Elementary symmetric polynomial: Information from Answers.com

In mathematics, specifically in commutative algebra, the elementary symmetric polynomials are one type of basic building block for symmetric polynomials, in the sense that any symmetric polynomial P can be expressed as a polynomial in elementary symmetric polynomials: P can be given by an expression involving only additions and multiplication of constants and elementary symmetric polynomials. There is one elementary symmetric polynomial of degree d in n variables for any d ≤ n, and it is formed by adding together all distinct products of d distinct variables.

1 Definition
2 Examples
3 Properties
4 The fundamental theorem of symmetric polynomials
- 4.1 Proof sketch
- 4.2 An alternative proof
5 See also
6 References

Definition

The elementary symmetric polynomials in $n$ variables X₁, …, X_n, written e_k(X₁, …, X_n) for k = 0, 1, ..., n, can be defined as

$\begin{align} e_0 (X_1, X_2, \dots,X_n) &= 1,\\ e_1 (X_1, X_2, \dots,X_n) &= \textstyle\sum_{1 \leq j \leq n} X_j,\\ e_2 (X_1, X_2, \dots,X_n) &= \textstyle\sum_{1 \leq j < k \leq n} X_j X_k,\\ e_3 (X_1, X_2, \dots,X_n) &= \textstyle\sum_{1 \leq j < k < l \leq n} X_j X_k X_l,\\ \end{align}$

and so forth, down to

$e_n (X_1, X_2, \dots,X_n) = X_1 X_2 \cdots X_n$

(sometimes the notation σ_k is used instead of e_k). In general, for k ≥ 0 we define

$e_k (X_1 , \ldots , X_n )=\sum_{1\le j_1 < j_2 < \cdots < j_k \le n} X_{j_1} \cdots X_{j_k}.$

Thus, for each positive integer $k,$ less than or equal to $n$ , there exists exactly one elementary symmetric polynomial of degree $k$ in $n$ variables. To form the one which has degree $k$ , we form all products of $k$ -tuples of the $n$ variables and add up these terms.

The fact that $X 1 X 2 = X 2 X 1$ and so forth is the defining feature of commutative algebra. That is, the polynomial ring formed by taking all linear combinations of products of the elementary symmetric polynomials is a commutative ring.

Examples

The following lists the n elementary symmetric polynomials for the first four positive values of n. (In every case, e₀ = 1 is also one of the polynomials.)

For n = 1:

$e_1(X_1) = X_1.\,$

For n = 2:

$\begin{align} e_1(X_1,X_2) &= X_1 + X_2,\\ e_2(X_1,X_2) &= X_1X_2.\,\\ \end{align}$

For n = 3:

$\begin{align} e_1(X_1,X_2,X_3) &= X_1 + X_2 + X_3,\\ e_2(X_1,X_2,X_3) &= X_1X_2 + X_1X_3 + X_2X_3,\\ e_3(X_1,X_2,X_3) &= X_1X_2X_3.\,\\ \end{align}$

For n = 4:

$\begin{align} e_1(X_1,X_2,X_3,X_4) &= X_1 + X_2 + X_3 + X_4,\\ e_2(X_1,X_2,X_3,X_4) &= X_1X_2 + X_1X_3 + X_1X_4 + X_2X_3 + X_2X_4 + X_3X_4,\\ e_3(X_1,X_2,X_3,X_4) &= X_1X_2X_3 + X_1X_2X_4 + X_1X_3X_4 + X_2X_3X_4,\\ e_4(X_1,X_2,X_3,X_4) &= X_1X_2X_3X_4.\,\\ \end{align}$

Properties

The elementary symmetric polynomials appear when we expand a linear factorization of a monic polynomial: we have the identity

$\prod_{j=1}^n ( \lambda-X_j)=\lambda^n-e_1(X_1,\ldots,X_n)\lambda^{n-1}+e_2(X_1,\ldots,X_n)\lambda^{n-2}-\cdots+(-1)^n e_n(X_1,\ldots,X_n).$

That is, when we substitute numerical values for the variables $X_1,X_2,\dots,X_n$ , we obtain the monic univariate polynomial (with variable λ) whose roots are the values substituted for $X_1,X_2,\dots,X_n$ and whose coefficients are the elementary symmetric polynomials.

The characteristic polynomial of a linear operator is an example of this. The roots are the eigenvalues of the operator. When we substitute these eigenvalues into the elementary symmetric polynomials, we obtain the coefficients of the characteristic polynomial, which are numerical invariants of the operator. This fact is useful in linear algebra and its applications and generalizations, like tensor algebra and disciplines which extensively employ tensor fields, such as differential geometry.

