Polarization identity

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In mathematics, and more specifically in the theory of normed spaces and pre-Hilbert spaces in functional analysis, a vector space over the real numbers (the formula for the complex case is given in the article on Banach spaces) whose norm is defined in terms of its inner product satisfies as a necessary condition the polarization identity:

$\| x + y \|^2 = \|x\|^2 + \|y\|^2 + 2\langle x, y\rangle. \qquad \qquad (1)$

This identity is analogous to the formula for the square of a binomial:

$(x + y)^2 = x^2 + y^2 + 2 x y. \qquad \qquad (2)$

If y in equation (1) is replaced by −y the result is

$\| x - y \|^2 = \|x\|^2 + \|y\|^2 - 2\langle x, y\rangle. \qquad \qquad (3)$

which corresponds to the cosine law and is analogous to equation (2) with y replaced by −y:

$(x - y)^2 = x^2 + y^2 - 2 x y. \qquad \qquad (4)$

Adding equations (1) and (3) yields

$\|x + y\|^2 + \|x - y\|^2 = 2\|x\|^2 + 2\|y\|^2$

which corresponds to the parallelogram law and is analogous to the sum of equations (2) and (4):

(x + y) 2 + (x - y) 2 = 2 x 2 + 2 y 2 .

[edit] Derivation

Let the norm of a vector be defined as the square root of the inner product of a vector with itself, like so

$\|x\| = \sqrt{\langle x, x\rangle}. \qquad \qquad (5)$

Now find the inner product of x + y with itself:

$\langle x + y, x + y\rangle = \langle x + y, x\rangle + \langle x + y, y\rangle \qquad \qquad (6)$

is the result of distributing the first factor with respect to the sum of the second factor, which is possible due to linearity of the inner product. Distributing the second factors with respect to the sums of the first factors on the right side of equation (6) yields

$\langle x + y, x + y\rangle = \langle x, x\rangle + \langle y, x\rangle + \langle x, y\rangle + \langle y, y\rangle \qquad \qquad (7)$