be linearly independent.

I really don't know how I would represent a basis for this subspace; I am inclined to consider x(x-1) as an element of degree 2. But it just seems wrong to me to fix that as an element in the basis: (z, z2-z, z3, z4).

But are not the roots of polynomials determined by their coefficients? Seems to me that I need to find some real coefficients so that f(0)=f(1)=0 is guaranteed.

So not only am I uncertain about how to express a basis of this subspace, but I am not even sure of this approach.

This subspace is defined by polynomials whose roots are 0, 1, and beyond that, any real number. As a subspace is closed under scalar multiplication, multiplying the elements of the basis (which I have yet to obtain) must produce elements with roots that include 0 and 1. But given how the roots are determined by the coefficients, I really don't know how to fix this relationship.

I would greatly appreciate any help and direction, since I'm probably confusing a lot of things at once.

Thanks in advance.
 

Answers and Replies

  • #2
radou
Homework Helper
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If f(x) = ax + bx^2 + cx^3 +dx^4 (after applying f(0) = 0) then, for f(1) = 0 you have a + b + c + d = 0. What does this tell you?
 
  • #3
If f(x) = ax + bx^2 + cx^3 +dx^4 (after applying f(0) = 0) then, for f(1) = 0 you have a + b + c + d = 0. What does this tell you?

Thanks for the quick reply, btw. I think I catch your meaning. That means the coefficients must sum to 0, for every element in the subspace(?).

Given your helpful reply, I don't feel inclined anymore to fix x(x-1) as an element. That is, the fact that 0 is a root follows from there being no constant term and likewise for f(1)=0 "restricting" the coefficients. If I'm right, would it be correct to express the basis as:

(z, z2, z3, z4), with the restriction that the coefficients a1 + a2 + a3 + a4 = 0

I don't know if I am correctly expressing the "restriction" on the coefficients. Or even if this is what you meant, since wouldn't closure under scalar multiplication forbid such restrictions on the coefficients? I'll have to consult the definition of that, again; but I'm just thinking out loud so that I don't forget.
 
Last edited:
  • #4
radou
Homework Helper
3,120
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a + b + c + d = 0 means that a = -b -c -d. Hence, after plugging in, b(x^2 - x) + c (x^3 - x) + d(x^4 - x) = f(x). What is {(x^2 - x), (x^2 - x), (x^4 - x)}?
 
  • #5
a + b + c + d = 0 means that a = -b -c -d. Hence, after plugging in, b(x^2 - x) + c (x^3 - x) + d(x^4 - x) = f(x). What is {(x^2 - x), (x^2 - x), (x^4 - x)}?

Ohh. I think I follow; by plugging in for a, the elements we obtained form a basis? That is, in each case of {(x2-x), (x3-x), (x4-x)}, 0 and 1 are roots: e.g., x2-x = x(x-1).
 
  • #6
radou
Homework Helper
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More precisely, for any real number b, c, d the polynomial is an element of your subspace. So, ({(x^2-x), (x^3-x), (x^4-x)}) spans the subspace, and if it's linearly independent (which it is, you should check), it's a basis for your subspace.
 

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