Prove Lagrange Property w/ Algebra - A Hint for You!

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    Lagrange Property
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Discussion Overview

The discussion revolves around proving the Lagrange property using algebraic methods. Participants explore the relationship between sums of products of real numbers and their squares, focusing on the algebraic identity involving sums of products and inner products.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant requests a hint for proving the Lagrange property involving sums of products of real numbers.
  • Another participant suggests using inner products and mentions the potential utility of polarization identities.
  • A subsequent post reiterates the suggestion to use inner products and clarifies the notation for sums of squares.
  • A detailed algebraic manipulation is presented, breaking down the expression for the sum of squares and exploring the relationship between the terms involved.
  • The final algebraic expression leads to a formulation of the Lagrange property, indicating a connection between the sums of squares and the product of sums.

Areas of Agreement / Disagreement

Participants present various approaches and hints, but there is no consensus on a definitive method or conclusion regarding the proof of the Lagrange property.

Contextual Notes

The discussion includes complex algebraic manipulations that may depend on specific interpretations of inner products and the definitions of the sums involved. Some steps in the algebraic reasoning remain unresolved or may require additional assumptions.

evinda
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Hello! (Wave)

With use of algebra I want to prove the Lagrange property:For any real numbers $x_1, \dots, x_n$ and $y_1, \dots, y_n$, $$\left( \sum_{i=1}^n x_i y_i\right)^2=\left(\sum_{i=1}^n x_i^2 \right)\left(\sum_{i=1}^n y_i^2 \right)- \sum_{i<j} (x_i y_j-x_j y_i)^2$$

Could you give me a hint how we could show the above property?
 
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Hint: try using inner products (you may find the polarization identities useful).
 
Deveno said:
Hint: try using inner products (you may find the polarization identities useful).

We have that $\left( \sum_{i=1}^n x_i y_i \right)^2=(x \cdot y)^2$ and $\sum_{i=1}^n x_i^2=||x||^2,\sum_{i=1}^n y_i^2=||y||^2 $, right?

How can we write the other sum, which is not till n? (Thinking)
 
$ \begin{aligned} & \begin{aligned} ~~~~~~~~~~ \mathcal{S}: &= \sum_{1 \le i<j \le n} (x_i y_j-x_j y_i)^2 \\& =\sum_{1 \le i \le j \le n} (x_i y_j-x_j y_i)^2 -\sum_{1 \le j \le n} (x_jy_j-x_jy_j)^2 \\& =\sum_{1 \le i \le j \le n}x_i^2 y_j^2-2 \sum_{1 \le i \le j \le n}x_ix_jy_i y_j+\sum_{1 \le i \le j \le n}x_j^2 y_i^2 \\& =\sum_{1 \le j \le n}\sum_{1 \le i \le j} x_i^2 y_j^2-2 \sum_{1 \le j \le n}\sum_{1 \le i \le j}x_ix_jy_i y_j+\sum_{1 \le j \le n}\sum_{1 \le i \le j}x_j^2 y_i^2 \\& = \sum_{1 \le i \le n}\sum_{i \le j \le n} x_i^2 y_j^2-2 \sum_{1 \le i \le n}\sum_{i \le j \le n}x_ix_jy_i y_j+\sum_{1 \le i \le n}\sum_{i \le j \le n}x_j^2 y_i^2 \\& = \sum_{1 \le i \le n}\sum_{1 \le j \le n} x_i^2 y_j^2-2 \sum_{1 \le i \le n}\sum_{1 \le j \le n}x_ix_jy_i y_j+\sum_{1 \le i \le n}\sum_{1 \le j \le n}x_j^2 y_i^2 \\& - \sum_{1 \le i \le n}\sum_{1 \le j \le i-1} x_i^2 y_j^2+2 \sum_{1 \le i \le n}\sum_{1 \le j \le i-1}x_ix_jy_i y_j-\sum_{1 \le i \le n}\sum_{1 \le j \le i-1}x_j^2 y_i^2 \\& = \sum_{1 \le i \le n}x_i^2 \sum_{1 \le j \le n}y_j^2-2 \sum_{1 \le i \le n}x_iy_i \sum_{1 \le j \le n}x_jy_j+\sum_{1 \le i \le n} y_i^2\sum_{1 \le j \le n}x_j^2 -\sum_{1 \le j <i \le n} (x_i y_j-x_j y_i)^2 \\& =\bigg(\sum_{1 \le i \le n}x_i^2\bigg)\bigg(\sum_{1 \le i \le n}y_i^2\bigg)-2 \bigg(\sum_{1 \le i \le n}x_iy_i \bigg)^2+\bigg(\sum_{1 \le i \le n} y_i^2\bigg)\bigg(\sum_{1 \le i \le n}x_i^2\bigg) -\mathcal{S} \\& =2\bigg(\sum_{1 \le i \le n}x_i^2\bigg)\bigg(\sum_{1 \le i \le n}y_i^2\bigg)-2 \bigg(\sum_{1 \le i \le n}x_iy_i \bigg)^2 -\mathcal{S} \end{aligned} \\& \implies 2 \mathcal{S} = 2\bigg(\sum_{1 \le i \le n}x_i^2\bigg)\bigg(\sum_{1 \le i \le n}y_i^2\bigg)-2 \bigg(\sum_{1 \le i \le n}x_iy_i \bigg)^2 \\& \implies ~~\mathcal{S} =\bigg(\sum_{1 \le i \le n}x_i^2\bigg)\bigg(\sum_{1 \le i \le n}y_i^2\bigg)- \bigg(\sum_{1 \le i \le n}x_iy_i \bigg)^2. \end{aligned} $
 
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