Lagrange's identity

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In mathematics, Lagrange's identity is the algebraic equation

\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(= {1 \over 2} \sum_{i=1}^n \sum_{j=1}^n (a_i b_j - a_j b_i)^2\biggr),

which applies to any two sets {a1, a2, . . ., an} and {b1, b2, . . ., bn} of real or complex numbers (or more generally, elements of a commutative ring). This identity is a special form of the Binet–Cauchy identity. For complex numbers it can also be written in the form

\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

involving the absolute value.

Since the right-hand side of the identity is clearly non-negative, it implies Cauchy's inequality in the finite-dimensional real coordinate space \mathbb{R}^n and its complex counterpart \mathbb{C}^n.

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[edit] Lagrange's identity and exterior algebra

In terms of the wedge product, Lagrange's identity can be written

(a \cdot a)(b \cdot b) - (a \cdot b)^2 = (a \wedge b) \cdot (a \wedge b).

Hence, it can be seen as a formula which gives the length of the wedge product of two vectors, which is the area of the paralleogram they define, in terms of the dot products of the two vectors, as

\|a \wedge b\| = \sqrt{(\|a\|\ \|b\|)^2 - \|a \cdot b\|^2}.

[edit] Lagrange's identity and vector calculus

If a and b are vectors in \mathbb{R}^3, Lagrange's identity can be also written in terms of the cross product and dot product:

|a \times b|^2 + (a \cdot b)^2 = |a|^2 |b|^2 = (a \cdot a)(b \cdot b).

This is a special case of the multiplicativity of the norm in the quaternion algebra:

|vw| = |v| |w|. \,

Or more generally,

(v \times w) \cdot (a \times b) = (v \cdot a)(w \cdot b) - (v \cdot b)(w \cdot a)

[edit] Lagrange's identity and calculus

In terms of the Sturm-Liouville theory, Lagrange's identity can be written

\int_0^1(Lu)v-u(Lv)\,dx=-p(u'v-uv')\bigg|_0^1 [1] 

 

 (1)

 

where p = P(x), q = Q(x), u = U(x) and v = V(x) are functions of x. u and v having continuous second derivatives on the interval [0,1]. L is Sturm-Liouville differential operators defined by

Lu = − (pu')' + qu 

 

 (2)

 

[edit] Proof

[edit] Algebraic form

The first version 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 of the first version. The expansion of the first term on the left side is

\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 

 

 (3)

 

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 like so

\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 

 

 (4)

 

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 - 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 .

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

\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 . 

 

 (5)

 

Back to the left side of Lagrange's identity: it has two terms, given in expanded form by Equations (3) and (4). The first term on the right side of Equation (4) ends up cancelling out the first term on the right side of Equation (3), yielding

(3) - (4) = \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 (5), so Lagrange's identity is indeed an identity, q. e. d..

[edit] Calculus form[1]

Replace f(x) = pu', g(x) = v, a = 0 and b = 1 into the rule integration by parts

\int_a^bf'(x)\,g(x)\,dx=f(x)\,g(x)\bigg|_a^b-\int_a^bf(x)\,g'(x)\,dx 

 

 (6)

 

we have

\int_0^1(pu')'v\,dx=(pu')v\bigg|_0^1-\int_0^1(pu')v'\,dx=p(u'v)\bigg|_0^1-\int_0^1(pu')v'\,dx 

 

 (7)

 

Replace f(x) = u, g(x) = pv', a = 0 and b = 1 into the rule (6) again, we have

\int_0^1u'(pv')\,dx=u(pv')\bigg|_0^1-\int_0^1u(pv')'\,dx

-\int_0^1(pu')v'\,dx=-p(uv')\bigg|_0^1+\int_0^1u(pv')'\,dx 

 

 (8)

 

Replace (8) into (7), we get

\int_0^1(pu')'v\,dx=p(u'v)\bigg|_0^1-p(uv')\bigg|_0^1+\int_0^1u(pv')'\,dx

-\int_0^1(pu')'v\,dx=-p(u'v-uv')\bigg|_0^1-\int_0^1u(pv')'\,dx 

 

 (9)

 

From the definition (2), we can get

\int_0^1(Lu)v\,dx=\int_0^1[-(pu')'+qu]v\,dx=-\int_0^1(pu')'v\,dx+\int_0^1uqv\,dx 

 

 (10)

 

Replace (9) into (10), we have

\int_0^1(Lu)v\,dx=-p(u'v-uv')\bigg|_0^1-\int_0^1u(pv')'\,dx+\int_0^1uqv\,dx

\int_0^1(Lu)v\,dx=-p(u'v-uv')\bigg|_0^1+\int_0^1u(Lv)\,dx 

 

 (11)

 

Rearrange terms of (11) then (1) is obtained. q.e.d.

[edit] See also

Brahmagupta-Fibonacci_identity

[edit] References

  1. ^ a b Boyce, William E.; Richard C. DiPrima (2001). "Boundary Value Problems and Sturm–Liouville Theory", Elementary Differential Equations and Boundary Value Problems (PDF), 7th ed. (in English), New York: John Wiley & Sons, pp. 630. ISBN 0-471-31999-6. OCLC 64431691.  (for the two sections Lagrange's identity and calculus and Calculus form of this article)