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Existence of a Basis of a Vector Space

crypt50

New member
Jun 29, 2013
21
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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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).....
 

crypt50

New member
Jun 29, 2013
21
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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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 $f_i$?)