Existence of a Basis of a Vector Space

[toni07]

Let n be a positive integer, and for each $j = 1,..., n$ define the polynomial $f_j(x)$ by f_j(x) = $\prod_{i=1,i \ne j}^n(x-a_i)$

The factor $x−a_j$ is omitted, so $f_j$ has degree n-1

a) Prove that the set $f_1(x),...,f_n(x)$ is a basis of the vector space of all polynomials of degree ≤ n - 1 in x with coefficients in F.

b) Let $b_1,...,b_n$ in F be arbitrary (not necessarily distinct). Prove that there exists a unique polynomial g(x) of degree ≤ n - 1 in x with coefficients in F.

I don't know how to go about this question. Any help would be appreciated.

 

[Deveno]

Re: Assume that the field F has at least n distinct elements $a_1, …, a_n$

Since the set $\{f_1,\dots,f_n\}$ has $n$ elements, and $\text{dim}(P_{n-1}) = n$, it suffices to prove this set is linearly independent to show it is a basis.

So suppose we have $c_1,\dots,c_n \in F$ with:

$g = c_1f_1 + \cdots + c_nf_n = 0$ (the 0-polynomial).

Since for ALL $a \in F$, $g(a) = 0$, in particular, we must have $g(a_1) = 0$.

Now $a_1$ is a root of $f_2,\dots,f_n$, so:

$0 = g(a_1) = c_1f_1(a_1) + c_2f_2(a_2) + \cdots + c_nf_n(a_n)$

$= c_1f_1(a_1) + 0 + \cdots + 0 = c_1f(a_1)$.

Since the $a_i$ are distinct, we have that $f(a_1)$ is the product of $n-1$ non-zero elements of $F$, and thus is non-zero. Hence it must be the case that $c_1 = 0$.

By similarly considering $g(a_i)$ for each $i$, we see that all the $c_i = 0$, which establishes linear independence of the $f_i$.

Something is missing from your statement of part b)...

 

[toni07]

Re: Assume that the field F has at least n distinct elements $a_1, …, a_n$

Since the set $\{f_1,\dots,f_n\}$ has $n$ elements, and $\text{dim}(P_{n-1}) = n$, it suffices to prove this set is linearly independent to show it is a basis.

So suppose we have $c_1,\dots,c_n \in F$ with:

$g = c_1f_1 + \cdots + c_nf_n = 0$ (the 0-polynomial).

Since for ALL $a \in F$, $g(a) = 0$, in particular, we must have $g(a_1) = 0$.

Now $a_1$ is a root of $f_2,\dots,f_n$, so:

$0 = g(a_1) = c_1f_1(a_1) + c_2f_2(a_2) + \cdots + c_nf_n(a_n)$

$= c_1f_1(a_1) + 0 + \cdots + 0 = c_1f(a_1)$.

Since the $a_i$ are distinct, we have that $f(a_1)$ is the product of $n-1$ non-zero elements of $F$, and thus is non-zero. Hence it must be the case that $c_1 = 0$.

By similarly considering $g(a_i)$ for each $i$, we see that all the $c_i = 0$, which establishes linear independence of the $f_i$.

Something is missing from your statement of part b)...

Sorry, I didn't realize I omitted the last part.
b) Let $b_1,...,b_n$ in F be arbitrary (not necessarily distinct). Prove that there exists a unique polynomial g(x) of degree ≤ n - 1 in x with coefficients in F such that $g(a_i) = b_i$ for i = $1,..., n$ .

 

[Deveno]

Re: Assume that the field F has at least n distinct elements $a_1, …, a_n$

Suppose the $f_i$ are as in part (a).

We know that $f_i(a_i) \neq 0$, so because we are in a field $f_i(a_i)^{-1}$ exists.

Define:

$\displaystyle g(x) = \sum_{j = 1}^n \frac{b_j}{f_i(a_j)}f_j(x)$

Then $\displaystyle g(a_i) = \frac{b_i}{f_i(a_i)}f_i(a_i) = b_i$

(all the non-$i$ terms are 0).

