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

In summary, the Lagrange property states that for any real numbers $x_1, \dots, x_n$ and $y_1, \dots, y_n$, the sum of the products of the squares of the elements of $x_i, y_j$ is equal to the sum of the products of the squares of the elements of $x_i, y_j-x_j y_i$.
  • #1
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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  • #2
Hint: try using inner products (you may find the polarization identities useful).
 
  • #3
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)
 
  • #4
$ \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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FAQ: Prove Lagrange Property w/ Algebra - A Hint for You!

What is the Lagrange property and why is it important?

The Lagrange property, also known as the intermediate value theorem, states that if a function is continuous on an interval, then it must take on every value between its minimum and maximum values. This property is important because it allows us to prove the existence of solutions to equations, which is crucial in many areas of mathematics and science.

How do you prove the Lagrange property using algebra?

To prove the Lagrange property using algebra, we can start by setting up the equation f(x) = y, where y is a value between the minimum and maximum values of the function. Next, we can use the definition of continuity to show that there must exist a value c within the interval where f(c) = y. This proves that the function takes on every value between its minimum and maximum values.

What is the role of the hint in proving the Lagrange property with algebra?

The hint in proving the Lagrange property with algebra is to use the intermediate value theorem, which states that if a function is continuous on an interval, then it must take on every value between its minimum and maximum values. This helps guide our approach in setting up the equation and using algebra to prove the property.

Can the Lagrange property be applied to all functions?

The Lagrange property can be applied to all continuous functions on a closed interval. It is important to note that the function must be continuous for the property to hold.

How is the Lagrange property used in real-world applications?

The Lagrange property is used in various real-world applications, such as in economics, physics, and engineering. For example, it can be used to prove the existence of solutions to equations in economic models, to show the existence of roots in equations for physical laws, and to analyze the behavior of systems in engineering.

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