Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Lagrange's Identity"
m (comma at end turned into period, most important edit ever)
 
(6 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
In algebra, Lagrange's identity, named after Joseph Louis Lagrange, is:[1][2]
 
In algebra, Lagrange's identity, named after Joseph Louis Lagrange, is:[1][2]
  
−
<math>\begin{align}{\biggl (}\sum _{k=1}^{n}a_{k}^{2}{\biggr )}{\biggl (}\sum _{k=1}^{n}b_{k}^{2}{\biggr )}-{\biggl (}\sum _{k=1}^{n}a_{k}b_{k}{\biggr )}^{2}&=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}\\&{\biggl (}={\frac {1}{2}}\sum _{i=1}^{n}\sum _{j=1,j\neq i}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}{\biggr )},\end{align}}}
+
<cmath>\begin{align}{\biggl (}\sum _{k=1}^{n}a_{k}^{2}{\biggr )}{\biggl (}\sum _{k=1}^{n}b_{k}^{2}{\biggr )}-{\biggl (}\sum _{k=1}^{n}a_{k}b_{k}{\biggr )}^{2}&=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}\\&{\biggl (}={\frac {1}{2}}\sum _{i=1}^{n}\sum _{j=1,j\neq i}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}{\biggr )},\end{align} </cmath>
−
\begin{align}
+
 
−
\biggl( \sum_{k=1}^n a_k^2\biggr) \biggl(\sum_{k=1}^n b_k^2\biggr) - \biggl(\sum_{k=1}^n a_k b_k\biggr)^2 & = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 \\
+
 
−
& \biggl(= \frac{1}{2} \sum_{i=1}^n \sum_{j=1,j\neq i}^n (a_i b_j - a_j b_i)^2\biggr),
 
−
\end{align}</math>
 
 
which applies to any two sets <math>\{a_1, a_2, . . ., a_n\}</math> and <math>\{b_1, b_2, . . ., b_n\}</math> of real or complex numbers.
 
which applies to any two sets <math>\{a_1, a_2, . . ., a_n\}</math> and <math>\{b_1, b_2, . . ., b_n\}</math> of real or complex numbers.
  
  
 
Proof:
 
Proof:
−
The vector form follows from the Binet-Cauchy identity by setting ci = ai and di = bi. The second version follows by letting ci and di denote the complex conjugates of ai and bi, respectively,
+
The vector form follows from the Binet-Cauchy identity by setting <math>ci = ai</math> and <math>di = bi</math>. The second version follows by letting <math>ci</math> and <math>di</math> denote the complex conjugates of <math>ai</math> and <math>bi</math>, respectively,
  
 
Here is also a direct proof.[10] The expansion of the first term on the left side is:
 
Here is also a direct proof.[10] The expansion of the first term on the left side is:
  
−
(1)    {\left(\sum _{k=1}^{n}a_{k}^{2}\right)\left(\sum _{k=1}^{n}b_{k}^{2}\right)=\sum _{i=1}^{n}\sum _{j=1}^{n}a_{i}^{2}b_{j}^{2}=\sum _{k=1}^{n}a_{k}^{2}b_{k}^{2}+\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{j=1}^{n-1}\sum _{i=j+1}^{n}a_{i}^{2}b_{j}^{2}\ ,}  \left( \sum_{k=1}^n a_k^2\right) \left(\sum_{k=1}^n b_k^2\right) =  
+
(1)    <math>{\left(\sum _{k=1}^{n}a_{k}^{2}\right)\left(\sum _{k=1}^{n}b_{k}^{2}\right)=\sum _{i=1}^{n}\sum _{j=1}^{n}a_{i}^{2}b_{j}^{2}=\sum _{k=1}^{n}a_{k}^{2}b_{k}^{2}+\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{j=1}^{n-1}\sum _{i=j+1}^{n}a_{i}^{2}b_{j}^{2}\ ,}  \left( \sum_{k=1}^n a_k^2\right) \left(\sum_{k=1}^n b_k^2\right) =  
 
\sum_{i=1}^n \sum_{j=1}^n a_i^2 b_j^2  
 
\sum_{i=1}^n \sum_{j=1}^n a_i^2 b_j^2  
 
= \sum_{k=1}^n a_k^2 b_k^2  
 
= \sum_{k=1}^n a_k^2 b_k^2  
 
+ \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2  
 
+ \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2  
−
+ \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 \ ,
+
+ \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 \ ,</math>
 
which means that the product of a column of as and a row of bs yields (a sum of elements of) a square of abs, which can be broken up into a diagonal and a pair of triangles on either side of the diagonal.
 
which means that the product of a column of as and a row of bs yields (a sum of elements of) a square of abs, which can be broken up into a diagonal and a pair of triangles on either side of the diagonal.
  