This shows existence...can you show uniqueness?

(hint: what does being a basis mean in terms of the coefficients of the $f_i$?)

 

[blue_raver22]

a) To prove that the set $f_1(x),...,f_n(x)$ is a basis of the vector space of all polynomials of degree ≤ n - 1 in x with coefficients in F, we need to show that it is a linearly independent set and that it spans the vector space.

Linear Independence:
To show linear independence, we need to show that the only solution to the equation $c_1f_1(x)+...+c_nf_n(x)=0$ for $c_1,...,c_n \in F$ is $c_1=...=c_n=0$. Since $f_j(x) = \prod_{i=1,i \ne j}^n(x-a_i)$, we can rewrite the equation as $c_1\prod_{i=1,i \ne 1}^n(x-a_i)+...+c_n\prod_{i=1,i \ne n}^n(x-a_i)=0$. This can be further simplified to $c_1(x-a_2)...(x-a_n)+...+c_n(x-a_1)...(x-a_{n-1})=0$. Since $x-a_j$ is omitted from each term, the degree of each term is n-1. Therefore, the only way for this equation to be true is if $c_1=...=c_n=0$, which proves linear independence.

Span:
To show that the set spans the vector space, we need to show that for any polynomial $p(x)$ of degree ≤ n-1, there exist $c_1,...,c_n \in F$ such that $p(x)=c_1f_1(x)+...+c_nf_n(x)$. Since $p(x)$ has degree ≤ n-1, it can be written as $p(x)=a_0+a_1x+...+a_{n-1}x^{n-1}$. We can rewrite this as $p(x)=a_0+a_1(x-a_2)...(x-a_n)+...+a_{n-1}(x-a_1)...(x-a_{n-1})$. Comparing this to the equation in the linear independence proof, we can see that $c_j=a_j$ for $j=1,...,n$. Therefore, the set spans the vector space.

b) To prove that there exists

 

[blue_raver22]

As a scientist, it is important to approach problems with a logical and analytical mindset. In this case, we are dealing with polynomials and vector spaces, so it is important to have a solid understanding of these concepts before attempting to prove anything.

To begin, let us define the vector space of all polynomials of degree ≤ n-1 in x with coefficients in F. This means that each polynomial in this vector space has a maximum degree of n-1 and can have any coefficients from the field F.

a) To prove that the set $f_1(x),...,f_n(x)$ is a basis of this vector space, we must show that it satisfies two conditions: linear independence and spanning the vector space.

First, let us consider linear independence. This means that no polynomial in the set can be written as a linear combination of the others. In other words, there are no coefficients $c_1,...,c_n$ such that $c_1f_1(x)+...+c_nf_n(x)=0$.

We can prove this by considering the degree of each polynomial $f_j(x)$. As stated in the problem, each $f_j(x)$ has a degree of n-1, and since we are dealing with polynomials of degree ≤ n-1, this is the highest degree possible. Therefore, the only way for a linear combination of these polynomials to equal 0 is if all coefficients are 0. This proves linear independence.

Next, we must show that the set spans the vector space. This means that any polynomial of degree ≤ n-1 can be written as a linear combination of $f_1(x),...,f_n(x)$. To prove this, let us consider an arbitrary polynomial $p(x)$ of degree ≤ n-1. By the definition of $f_j(x)$, we can see that $p(x)$ can be written as $p(x) = \sum_{j=1}^n c_jf_j(x)$, where $c_j$ is the coefficient of $f_j(x)$ in the product. Therefore, the set $f_1(x),...,f_n(x)$ spans the vector space.

b) To prove that there exists a unique polynomial $g(x)$ of degree ≤ n-1 in x with coefficients in F, we must show that for any set of arbitrary coefficients $b_1,...,b_n$, there is only one possible polynomial