Can you find a basis without deg. 2 polynomials?

[tamintl]

Homework Statement

Can you find a basis {p₁, p₂, p₃, p₄} for the vector space ℝ[x]_<4 s.t. there does NOT exist any polynomials p_i of degree 2? Justify fully.

Homework Equations

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.

 

[ppham27]

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

 

[tamintl]

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

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 p_i that it will span the vector space?

Thanks

 

[D H]

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 p_i that it will span the vector space?

Thanks

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

 

[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.

 

[tamintl]

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

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

 

[D H]

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?

 

[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?

 

[tamintl]

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?

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.

 

[arildno]

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.
.

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