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Sum expansion of the elementary symmetric polynomials

Reference request on symmetric polynomialsIrreducibility over $mathbbC$ of symmetric polynomialsWhy is the inverse of this matrix expressible in terms of elementary symmetric polynomials?Does the fundamental theorem of symmetric polynomials hold in any ring?terms of taylor expansions of multiple variables at the originProve that every symmetric polynomial can be written in terms of the elementary symmetric polynomialsIntegral Representation of Symmetric Polynomial SumsElementary Symmetric Polynomials and DeterminantExpansion of Elementary Symmetric FunctionsWhat is the Computational Complexity of the Elementary Symmetric Polynomials

1

$begingroup$

Recently I stumbled upon the following equation:

$$e_k(Vcup W) = sum_i=0^e_k-i(V)e_i(W)$$

$V$ and $W$ are subsets of $x_0,x_1,...,x_n$, and $Vcap W = varnothing$. Where both $e_k(V)$ and $e_k-i(W)$ are the elementary symmetric polynomials.

I tried cases in my Java program, and it seems to hold. (Question) But how would one prove this equation? My first attempt would be induction, but can it be done without induction?

Thanks in advance.

proof-writing symmetric-polynomials

share|cite|improve this question

edited Mar 16 at 18:08

Max

asked Mar 15 at 17:34

MaxMax

9031318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There seems to quite some information that isn't contained in the question: what are $;V, W;$ , for example? How are we to understand that $;e_k(V);$ is a symmetric polynomial? Symmetric... on what? On the set $;V;$ , in case it is a set?
  $endgroup$
  – DonAntonio
  Mar 15 at 17:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I did edit the question. and included a link to the elementary symmetric polynomial wikipedia. Hopefully this clears things up.
  $endgroup$
  – Max
  Mar 15 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is analogous to Vandermonde's identity, and the same proof idea should work.
  $endgroup$
  – Jair Taylor
  Mar 15 at 18:10
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Recently I stumbled upon the following equation:

$$e_k(Vcup W) = sum_i=0^e_k-i(V)e_i(W)$$

$V$ and $W$ are subsets of $x_0,x_1,...,x_n$, and $Vcap W = varnothing$. Where both $e_k(V)$ and $e_k-i(W)$ are the elementary symmetric polynomials.

I tried cases in my Java program, and it seems to hold. (Question) But how would one prove this equation? My first attempt would be induction, but can it be done without induction?

Thanks in advance.

proof-writing symmetric-polynomials

share|cite|improve this question

edited Mar 16 at 18:08

Max

asked Mar 15 at 17:34

MaxMax

9031318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There seems to quite some information that isn't contained in the question: what are $;V, W;$ , for example? How are we to understand that $;e_k(V);$ is a symmetric polynomial? Symmetric... on what? On the set $;V;$ , in case it is a set?
  $endgroup$
  – DonAntonio
  Mar 15 at 17:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I did edit the question. and included a link to the elementary symmetric polynomial wikipedia. Hopefully this clears things up.
  $endgroup$
  – Max
  Mar 15 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is analogous to Vandermonde's identity, and the same proof idea should work.
  $endgroup$
  – Jair Taylor
  Mar 15 at 18:10
  
  

  
  
  
  

add a comment | 

$begingroup$

Recently I stumbled upon the following equation:

$$e_k(Vcup W) = sum_i=0^e_k-i(V)e_i(W)$$

$V$ and $W$ are subsets of $x_0,x_1,...,x_n$, and $Vcap W = varnothing$. Where both $e_k(V)$ and $e_k-i(W)$ are the elementary symmetric polynomials.

I tried cases in my Java program, and it seems to hold. (Question) But how would one prove this equation? My first attempt would be induction, but can it be done without induction?

