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Lagrange's identity (Theorem)

Let $ R$ be a commutative ring, and let $ x_1, \ldots, x_n, y_1, \ldots, y_n$ be arbitrary elements in $ R$. Then

$\displaystyle \left(\sum_{k=1}^n x_ky_k\right)^2 =\left(\sum_{k=1}^n x_k^2\right)\left(\sum_{k=1}^n y_k^2\right) - \sum_{1 \le k < i \le n} (x_ky_i -x_iy_k)^2$.
Proof. Since $ R$ is commutative, we can apply the binomial formula.We start out with
$\displaystyle \left(\sum_{i=1}^n x_iy_i\right)^2 =\sum_{i=1}^n (x_i^2y_i^2) +\sum_{1 \leq i< j\leq n} 2 x_iy_jx_jy_i$ (1)

Using the binomial formula, we see that
$\displaystyle (x_iy_j -x_jy_i)^2 =x_i^2y_j^2 -2x_ix_jy_iy_j +x_j^2y_i^2.$
So we get


$\displaystyle \left(\sum\limits_{i=1}^n x_iy_i\right)^2 +\sum\limits_{1 \leq i< j\leq n}^n(x_iy_j -x_jy_i)^2$ $\displaystyle =$ $\displaystyle \sum\limits_{i=1}^n \left(x_i^2y_i^2\right) +\sum\limits_{1 \leq i< j\leq n}^n \left(x_i^2y_j^2 +x_j^2y_i^2\right)$ (2)
  $\displaystyle =$ $\displaystyle \left(\sum\limits_{i=1}^n x_i^2\right)\left(\sum\limits_{i=1}^n y_i^2\right)$ (3)

Note that changing the roles of $ i$ and $ j$ in $ x_iy_j -x_jy_i$, we get
$\displaystyle x_jy_i -x_iy_j =-(x_iy_j -x_jy_i),$
but the negative sign will disappear when we square. So we can rewrite the last equation to
$\displaystyle \left(\sum\limits_{i=1}^n x_iy_i\right)^2 +\sum\limits_{1 \le i <... ...y_j -x_jy_i)^2 =\left(\sum_{i=1}^n x_i^2\right)\left(\sum_{i=1}^n y_i^2\right).$ (4)

This is equivalent to the stated identity. $ \qedsymbol$



"Lagrange's identity" is owned by mathcam. [ full author list (2) | owner history (1) ]
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Cross-references: identity, equivalent, equation, square, negative, binomial formula, commutative, commutative ring
There are 2 references to this entry.

This is version 18 of Lagrange's identity, born on 2002-12-21, modified 2007-02-25.
Object id is 3802, canonical name is LagrangesIdentity.
Accessed 7901 times total.

Classification:
AMS MSC13A99 (Commutative rings and algebras :: General commutative ring theory :: Miscellaneous)

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Proof of the Lagrange Identity? by spacechecker on 2005-06-04 09:13:01
Are you sure that equation (1) in the proof is correct?
My guess would be that the factor 2 inside the sum
has to be dropped (since the indexes are running
over ALL i /= j, NOT just over the lower/upper triangle).
This would then multiply the last term on the LHS and
also the last term on the RHS by 1/2, which makes sense
for getting (3). Finally, this gets rid of factor 2
in front of the last term on the RHS, otherwise this
would seem hard to bring in line with the claimed equivalence.
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