Unique Representation in Polynomial Forms/General Result
Theorem
Let $f$ be a polynomial form in the indeterminates $\set {X_j: j \in J}$ such that $f: \mathbf X^k \mapsto a_k$.
For $r \in \R$, $\mathbf X^k \in M$, let $r \mathbf X^k$ denote the polynomial form that takes the value $r$ on $\mathbf X^k$ and zero on all other monomials.
Let $Z$ denote the set of all multiindices indexed by $J$.
Then the sum representation:
- $\ds \hat f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
has only finitely many non-zero terms.
Moreover it is everywhere equal to $f$, and is the unique such sum.
Corollary
Dropping the zero terms from the sum we can write the polynomial $f$ as
- $f = a_{k_1} \mathbf X^{k_1} + \cdots + a_{k_r} \mathbf X^{k_r}$
for some $a_{k_i}\in R$, $i = 1, \ldots, r$.
Proof
Suppose that the sum has infinitely many non-zero terms.
Then infinitely many $a_k$ are non-zero, which contradicts the definition of a polynomial.
Therefore the sum consists of finitely many non-zero terms.
Let $\mathbf X^m \in M$ be arbitrary.
Then:
\(\ds \map {\hat f} {\mathbf X^m}\) | \(=\) | \(\ds \paren {\sum_{k \mathop \in Z} a_k \mathbf X^k} \paren {\mathbf X^m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a_m \mathbf X^m} \paren {\mathbf X^m} + \sum_{k \mathop \ne m \mathop \in Z} \paren {a_k \mathbf X^k} \paren {\mathbf X^m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a_m\) |
So $\hat f = f$.
Finally suppose that:
- $\ds \tilde f = \sum_{k \mathop \in Z} b_k \mathbf X^k$
is another such representation with $b_m \ne a_m$ for some $m \in Z$.
Then:
- $\map {\tilde f} {\mathbf X^m} = b_m \ne a_m = \map f {\mathbf X^m}$
Therefore $\hat f$ as defined above is the only such representation.
$\blacksquare$