 
The second term on the left side of Lagrange's identity can be expanded as:
 
The second term on the left side of Lagrange's identity can be expanded as:
  
−
(2)    {\left(\sum _{k=1}^{n}a_{k}b_{k}\right)^{2}=\sum _{k=1}^{n}a_{k}^{2}b_{k}^{2}+2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{i}a_{j}b_{j}\ ,}  \left(\sum_{k=1}^n a_k b_k\right)^2 =  
+
(2)    <math>{\left(\sum _{k=1}^{n}a_{k}b_{k}\right)^{2}=\sum _{k=1}^{n}a_{k}^{2}b_{k}^{2}+2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{i}a_{j}b_{j}\ ,}  \left(\sum_{k=1}^n a_k b_k\right)^2 =  
−
\sum_{k=1}^n a_k^2 b_k^2 + 2\sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j \ ,
+
\sum_{k=1}^n a_k^2 b_k^2 + 2\sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j \ ,</math>
 
which means that a symmetric square can be broken up into its diagonal and a pair of equal triangles on either side of the diagonal.
 
which means that a symmetric square can be broken up into its diagonal and a pair of equal triangles on either side of the diagonal.
  
 
To expand the summation on the right side of Lagrange's identity, first expand the square within the summation:
 
To expand the summation on the right side of Lagrange's identity, first expand the square within the summation:
  
−
{\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}^{2}b_{j}^{2}+a_{j}^{2}b_{i}^{2}-2a_{i}b_{j}a_{j}b_{i}).}  \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i^2 b_j^2 + a_j^2 b_i^2 - 2 a_i b_j a_j b_i).  
+
<math>{\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}^{2}b_{j}^{2}+a_{j}^{2}b_{i}^{2}-2a_{i}b_{j}a_{j}b_{i}).}  \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i^2 b_j^2 + a_j^2 b_i^2 - 2 a_i b_j a_j b_i). </math>
 
Distribute the summation on the right side,
 
Distribute the summation on the right side,
  
−
{\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{j}^{2}b_{i}^{2}-2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{j}a_{j}b_{i}.}  \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_j^2 b_i^2 - 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_j a_j b_i .
+
<math>{\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{j}^{2}b_{i}^{2}-2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{j}a_{j}b_{i}.}  \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_j^2 b_i^2 - 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_j a_j b_i .</math>
 
Now exchange the indices i and j of the second term on the right side, and permute the b factors of the third term, yielding:
 
Now exchange the indices i and j of the second term on the right side, and permute the b factors of the third term, yielding:
  
−
(3)    {\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{j=1}^{n-1}\sum _{i=j+1}^{n}a_{i}^{2}b_{j}^{2}-2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{i}a_{j}b_{j}\ .}  \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 - 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j \ .
+
(3)    <math>{\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{j=1}^{n-1}\sum _{i=j+1}^{n}a_{i}^{2}b_{j}^{2}-2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{i}a_{j}b_{j}\ .}  \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 - 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j \ .</math>
 
Back to the left side of Lagrange's identity: it has two terms, given in expanded form by Equations ('1') and ('2'). The first term on the right side of Equation ('2') ends up canceling out the first term on the right side of Equation ('1'), yielding
 
Back to the left side of Lagrange's identity: it has two terms, given in expanded form by Equations ('1') and ('2'). The first term on the right side of Equation ('2') ends up canceling out the first term on the right side of Equation ('1'), yielding
  
−
('1') - ('2') = {\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{j=1}^{n-1}\sum _{i=j+1}^{n}a_{i}^{2}b_{j}^{2}-2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{i}a_{j}b_{j}} \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2  
+
('1') - ('2') = <math>{\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{j=1}^{n-1}\sum _{i=j+1}^{n}a_{i}^{2}b_{j}^{2}-2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{i}a_{j}b_{j}} \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2  
−
+ \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 - 2\sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j  
+
+ \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 - 2\sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j </math>
−
which is the same as Equation ('3'), so Lagrange's identity is indeed an identity,
+
which is the same as Equation ('3'), so Lagrange's identity is indeed an identity.