Thanks in advance.

proof-writing symmetric-polynomials

share|cite|improve this question

edited Mar 16 at 18:08

Max

asked Mar 15 at 17:34

MaxMax

9031318

$endgroup$

share|cite|improve this question

edited Mar 16 at 18:08

Max

asked Mar 15 at 17:34

MaxMax

9031318

share|cite|improve this question

edited Mar 16 at 18:08

Max

asked Mar 15 at 17:34

MaxMax

9031318

asked Mar 15 at 17:34

MaxMax

9031318

asked Mar 15 at 17:34

MaxMax

9031318

1 Answer
1

active

oldest

votes

1

$begingroup$

This follows since every $S subseteq V cup W$ can be decomposed uniquely into $S = S_1 cup S_2$ where $S_1 subseteq V$ and $S_2 subseteq W$. Explicitly,

beginalign*
e_k(Vcup W) &= sum_S prod_x in S x\
&= sum_S_1 subseteq V, S_2 subseteq W, prod_x in S_1 xprod_y in S_2 y\
&= ldots\
endalign*

(steps left you to fill.) Note that setting all $x_i = 1$ recovers Vandermonde's Identity.

Alternatively: Consider the product $$prod_x in V cup W (1 + x) = left( prod_x in V(1+x) right) left( prod_y in W(1+y) right)$$

and consider the degree-$k$ homogeneous component of both sides.

share|cite|improve this answer

edited 2 days ago

answered Mar 15 at 20:27

Jair TaylorJair Taylor

9,19432144

$endgroup$

add a comment | 

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1

$begingroup$

This follows since every $S subseteq V cup W$ can be decomposed uniquely into $S = S_1 cup S_2$ where $S_1 subseteq V$ and $S_2 subseteq W$. Explicitly,

beginalign*
e_k(Vcup W) &= sum_S prod_x in S x\
&= sum_S_1 subseteq V, S_2 subseteq W, prod_x in S_1 xprod_y in S_2 y\
&= ldots\
endalign*

(steps left you to fill.) Note that setting all $x_i = 1$ recovers Vandermonde's Identity.

Alternatively: Consider the product $$prod_x in V cup W (1 + x) = left( prod_x in V(1+x) right) left( prod_y in W(1+y) right)$$

and consider the degree-$k$ homogeneous component of both sides.

share|cite|improve this answer

edited 2 days ago

answered Mar 15 at 20:27

Jair TaylorJair Taylor

9,19432144

$endgroup$

add a comment | 

1

$begingroup$

This follows since every $S subseteq V cup W$ can be decomposed uniquely into $S = S_1 cup S_2$ where $S_1 subseteq V$ and $S_2 subseteq W$. Explicitly,

beginalign*
e_k(Vcup W) &= sum_S prod_x in S x\
&= sum_S_1 subseteq V, S_2 subseteq W, prod_x in S_1 xprod_y in S_2 y\
&= ldots\
endalign*

(steps left you to fill.) Note that setting all $x_i = 1$ recovers Vandermonde's Identity.

Alternatively: Consider the product $$prod_x in V cup W (1 + x) = left( prod_x in V(1+x) right) left( prod_y in W(1+y) right)$$

and consider the degree-$k$ homogeneous component of both sides.

share|cite|improve this answer

edited 2 days ago

answered Mar 15 at 20:27

Jair TaylorJair Taylor

9,19432144

$endgroup$

add a comment | 

$begingroup$

This follows since every $S subseteq V cup W$ can be decomposed uniquely into $S = S_1 cup S_2$ where $S_1 subseteq V$ and $S_2 subseteq W$. Explicitly,

beginalign*
e_k(Vcup W) &= sum_S prod_x in S x\
&= sum_S_1 subseteq V, S_2 subseteq W, prod_x in S_1 xprod_y in S_2 y\
&= ldots\
endalign*

(steps left you to fill.) Note that setting all $x_i = 1$ recovers Vandermonde's Identity.

Alternatively: Consider the product $$prod_x in V cup W (1 + x) = left( prod_x in V(1+x) right) left( prod_y in W(1+y) right)$$

and consider the degree-$k$ homogeneous component of both sides.

share|cite|improve this answer

edited 2 days ago

answered Mar 15 at 20:27

Jair TaylorJair Taylor

9,19432144

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered Mar 15 at 20:27

Jair TaylorJair Taylor

9,19432144

answered Mar 15 at 20:27

Jair TaylorJair Taylor

9,19432144

answered Mar 15 at 20:27

Jair TaylorJair Taylor

9,19432144