Can this work as a basis for S?

  • #1
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Let ##S## be a set of all polynomials of degree equal to or less than ##n## (n is fixed) and ##p(0)=p(1)##.

Then, a sample element of ##S## would look like:
$$
p(t) = c_0 + c_1t +c_2t^2 + \cdots + c_nt^n
$$
Now, to satisfy ##p(0)=p(1)## we must have
$$
\sum_{i=1}^{n} c_i =0
$$

What could be the possible bases for S? I thought of one of them and it looks like this

$$
A = \{ 1, c_1t +c_2t^2, c_1t +c_2t^2+c_3t^3, \cdots c_1t +c_2t^2 ... +c_nt^n | \sum_{i=1}^{j} c_i =0 ; j=1,2,3 ... n\}
$$

Is A a basis for S? I mean, I'm unable to disprove it.
 

Answers and Replies

  • #2
PeroK
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That solution is not very explicit. You really want to find ##n## explicit, linearly independent polynomials.
 
Last edited:
  • #3
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What do you have to check, in order to prove that ##S## is a subspace?
 
  • #4
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What do you have to check, in order to prove that ##S## is a subspace?
S should be a subset of a linear space of all polynomials of degree equal to or less than n and S should satisfy closure axioms.
 
  • #5
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S should be a subset of a linear space of all polynomials of degree equal to or less than n and S should satisfy closure axioms.
Yes. You don't need a basis, you only have to show that ##p(x)+q(x)\, , \, \lambda p(x) \in S## whenever ##p(x),q(x) \in S## and ##\lambda \in \mathbb{F}## whatever the scalar field ##\mathbb{F}## is.
 
  • #6
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Yes. You don't need a basis, you only have to show that ##p(x)+q(x)\, , \, \lambda p(x) \in S## whenever ##p(x),q(x) \in S## and ##\lambda \in \mathbb{F}## whatever the scalar field ##\mathbb{F}## is.
Yes, but I wanted to find a basis for S.
 
  • #7
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Yes, but I wanted to find a basis for S.
In which case you must be more specific. E.g. you defined ##A=\{1\}## which is only a basis for ##n=1.##

And after you have found polynomials which are not all zero, then you still have to prove linear independence and determine the dimension of ##S.##
 
  • #8
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In which case you must be more specific. E.g. you defined ##A=\{1\}## which is only a basis for ##n=1.##

And after you have found polynomials which are not all zero, then you still have to prove linear independence and determine the dimension of ##S.##
I wrote that S is a set of all polynomials of degree ##\leq n## and ##p(0)=p(1)## for all p(t) belonging to S. And A is not a singleton set. ##p(t)=1## does satisfy the qualifications for belonging to S.
 
  • #9
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In which case you must be more specific. E.g. you defined ##A=\{1\}## which is only a basis for ##n=1.##
What could be the possible bases for S?
I thought of one of them and it looks like this
$$
A = \{ 1, c_1t +c_2t^2, c_1t +c_2t^2+c_3t^3, \cdots c_1t +c_2t^2 ... +c_nt^n | \sum_{i=1}^{j} c_i =0 ; j=1,2,3 ... n\}
$$
You must have misread post #1, in which he wrote the above.
 
  • #10
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You must have misread post #1, in which he wrote the above.
You must have failed to solve the linear equation system.
 
  • #11
PeroK
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I wrote that S is a set of all polynomials of degree ##\leq n## and ##p(0)=p(1)## for all p(t) belonging to S. And A is not a singleton set. ##p(t)=1## does satisfy the qualifications for belonging to S.
I'm not sure where this thread is going. A basis is a set of specific vectors. It's not a set of equations.
 
  • #12
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The definition of ##A## is ##A=\{1\}##.

What the OP wanted to define is
$$
A = \{ 1, c_1^{(1)}t +c_2^{(1)}t^2, c_1^{(2)}t +c_2^{(2)}t^2+c_3^{(2)}t^3, \cdots c_1^{(n-1)}t +c_2^{(n-1)}t^2 ... +c_n^{(n-1)}t^n | \sum_{i=1}^{j} c_i^{(k)} =0 ; j=1,2,3 ... n;k=1,\ldots,n-1\}
$$
which still is an infinite set of vectors.
 
  • #13
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I'm not sure where this thread is going. A basis is a set of specific vectors. It's not a set of equations.
All right, let me once again try to make myself clear.

##S## is a set of all polynomials of degree ##\leq n## which satisfy the following condition
$$
p(0)=p(1)
$$

Find a basis for S.
 
  • #14
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The definition of ##A## is ##A=\{1\}##.

What the OP wanted to define is
$$
A = \{ 1, c_1^{(1)}t +c_2^{(1)}t^2, c_1^{(2)}t +c_2^{(2)}t^2+c_3^{(2)}t^3, \cdots c_1^{(n-1)}t +c_2^{(n-1)}t^2 ... +c_n^{(n-1)}t^n | \sum_{i=1}^{j} c_i^{(k)} =0 ; j=1,2,3 ... n;k=1,\ldots,n-1\}
$$
which still is an infinite set of vectors.
I'm sorry if I was not very clear but in very beginning I wrote that "n is fixed"
 
  • #15
PeroK
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All right, let me once again try to make myself clear.

##S## is a set of all polynomials of degree ##\leq n## which satisfy the following condition
$$
p(0)=p(1)
$$

Find a basis for S.
I understand that. You need to find a suitable set of polynomials.
 
  • #16
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I'm not sure where this thread is going. A basis is a set of specific vectors. It's not a set of equations.
Just wanted to clear, "A basis is a set of specific vectors" or "A basis is a set of specific mathematical objects" ?
 
  • #17
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I'm sorry if I was not very clear but in very beginning I wrote that "n is fixed"
##A## is ill-defined! See post #12. You defined all ##c_j=0##.
 
  • #18
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##A## is ill-defined! See post #12. You defined all ##c_j=0##.
I wrote the sum of c_i's to be zero.
 
  • #19
PeroK
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Just wanted to clear, "A basis is a set of specific vectors" or "A basis is a set of specific mathematical objects" ?
If we are talking about a basis for a vector space, then the appropriate mathematical objects are vectors. Although, in this case the vectors are polynomials.
 
  • #20
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If we are talking about a basis for a vector space, then the appropriate mathematical objects are vectors. Although, in this case the vectors are polynomials.
Vectors are polynomials?
 
  • #22
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Let's leave my A, and start afresh for finding a basis for S.
 
  • #24
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Let's leave my A, and start afresh for finding a basis for S.
You must understand how sets are defined! You used the same set of coefficients for all of your polynomials and that is a big mistake. Furthermore, you have to specify all ##2+3+\ldots + n## coefficients in order to specify a certain basis.
 
  • #25
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Vectors are polynomials?
They can be. Specific examples of vector spaces are ##\mathbb R^n, \mathbb C^n##, sets of matrices, sets of linear operators, sets of polynomials, sets of functions. Sometimes a vector space generally is called a linear space. I would have thought that was all clear given your attempt at this problem!
 

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