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Can you find a basis without deg. 2 polynomials?

  • Thread starter tamintl
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  • #1
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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 pi 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.
 

Answers and Replies

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

Thanks
 
  • #4
D H
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Nice. Can't believe I didn't see that. It is clearly linearly independant. Is it enough to say that since [itex]\{1,x,x^3+x^2,x^3\}[/itex] is a linear combination of pi 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?
 
  • #5
arildno
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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
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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.
 
  • #7
D H
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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
arildno
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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:
  • #9
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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.
 
  • #10
arildno
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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.
 

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