Linear Inependence in polynomial space

1. Nov 25, 2008

phil ess

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

Determine whether the following set is linearly independent in P4

{1+(x2/2) , 1-(x2/2) , x+(x3/6) , x+(x3/6)}

2. Relevant equations

3. The attempt at a solution

Well I know that if any of the entries in the set can be written as the linear combination of the others, then the set is NOT linearly independent. But how can I show this without just trying to put together random combinations hoping to find one that results in another entry in the set?

2. Nov 25, 2008

JG89

In this situation I think it's best to see if a linear relationship has any non trivial solutions.

3. Nov 25, 2008

Dick

You know the functions 1,x,x^2 and x^3 are linearly independent. Use that.

4. Nov 25, 2008

phil ess

Im not sure what youre getting at Dick. From what youve said, I can see that each entry in the set can be expressed as a linear combination of {1,x,x2,x3}, so that means they are not linearly independent? {1,x,x2,x3} is a basis for P4 i think, so is that all there is to it? Im still kind of confused about all of this.

5. Nov 25, 2008

JG89

phil, in the set {1+(x2/2) , 1-(x2/2) , x+(x3/6) , x+(x3/6)}

if you do 1*(x+(x^3/6)) - 1*(x+(x^3/6)), that equals 0. By choosing 0 for both scalars for the first two vectors, you have shown that there is a non-trivial relationship, thus the set is linearly dependent.

6. Nov 25, 2008

Dick

To do it systematically, write down a*(1+x^2/2)+b*(1-x^2/2)+c*(x+x^3/6)+d*(x+x^3/6)=0 and see if it has a nontrivial solution. As JG89 points out that's pretty easy here, since the last two vectors are the same. a=b=0 and c=1 d=-1. If that hadn't been the case then group them by powers of x. Since 1,x,x^2 and x^3 are linearly independent, the coefficient of each must vanish. That's four equations in four unknowns.

7. Nov 25, 2008

HallsofIvy

The set you give is trivially dependent, as JG89 said, because two of the vectors are equal. I wonder if you didn't mean x- x3/6.

A set of vectors $\left{v_1, v_2, v_3, v_4}$ is independent if and only if $a_1v_+ a_2v_2+ a_3v_3+ a_4v_4= 0$ implies $a_1= a_2= a_3= a_4= 0$.
The set ${1+ x^2/2, 1- x^2/2, x+ x^3/6, x- x^3/6}[/tex] is independent if and only if [itex]a_1(1+ x^2/2)+ a_2(1- x^2/2)+ a_3(x+ x^3/6)+ a_4(x- x^3/6)= 0$ for all x implies that $a_1= a_2= a_3= a_4$. You can reduce that to a system of numerical equations by combining "like terms" and setting the coefficients equal to 0 or just setting x equal to 4 distinct values.

Last edited by a moderator: Nov 26, 2008
8. Nov 25, 2008

phil ess

Upon further inspection that last term is actually x-(x3/6) , just like you said HallsofIvy. You guys mention grouping by powers, does that mean I group the first two, which are degree 2, and the second two which are degree 3?

9. Nov 25, 2008

Dick

There are four groups, corresponding the powers 1,x,x^2 and x^3.

10. Nov 25, 2008

D H

Staff Emeritus
Surely you mean a1=a2=a3=a4=0, not just that a1=a2=a3=a4.

11. Nov 25, 2008

Johnson04

You may also use the equivalence between the P4 and R4. That is to say, you may construct a relation (function): l(x)=a1 + a2*x + a3*x2 + a4*x3, where x = (a1, a2, a3, a4). So in your case the set of vectors is actually equivalent to {(1,0,1/2,0), (1,0,-1/2,0), (0,1,0,1/6), (0,1,0,-1/6)}. Now it might be straightforward to determine if this group of vectors are linear independent.

12. Nov 26, 2008

HallsofIvy

Yes, thank you for catching that. I have corrected my response.