# Can you find a basis without deg. 2 polynomials?

1. Oct 1, 2013

### tamintl

1. The problem statement, all variables and given/known data
Can you find a basis {p1, p2, p3, p4} for the vector space ℝ[x]<4 s.t. there does NOT exist any polynomials pi of degree 2? Justify fully.

2. Relevant equations

3. The attempt at a solution
We know a basis must be linearly independant and must span ℝ[x]<4. So intuitively if there are no polynomials of degree 2 we cannot span V. It just isn't possible.

I'm just struggling to justify it rigorously.

Many thanks - I appreciate the time and help.

2. Oct 1, 2013

### ppham27

Yes. Try $\{1,x,x^3+x^2,x^3\}$.

3. Oct 1, 2013

### tamintl

Nice. Can't believe I didn't see that. It is clearly linearly independant. Is it enough to say that since $\{1,x,x^3+x^2,x^3\}$ is a linear combination of pi that it will span the vector space?

Thanks

4. Oct 1, 2013

### D H

Staff Emeritus
Consider {1,x,ax3+bx2,cx3+dx2}. That's obviously a linear combination of {1,x,x2,x3}. What happens if ad=bc?

5. Oct 1, 2013

### arildno

Another simple choice for a basis is, for 4 different constants a,b,c,d to look at

(x-b)(x-c)(x-d), (x-a)(x-c)(x-d), (x-a)(x-b)(x-d), (x-a)(x-b)(x-c)

These polynomials are closely related to those we typically call Lagrange interpolation polynomials.

6. Oct 1, 2013

### tamintl

If ad=bc, then we have linear independance, thus a basis.

7. Oct 1, 2013

### D H

Staff Emeritus
Try that again. Consider a=b=c=d=1 as a simple example of ad=bc. Do you really think that {1,x,x3+x2,x3+x2} is a basis?

8. Oct 1, 2013

### arildno

To transform one valid basis to another valid basis, think of each function in the base as a vector component. Say we've got n basis functions.
Now, we may transform that n-vector into another n-vector by having it multiplied with an n*n matrix with constant coefficients.

What do you think will be a condition under which the new vector of function components will necessarily be a linear independent set of functions?

Last edited: Oct 1, 2013
9. Oct 1, 2013

### tamintl

ok - taking a=b=c=d=1. our basis would be:{1,x,x3+x2}

I am not sure where this is going? the question asked for a basis consisting of 4 vectors.

Would an alternative method to show that we have a basis be the following:

Take {1,x,x3+x2,x3} which is linear independant.

Now we have e1, e2, e3={x3+x2}-{x3}, e4 and so we have the standard basis. It follows that since the span contains the standard basis, it contains all of ℝ[x]<4.

10. Oct 1, 2013

### arildno

Incorrect.
DH's set is {1,x,x3+x2, x3+x2}
That set is a linearly dependent set.