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

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SUMMARY

The discussion focuses on proving the Lagrange property using algebraic methods. The property states that for any real numbers \(x_1, \dots, x_n\) and \(y_1, \dots, y_n\), the equation \(\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 PREREQUISITES

  • Understanding of inner products in vector spaces
  • Familiarity with polarization identities
  • Basic knowledge of algebraic manipulation
  • Concept of sums and indices in mathematical notation
NEXT STEPS
  • Study the properties of inner products in linear algebra
  • Explore polarization identities and their applications
  • Practice algebraic manipulation techniques for complex equations
  • Investigate the geometric interpretation of the Lagrange property
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Mathematicians, students studying linear algebra, and anyone interested in algebraic proofs and properties of inner products will benefit from this discussion.

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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