I Is This a Valid Basis for the Set of Polynomials with \( p(0) = p(1) \)?

[Hall]

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.

 

[PeroK]

That solution is not very explicit. You really want to find $n$ explicit, linearly independent polynomials.

 

[fresh_42]

What do you have to check, in order to prove that $S$ is a subspace?

 

[Hall]

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.

 

[fresh_42]

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.

 

[Hall]

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.

 

[fresh_42]

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 $S.$

 

[Hall]

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.

 

[Mark44]

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.

 

[fresh_42]

You must have misread post #1, in which he wrote the above.

You must have failed to solve the linear equation system.

 

[PeroK]

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.

 

[fresh_42]

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.

 

[Hall]

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.

 

[Hall]

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"

 

[PeroK]

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.

 

[Hall]

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" ?

 

[fresh_42]

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 $c_j=0$.

 

[Hall]

$A$ is ill-defined! See post #12. You defined all $c_j=0$.

I wrote the sum of c_i's to be zero.

 

[PeroK]

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.

 

[Hall]

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?

 

[fresh_42]

I wrote the sum of c_i's to be zero.

Read post number 12.

 

[Hall]

Let's leave my A, and start afresh for finding a basis for S.

 

[fresh_42]

Vectors are polynomials?

The other way around: polynomials can be vectors.

 

[fresh_42]

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.

 

[PeroK]

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!

 

[Hall]

I would have thought that was all clear given your attempt at this problem!

Tom Apostol never uses the word vector space in his Calculus Vol 2, and I'm learning from that. He says linear space is a set of any mathematical objects that satisfy the axioms.

 

[Hall]

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.

Okay. Now, I see the point of post 12. Is the treated form of A a basis for S?

 

[PeroK]

Tom Apostol never uses the word vector space in his Calculus Vol 2, and I'm learning from that. He says linear space is a set of any mathematical objects that satisfy the axioms.

https://en.wikipedia.org/wiki/Vector_space

 

[fresh_42]

Okay. Now, I see the point of post 12. Is the treated form of A a basis for S?

Furthermore, you have to specify all $2+3+…+n$ coefficients in order to specify a certain basis.

What are the $c_i$? E.g. all $c_i=0$ satisfy your choices but are not a basis.

Then you have to prove that $\dim S =n$ and linear independence of your selected polynomials.

 

[Mark44]

You must have failed to solve the linear equation system.

A is ill-defined! See post #12. You defined all cj=0.

$\sum_{i = 1}^n c_i = 0$ doesn't necessarily mean that all of the constants $c_i$ must be zero.
For example, if n = 2, $p(x) = 1x^2 - 1x$ satisfies p(0) = 0 and p(1) = 0, and $\sum_{i = 1}^n c_i = 0$.

 

[fresh_42]

$\sum_{i = 1}^n c_i = 0$ doesn't necessarily mean that all of the constants $c_i$ must be zero.
For example, if n = 2, $p(x) = 1x^2 - 1x$ satisfies p(0) = 0 and p(1) = 0, and $\sum_{i = 1}^n c_i = 0$.

That was not what he wrote (see post #1). His definition implied all $c_i=0.$ If $j=1,2,\ldots,n$ is written in the definition of a set, it usually means for all $j.$

 

[Mark44]

That was not what he wrote (see post #1). His definition implied all $c_i=0.$ If $j=1,2,\ldots,n$ is written in the definition of a set, it usually means for all $j.$

I disagree. Here's what he wrote in post #1.

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

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

He did ***not** write $c_i = 0, i \in \{1, \dots, j\}$. Rather it was the sum of the $c_i$.

 

[fresh_42]

I disagree. Here's what he wrote in post #1.

He did ***not** write $c_i = 0, i \in \{1, \dots, j\}$. Rather it was the sum of the $c_i$.

It does not matter how often you repeat it. Wrong remains wrong. The upper limit of the sum is $j$, not $n$. Please be precise if you quote others.

\begin{align*}
\begin{bmatrix}
1&1&\ldots&1\\
0&1&\ldots&1\\
0&0&\ldots&1\\
&\vdots&\ddots&\\
0&0&\ldots&1\\
\end{bmatrix}
\end{align*}
is a regular matrix.

 

[Mark44]

It does not matter how often you repeat it. Wrong remains wrong. The upper limit of the sum is $j$, not $n$. Please be precise if you quote others.

Then what the OP wrote was at least ambiguous. The first summation he wrote was $\sum_{i = 1}^n c_i = 0$, which is what I was focused on. In the second summation, I don't believe he intended for all of the constants to be zero.
What I believe he was trying to get across was a set something like this: $\{x^2 - x, x^3 - 2x^2 + x, \dots \}$. Including 1 in the set as he did violates the conditions that p(0) = p(1) = 0

 

[fresh_42]

What I believe he was trying ...

This is irrelevant. He made a definition error by defining $A$ and I pointed this out. It is important to learn how to write sets. I even explained it in detail in post #12. Guessing what could have been meant is ineffective.

 

[PeroK]

@Hall I advise posting future problems in the Homework section, where the rules and guidelines focus the helpers on helping you!

 

[fresh_42]

@Hall I advise posting future problems in the Homework section, where the rules and guidelines focus the helpers on helping you!

Sorry. I thought I would have helped by correcting the definition of $A$. I guess someone wants to scare me away.

 

[PeterDonis]

He did ***not** write $c_i = 0, i \in \{1, \dots, j\}$. Rather it was the sum of the $c_i$.

He wrote that the sum of $c_i$ for $i$ from 1 to $j$ is zero, for all values of $j$ from $1$ to $n$. That means that $c_1 = 0$ (since that's the only term in the sum for $j = 1$), and then, by induction, all of the other $c_i$ for $i$ from $1$ to $n$ must also be zero (because each time we increase $j$ by $1$ we include just one additional term, and all the previous terms are already known to be $0$ from the previous values of $j$).

That might not be what he intended to write, but that's what he wrote. If he intended to write something else, he should tell us what.

 

[PeterDonis]

a sample element of $S$ would look like:
$$
p(t) = c_0 + c_1t +c_2t^2 + \cdots + c_nt^n
$$

A sample element, yes. But for each different element of $S$, all of the $c_i$ will in general be different.

So if you are trying to find a basis for $S$, you should be thinking in terms of picking $n$ sets of values for the $c_i$, such that the resulting polynomials are all linearly independent and that any member of $S$ can be expressed as a linear combination of them.

The fact that each member of $S$ can be described by a set of $n$ numbers (the $c_i$) is what suggests the "vector" terminology. (Strictly speaking, to show that these sets of numbers actually are vectors requires proving that they satisfy the vector space axioms.)

 

[Office_Shredder]

I think fresh is right here. The op should pick the set A that they want, the set described in the original post is wrong.

Even if you pick the "right" constraint on the coefficients (where they don't all draw from the same coefficients), setting all of them equal to zero still satisfies the criteria, so no, the set A is not guaranteed to be a basis without additional constraints.