Parallelogram law

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The sides of parallelogram ABCD are shown in blue and the diagonals in red. The sum of the areas of the blue squares equal that of the red ones.

In mathematics, the simplest form of the parallelogram law (also called the parallelogram identity) belongs to elementary geometry. It states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. We use these notations for the sides: AB, BC, CD, DA. But since in Euclidean geometry a parallelogram necessarily has opposite sides equal, that is, AB = CD and BC = DA, the law can be stated as <math display=block>2AB^2 + 2BC^2 = AC^2 + BD^2\,</math>

If the parallelogram is a rectangle, the two diagonals are of equal lengths AC = BD, so <math display=block>2AB^2 + 2BC^2 = 2AC^2</math> and the statement reduces to the Pythagorean theorem. For the general quadrilateral (with four sides not necessarily equal) Euler's quadrilateral theorem states <math display=block>AB^2 + BC^2 + CD^2+DA^2 = AC^2+BD^2 + 4x^2,</math> where <math>x</math> is the length of the line segment joining the midpoints of the diagonals. It can be seen from the diagram that <math>x = 0</math> for a parallelogram, and so the general formula simplifies to the parallelogram law.

ProofEdit

File:Color parallelogram.svg

In the parallelogram on the right, let AD = BC = a, AB = DC = b, <math>\angle BAD = \alpha.</math> By using the law of cosines in triangle <math>\triangle BAD,</math> we get: <math display=block>a^2 + b^2-2ab\cos(\alpha) = BD^2.</math>

In a parallelogram, adjacent angles are supplementary, therefore <math>\angle ADC = 180^{\circ} - \alpha.</math> Using the law of cosines in triangle <math>\triangle ADC,</math> produces: <math display=block>a^2 + b^2 - 2ab\cos(180^{\circ}-\alpha) = AC^2.</math>

By applying the trigonometric identity <math>\cos(180^{\circ} - x) = -\cos x</math> to the former result proves: <math display=block>a^2 + b^2 + 2ab\cos(\alpha) = AC^2.</math>

Now the sum of squares <math>BD^2 + AC^2</math> can be expressed as: <math display=block>BD^2 + AC^2 = a^2 + b^2 -2ab\cos(\alpha) + a^2 + b^2 +2ab\cos(\alpha).</math>

Simplifying this expression, it becomes: <math display=block>BD^2 + AC^2 = 2a^2 + 2b^2.</math>

The parallelogram law in inner product spacesEdit

File:Parallelogram law.svg

Vectors involved in the parallelogram law.

In a normed space, the statement of the parallelogram law is an equation relating norms: <math display=block>2\|x\|^2 + 2\|y\|^2 = \|x+y\|^2 + \|x-y\|^2 \quad \text{ for all } x, y.</math>

The parallelogram law is equivalent to the seemingly weaker statement: <math display=block>2\|x\|^2 + 2\|y\|^2 \leq \|x + y\|^2 + \|x - y\|^2 \quad \text{ for all } x, y</math> because the reverse inequality can be obtained from it by substituting <math display=inline>\frac{1}{2}\left( x + y \right)</math> for <math>x,</math> and <math display=inline>\frac{1}{2}\left( x - y \right)</math> for <math>y,</math> and then simplifying. With the same proof, the parallelogram law is also equivalent to: <math display=block>\|x + y\|^2 + \|x - y\|^2 \leq 2\|x\|^2 + 2\|y\|^2 \quad \text{ for all } x, y.</math>

In an inner product space, the norm is determined using the inner product: <math display=block>\|x\|^2 = \langle x, x\rangle.</math>

As a consequence of this definition, in an inner product space the parallelogram law is an algebraic identity, readily established using the properties of the inner product: <math display=block>\|x+y\|^2 = \langle x+y, x+y\rangle = \langle x, x\rangle + \langle x, y\rangle + \langle y, x\rangle + \langle y, y\rangle,</math> <math display=block>\|x-y\|^2 = \langle x-y, x-y\rangle = \langle x, x\rangle - \langle x, y\rangle - \langle y, x\rangle + \langle y, y\rangle.</math>

Adding these two expressions: <math display=block>\|x+y\|^2 + \|x-y\|^2 = 2\langle x, x\rangle + 2\langle y, y\rangle = 2\|x\|^2 + 2\|y\|^2,</math> as required.

If <math>x</math> is orthogonal to <math>y,</math> meaning <math>\langle x ,\ y \rangle = 0,</math> and the above equation for the norm of a sum becomes: <math display=block>\|x+y\|^2 = \langle x, x\rangle + \langle x, y\rangle + \langle y, x\rangle + \langle y, y\rangle = \|x\|^2 + \|y\|^2,</math> which is Pythagoras' theorem.

Normed vector spaces satisfying the parallelogram lawEdit

Most real and complex normed vector spaces do not have inner products, but all normed vector spaces have norms (by definition). For example, a commonly used norm for a vector <math>x = (x_1, x_2, \ldots, x_n)</math> in the real coordinate space <math>\R^n</math> is the <math>p</math>-norm: <math display=block>\|x\| _p = \left(|x_1|^p + |x_2|^p + \dotsb + |x_n|^p\right)^{1/p}.</math>

Given a norm, one can evaluate both sides of the parallelogram law above. A remarkable fact is that if the parallelogram law holds, then the norm must arise in the usual way from some inner product. In particular, it holds for the <math>p</math>-norm if and only if <math>p = 2,</math> the so-called Template:Em norm or Template:Em norm.<ref name="Pelinovsky">Template:Cite book</ref><ref name=Saxe>Template:Cite book</ref>

For any norm satisfying the parallelogram law (which necessarily is an inner product norm), the inner product generating the norm is unique as a consequence of the polarization identity. In the real case, the polarization identity is given by any of the expressions: <math display=block>\begin{align} \langle x, y \rangle &= \tfrac14\bigl(\|x+y\|^2 - \|x-y\|^2\bigr) \\[5mu] &= \tfrac12\bigl(\|x+y\|^2 - \|x\|^2 - \|y\|^2\bigr) \\[5mu] &= \tfrac12\bigl(\|x\|^2 + \|y\|^2 - \|x-y\|^2\bigr). \end{align}</math>

In the complex case it is given by: <math display=block>\langle x, y \rangle = \tfrac14\bigl(\|x+y\|^2 - \|x-y\|^2\bigr) + \tfrac14i\bigl(\|ix-y\|^2 - \|ix+y\|^2\bigr).</math>

For example, using the <math>p</math>-norm with <math>p = 2</math> and real vectors <math>x</math> and <math>y,</math> the evaluation of the inner product proceeds as follows: <math display=block>\begin{align} \langle x, y \rangle &= \tfrac14\bigl(\|x+y\|^2 - \|x-y\|^2\bigr)\\[8mu] &= \tfrac{1}{4} \Bigl(\sum_i |x_i +y_i|^2 - \sum_i |x_i-y_i|^2\Bigr)\\[2mu] &= \tfrac{1}{4} \Bigl(4 \sum_i x_i y_i\Bigr)\\ &= x \cdot y,\\ \end{align}</math> which is the standard dot product of two vectors.

Another necessary and sufficient condition for there to exist an inner product that induces the given norm <math>\|\cdot\|</math> is for the norm to satisfy Ptolemy's inequality: For any three vectors Template:Tmath, Template:Tmath, and Template:Tmath,<ref>Template:Cite journal</ref> <math display=block>\|x - y\| \, \|z\| ~+~ \|y - z\| \, \|x\| ~\geq~ \|x - z\| \, \|y\|.</math>

ReferencesEdit

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External linksEdit

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