Linear Inependence in polynomial space

[phil ess]

Homework Statement

Determine whether the following set is linearly independent in P₄

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

Homework Equations

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?

 

[JG89]

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

 

[Dick]

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

 

[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? I am still kind of confused about all of this.

 

[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.

 

[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.

 

[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 [itex]{1+ x^2/2, 1- x^2/2, x+ x^3/6, x- x^3/6}[/tex] is independent if and only if $a_1(1+ x^2/2)+ a_2(1- x^2/2)+ a_3(x+ x^3/6)+ a_4(x- x^3/6)= 0$ 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.

 

[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?

 

[Dick]

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

 

[D H]

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 $a_1= a_2= a_3= a_4$.

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

 

[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.

 

[HallsofIvy]

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

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