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{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> |
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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. | ||
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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: | ||
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('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 | ('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 </math> | + \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 17:27, 31 August 2022
In algebra, Lagrange's identity, named after Joseph Louis Lagrange, is:[1][2]
which applies to any two sets and
of real or complex numbers.
Proof:
The vector form follows from the Binet-Cauchy identity by setting and
. The second version follows by letting
and
denote the complex conjugates of
and
, respectively,
Here is also a direct proof.[10] The expansion of the first term on the left side is:
(1)
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)
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:
Distribute the summation on the right side,
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)
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') =
which is the same as Equation ('3'), so Lagrange's identity is indeed an identity.