Latest revision as of 16:27, 31 August 2022

In algebra, Lagrange's identity, named after Joseph Louis Lagrange, is:[1][2]

\begin{align}{\biggl (}\sum _{k=1}^{n}a_{k}^{2}{\biggr )}{\biggl (}\sum _{k=1}^{n}b_{k}^{2}{\biggr )}-{\biggl (}\sum _{k=1}^{n}a_{k}b_{k}{\biggr )}^{2}&=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}\\&{\biggl (}={\frac {1}{2}}\sum _{i=1}^{n}\sum _{j=1,j\neq i}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}{\biggr )},\end{align}


which applies to any two sets $\{a_1, a_2, . . ., a_n\}$ and $\{b_1, b_2, . . ., b_n\}$ of real or complex numbers.


Proof: The vector form follows from the Binet-Cauchy identity by setting $ci = ai$ and $di = bi$. The second version follows by letting $ci$ and $di$ denote the complex conjugates of $ai$ and $bi$, respectively,

Here is also a direct proof.[10] The expansion of the first term on the left side is:

(1)    ${\left(\sum _{k=1}^{n}a_{k}^{2}\right)\left(\sum _{k=1}^{n}b_{k}^{2}\right)=\sum _{i=1}^{n}\sum _{j=1}^{n}a_{i}^{2}b_{j}^{2}=\sum _{k=1}^{n}a_{k}^{2}b_{k}^{2}+\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{j=1}^{n-1}\sum _{i=j+1}^{n}a_{i}^{2}b_{j}^{2}\ ,} \left( \sum_{k=1}^n a_k^2\right) \left(\sum_{k=1}^n b_k^2\right) = \sum_{i=1}^n \sum_{j=1}^n a_i^2 b_j^2 = \sum_{k=1}^n a_k^2 b_k^2 + \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 \ ,$ which means that the product of a column of as and a row of bs yields (a sum of elements of) a square of abs, which can be broken up into a diagonal and a pair of triangles on either side of the diagonal.

The second term on the left side of Lagrange's identity can be expanded as:

(2)    ${\left(\sum _{k=1}^{n}a_{k}b_{k}\right)^{2}=\sum _{k=1}^{n}a_{k}^{2}b_{k}^{2}+2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{i}a_{j}b_{j}\ ,} \left(\sum_{k=1}^n a_k b_k\right)^2 = \sum_{k=1}^n a_k^2 b_k^2 + 2\sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j \ ,$ which means that a symmetric square can be broken up into its diagonal and a pair of equal triangles on either side of the diagonal.

To expand the summation on the right side of Lagrange's identity, first expand the square within the summation:

${\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}^{2}b_{j}^{2}+a_{j}^{2}b_{i}^{2}-2a_{i}b_{j}a_{j}b_{i}).} \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i^2 b_j^2 + a_j^2 b_i^2 - 2 a_i b_j a_j b_i).$ Distribute the summation on the right side,

${\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{j}^{2}b_{i}^{2}-2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{j}a_{j}b_{i}.} \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_j^2 b_i^2 - 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_j a_j b_i .$ Now exchange the indices i and j of the second term on the right side, and permute the b factors of the third term, yielding:

(3)    ${\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{j=1}^{n-1}\sum _{i=j+1}^{n}a_{i}^{2}b_{j}^{2}-2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{i}a_{j}b_{j}\ .} \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 - 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j \ .$ Back to the left side of Lagrange's identity: it has two terms, given in expanded form by Equations ('1') and ('2'). The first term on the right side of Equation ('2') ends up canceling out the first term on the right side of Equation ('1'), yielding

('1') - ('2') = ${\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}^{2}b_{j}^{2}+\sum _{j=1}^{n-1}\sum _{i=j+1}^{n}a_{i}^{2}b_{j}^{2}-2\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}a_{i}b_{i}a_{j}b_{j}} \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 - 2\sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j$ which is the same as Equation ('3'), so Lagrange's identity is indeed an identity.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Lagrange%27s_Identity&oldid=177800"