Lagrange's identity

In algebra, Lagrange's identity is a special case of the Binet–Cauchy identity, namely:

${\biggl (}\sum _{k=1}^{n}a_{k}^{2}{\biggr )}{\biggl (}\sum _{k=1}^{n}b_{k}^{2}{\biggr )}-{\biggl (}\sum _{k=1}^{n}a_{k}b_{k}{\biggr )}^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}{\biggl (}={1 \over 2}\sum _{i=1}^{n}\sum _{j=1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}{\biggr )},$

which applies to any two sets {a₁, a₂, . . ., a_n} and {b₁, b₂, . . ., b_n} 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

${\biggl (}\sum _{k=1}^{n}|a_{k}|^{2}{\biggr )}{\biggl (}\sum _{k=1}^{n}|b_{k}|^{2}{\biggr )}-{\biggl |}\sum _{k=1}^{n}a_{k}b_{k}{\biggr |}^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}|a_{i}{\overline {b}}_{j}-a_{j}{\overline {b}}_{i}|^{2}$

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 ℝⁿ and its complex counterpart ℂⁿ.

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

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

Lagrange's identity and vector calculus

If a and b are vectors in ℝ³, Lagrange's identity can be also written in terms of the cross product and dot product:

$|\mathbf {a\times b} |^{2}+(\mathbf {a\cdot b} )^{2}=|\mathbf {a} |^{2}|\mathbf {b} |^{2}=(\mathbf {a\cdot a} )(\mathbf {b\cdot b} ).$

This result can be related to quaternions, defined as the sum of a scalar and a vector:

$q=p+x\ \mathbf {i} +y\ \mathbf {j} +z\ \mathbf {k} \ ,$

with a squared norm given by:

$|q|^{2}=p^{2}\ +\ x^{2}+\ y^{2}\ +\ z^{2}\ .$

The multiplicativity of the norm in the quaternion algebra provides, for quaternions v and w:

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

Or, to generalize to four quaternions v and w and s and t:

$(v\times w)\cdot (s\times t)=(v\cdot s)(w\cdot t)-(v\cdot t)(w\cdot s).$

Seven dimensions

Main article: Seven-dimensional cross product

For a and b as vectors in ℝ⁷, Lagrange's identity:

${\biggl (}\sum _{k=1}^{n}a_{k}^{2}{\biggr )}{\biggl (}\sum _{k=1}^{n}b_{k}^{2}{\biggr )}-{\biggl (}\sum _{k=1}^{n}a_{k}b_{k}{\biggr )}^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2},$

takes on the particular form involving the cross product sometimes referred to as the Pythagorean theorem:

$|\mathbf {a} |^{2}|\mathbf {b} |^{2}-|\mathbf {a} \cdot \mathbf {b} |^{2}=|\mathbf {a} \times \mathbf {b} |^{2}\ ,$

that is, the same form found in ℝ³. However, the cross product in 7 dimensions does not share all the properties of the cross product in 3 dimensions. For example, in 3 dimensions if a × b = c × d, then a, b and c, d are in the same plane, but in 7 dimensions there are other planes with the same direction as a × b.

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)

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)

Proof

Algebraic form

The first version follows from the Binet-Cauchy identity by setting c_i = a_i and d_i = b_i. The second version follows by letting c_i and d_i denote the complex conjugates of a_i and b_i, 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:

$\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}^{2}b_{j}^{2}+a_{j}^{2}b_{i}^{2}-2a_{i}b_{j}a_{j}b_{i}).$

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..

Calculus form

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

$\int _{a}^{b}f'(x)\,g(x)\,dx=f(x)\,g(x){\bigg |}_{a}^{b}-\int _{a}^{b}f(x)\,g'(x)\,dx$ (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

$\ \int _{0}^{1}u'(pv')\,dx=u(pv'){\bigg |}_{0}^{1}-\int _{0}^{1}u(pv')'\,dx$

$\ -\int _{0}^{1}(pu')v'\,dx=-p(uv'){\bigg |}_{0}^{1}+\int _{0}^{1}u(pv')'\,dx$ (8)

Replace ('8') into ('7'), we get

$\ \int _{0}^{1}(pu')'v\,dx=p(u'v){\bigg |}_{0}^{1}-p(uv'){\bigg |}_{0}^{1}+\int _{0}^{1}u(pv')'\,dx$

$\ -\int _{0}^{1}(pu')'v\,dx=-p(u'v-uv'){\bigg |}_{0}^{1}-\int _{0}^{1}u(pv')'\,dx$ (9)

From the definition ('2'), we can get

$\ \int _{0}^{1}(Lu)v\,dx=\int _{0}^{1}[-(pu')'+qu]v\,dx=-\int _{0}^{1}(pu')'v\,dx+\int _{0}^{1}uqv\,dx$ (10)

Replace ('9') into ('10'), we have

$\ \int _{0}^{1}(Lu)v\,dx=-p(u'v-uv'){\bigg |}_{0}^{1}-\int _{0}^{1}u(pv')'\,dx+\int _{0}^{1}uqv\,dx$

$\ \int _{0}^{1}(Lu)v\,dx=-p(u'v-uv'){\bigg |}_{0}^{1}+\int _{0}^{1}u(Lv)\,dx$ (11)

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

See also

• Brahmagupta-Fibonacci identity