The set of elementary symmetric polynomials in $n$ variables generates the ring of symmetric polynomials in $n$ variables. More specifically, the ring of symmetric polynomials with integer coefficients equals the integral polynomial ring $\mathbb Z[e_1(X_1,\ldots,X_n),\ldots,e_n(X_1,\ldots,X_n)].$ (See below for a more general statement and proof.) This fact is one of the foundations of invariant theory. For other systems of symmetric polynomials with a similar property see power sum symmetric polynomials and complete homogeneous symmetric polynomials.

The fundamental theorem of symmetric polynomials

For any ring A denote the ring of symmetric polynomials in the variables $X_1,\ldots,X_n$ with coefficients in A by $A[X_1,\ldots,X_n]^{S_n}$ .

$A[X_1,\ldots,X_n]^{S_n}$ is a polynomial ring in the n elementary symmetric polynomials $e_k (X_1 , \ldots ,X_n )$ for k = 1, ..., n.

(Note that $e 0$ is not among these polynomials; since $e 0 = 1$ , it cannot be member of any set of algebraically independent elements.)

This means that every symmetric polynomial $P(X_1,\ldots, X_n) \in A[X_1,\ldots,X_n]^{S_n}$ has a unique representation

$P(X_1,\ldots, X_n)=Q(e_1(X_1 , \ldots ,X_n), \ldots, e_n(X_1 , \ldots ,X_n))$

for some polynomial $Q \in A[Y_1,\ldots,Y_n]$ . Another way of saying the same thing is that $A[X_1,\ldots,X_n]^{S_n}$ is isomorphic to the polynomial ring $A[Y_1,\ldots,Y_n]$ through an isomorphism that sends $Y k$ to $e_k(X_1 , \ldots ,X_n)$ for $k=1,\ldots,n$ .

Proof sketch

The theorem may be proved for symmetric homogeneous polynomials by a double mathematical induction with respect to the number of variables n and, for fixed n, with respect to the degree of the homogeneous polynomial. The general case then follows by splitting an arbitrary symmetric polynomial into its homogeneous components (which are again symmetric).

In the case n = 1 the result is obvious because every polynomial in one variable is automatically symmetric.

Assume now that the theorem has been proved for all polynomials for $m < n$ variables and all symmetric polynomials in n variables with degree < d. Every homogeneous symmetric polynomial P in $A[X_1,\ldots,X_n]^{S_n}$ can be decomposed as a sum of homogeneous symmetric polynomials

$P(X_1,\ldots,X_n)= P_{\mbox{lacunary}} (X_1,\ldots,X_n) + X_1 \cdots X_n \cdot Q(X_1,\ldots,X_n).$

Here the "lacunary part" $P lacunary$ is defined as the sum of all monomials in P which depend only on a proper subset of the n variables X₁, ..., X_n, i.e., where one variable X_j is missing.

Because P is symmetric the lacunary part is determined by the coefficients of all monomials which only depend on the variables X₁, ..., X_n−1. The sum of all these monomials is equal to a polynomial $\tilde{P}(X_1, \ldots, X_{n-1})=P(X_1, \ldots,X_{n-1},0)$ which is symmetric in n−1 variables. According to the induction assumption $\tilde{P}(X_1, \ldots. X_{n-1})$ can be written as

$\tilde{P}(X_1, \ldots, X_{n-1})=\tilde{Q}(\sigma_{1,n-1}, \ldots, \sigma_{n-1,n-1})$

for some $\tilde{Q}$ . Here the doubly-indexed $σ j, n - 1$ denote the elementary symmetric polynomials in n−1 variables.

Consider now the polynomial

$R(X_1, \ldots, X_{n}):= \tilde{Q}(\sigma_{1,n}, \ldots, \sigma_{n-1,n}) \ .$

Then $R(X_1, \ldots, X_{n})$ is a symmetric polynomial whose lacunary part coincides with that of the original polynomial P (note here that the difference $σ j, n - σ j, n - 1$ is divisible by X_n, therefore in the three polynomials R, $\tilde{P}$ and finally $P lacunary$ all monomials involving only X₁, ... ,X_n−1 are identical). Therefore the difference P−R has zero lacunary part and is of the form $X_1 \cdots X_n \cdot Q(X_1, \ldots, X_n)$ . The first factor coincides with the elementary symmetric polynomial $σ n, n$ , the second factor Q is a homogeneous symmetric polynomial of lower degree $d < n$ which by the induction assumption can be expressed as a polynomial in the elementary symmetric functions. Combining the representations for P−R and R one finds a polynomial representation for P.

