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Cauchy-Lagrange Identity

$({a_1}^2+{a_2}^2+\ldots+{a_n}^2)({b_1}^2+{b_2}^2+\ldots+{b_n}^2)$
$ =(a_1b_2-a_2b_1)^2+(a_1b_3-a_3b_1)^2+\ldots+(a_{n-1}b_n-a_nb_{n-1})^2.$
From this identity, the $n$-D Cauchy Inequality follows.




© 1996-9 Eric W. Weisstein
1999-05-26