Difference between revisions of "Lagrange's Identity"

 
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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]
  
<cmath>{{\begin{aligned}{\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{aligned}}} </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}
+
\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( \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),
 
& \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>
+
\end{align}
 
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.
  

Latest revision as of 02:34, 15 September 2020

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}}} \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,

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