The uniqueness of the representation can be proved inductively in a similar way. (It is equivalent to the fact that the n polynomials $e_1, \ldots, e_n$ are algebraically independent over the quotient field of A.) The fact that the polynomial representation is unique implies that $A[X_1,\ldots,X_n]^{S_n}$ is isomorphic to $A[Y_1,\ldots,Y_n]$ .

An alternative proof

The following proof is also inductive, but does not involve other polynomials than those symmetric in $X_1,\ldots,X_n$ , and also leads to a fairly direct procedure to effectively write a symmetric polynomial as a polynomial in the elementary symmetric ones. Assume the symmetric polynomial to be homogenous of degree d; different homogeneous components can be decomposed separately. Order the monomials in the variables X_i lexicographically, where the individual variables are ordered $X_1>\cdots>X_n$ , in other words the dominant term of a polynomial is one with the highest occurring power of X₁, and among those the one with the highest power of X₂, etc. Furthermore parametrize all products of elementary symmetric polynomials that have degree d (they are in fact homogeneous) as follows by partitions of d. Order the individual elementary symmetric polynomials $e_i(X_1,\ldots,X_n)$ in the product so that those with larger indices i come first, then build for each such factor a column of i boxes, and arrange those columns from left to right to form a Young diagram containing d boxes in all. The shape of this diagram is a partition of d, and each partition λ of d arises for exactly one product of elementary symmetric polynomials, which we shall denote by e_λ^t (X₁,…,X_n) (the "t" is present only because traditionally this product is associated to the transpose partition of λ). The essential ingredient of the proof is the following simple property, which uses multi-index notation for monomials in the variables X_i.

Lemma. The leading term of e_λ^t (X₁,…,X_n) is X^λ.

Proof. To get the leading term of the product one must select the leading term in each factor $e_i(X_1,\ldots,X_n)$ , which is clearly $X_1X_2\cdots X_i$ , and multiply these together. To count the occurrences of the individual variables in the resulting monomial, fill the column of the Young diagram corresponding to the factor concerned with the numbers 1,…,i of the variables, then all boxes in the first row contain 1, those in the second row 2, and so forth, which means the leading term is X^λ (its coefficient is 1 because there is only one choice that leads to this monomial).

Now one proves by induction on the leading monomial in lexicographic order, that any nonzero homogenous symmetric polynomial P of degree d can be written as polynomial in the elementary symmetric polynomials. Since P is symmetric, its leading monomial has weakly decreasing exponents, so it is some X^λ with λ a partition of d. Let the coefficient of this term be c, then P − ce_λ^t (X₁,…,X_n) is either zero or a symmetric polynomial with a strictly smaller leading monomial. Writing this difference inductively as a polynomial in the elementary symmetric polynomials, and adding back ce_λ^t (X₁,…,X_n) to it, one obtains the sought for polynomial expression for P.

The fact that this expression is unique, or equivalently that all the products (monomials) e_λ^t (X₁,…,X_n) of elementary symmetric polynomials are linearly independent, is also easily proved. The lemma shows that all these products have different leading monomials, and this suffices: if a nontrivial linear combination of the e_λ^t (X₁,…,X_n) were zero, one focusses on the contribution in the linear combination with nonzero coefficient and with (as polynomial in the variables X_i) the largest leading monomial; the leading term of this contribution cannot be cancelled by any other contribution of the linear combination, which gives a contradiction.

References

Macdonald, I.G. (1995), Symmetric Functions and Hall Polynomials, second ed. Oxford: Clarendon Press. ISBN 0-19-850450-0 (paperback, 1998).
Richard P. Stanley (1999), Enumerative Combinatorics, Vol. 2. Camridge: Cambridge University Press. ISBN 0-521-56069-1

This entry is from Wikipedia, the leading user-contributed encyclopedia. It may not have been reviewed by professional editors (see full disclaimer)

Donate to Wikimedia

	How is dividing a polynomial by a binomial similar to or different from the long division you learned in elementary school? Read answer...
	How is dividing a polynomial by a binomial similar to or different from the long division you learned in elementary school Can understanding how to do one kind of division help you with the other? Read answer...
	Is a log of a polynomial still a polynomial? Read answer...

	What are the similarities of elementary school long division to the dividing of polynomial?
	Division of polynomial by a polynomial?
	How do you multiply polynomial by anther polynomial?

Elementary symmetric polynomial

Contents