Parallelogram law

In the parallelogram on the right, let AD = BC = a, AB = DC = b, ${\displaystyle \angle BAD=\alpha .}$ By using the law of cosines in triangle ${\displaystyle \triangle BAD,}$ we get:

$${\displaystyle a^{2}+b^{2}-2ab\cos(\alpha )=BD^{2}.}$$

In a parallelogram, adjacent angles are supplementary, therefore ${\displaystyle \angle ADC=180^{\circ }-\alpha .}$ Using the law of cosines in triangle ${\displaystyle \triangle ADC,}$ produces:

$${\displaystyle a^{2}+b^{2}-2ab\cos(180^{\circ }-\alpha )=AC^{2}.}$$

By applying the trigonometric identity ${\displaystyle \cos(180^{\circ }-x)=-\cos x}$ to the former result proves:

$${\displaystyle a^{2}+b^{2}+2ab\cos(\alpha )=AC^{2}.}$$

Now the sum of squares ${\displaystyle BD^{2}+AC^{2}}$ can be expressed as:

$${\displaystyle BD^{2}+AC^{2}=a^{2}+b^{2}-2ab\cos(\alpha )+a^{2}+b^{2}+2ab\cos(\alpha ).}$$

Simplifying this expression, it becomes:

$${\displaystyle BD^{2}+AC^{2}=2a^{2}+2b^{2}.}$$

Vectors involved in the parallelogram law.

In a normed space, the statement of the parallelogram law is an equation relating norms:

$${\displaystyle 2\|x\|^{2}+2\|y\|^{2}=\|x+y\|^{2}+\|x-y\|^{2}\quad {\text{ for all }}x,y.}$$

The parallelogram law is equivalent to the seemingly weaker statement:

$${\displaystyle 2\|x\|^{2}+2\|y\|^{2}\leq \|x+y\|^{2}+\|x-y\|^{2}\quad {\text{ for all }}x,y}$$

because the reverse inequality can be obtained from it by substituting ${\textstyle {\frac {1}{2}}\left(x+y\right)}$ for ${\displaystyle x,}$ and ${\textstyle {\frac {1}{2}}\left(x-y\right)}$ for ${\displaystyle y,}$ and then simplifying. With the same proof, the parallelogram law is also equivalent to:

$${\displaystyle \|x+y\|^{2}+\|x-y\|^{2}\leq 2\|x\|^{2}+2\|y\|^{2}\quad {\text{ for all }}x,y.}$$

In an inner product space, the norm is determined using the inner product:

$${\displaystyle \|x\|^{2}=\langle x,x\rangle .}$$

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:

$${\displaystyle \|x+y\|^{2}=\langle x+y,x+y\rangle =\langle x,x\rangle +\langle x,y\rangle +\langle y,x\rangle +\langle y,y\rangle ,}$$

$${\displaystyle \|x-y\|^{2}=\langle x-y,x-y\rangle =\langle x,x\rangle -\langle x,y\rangle -\langle y,x\rangle +\langle y,y\rangle .}$$

Adding these two expressions:

$${\displaystyle \|x+y\|^{2}+\|x-y\|^{2}=2\langle x,x\rangle +2\langle y,y\rangle =2\|x\|^{2}+2\|y\|^{2},}$$

as required.

If ${\displaystyle x}$ is orthogonal to ${\displaystyle y,}$ meaning ${\displaystyle \langle x,\ y\rangle =0,}$ and the above equation for the norm of a sum becomes:

$${\displaystyle \|x+y\|^{2}=\langle x,x\rangle +\langle x,y\rangle +\langle y,x\rangle +\langle y,y\rangle =\|x\|^{2}+\|y\|^{2},}$$

which is Pythagoras' theorem.

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 ${\displaystyle x=(x_{1},x_{2},\ldots ,x_{n})}$ in the real coordinate space ${\displaystyle \mathbb {R} ^{n}}$ is the ${\displaystyle p}$-norm:

$${\displaystyle \|x\|_{p}=\left(|x_{1}|^{p}+|x_{2}|^{p}+\dotsb +|x_{n}|^{p}\right)^{1/p}.}$$

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 ${\displaystyle p}$-norm if and only if ${\displaystyle p=2,}$ the so-called Euclidean norm or standard norm.[1][2]

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:

$${\displaystyle \langle x,y\rangle ={\frac {\|x+y\|^{2}-\|x-y\|^{2}}{4}},}$$

or equivalently by

$${\displaystyle {\frac {\|x+y\|^{2}-\|x\|^{2}-\|y\|^{2}}{2}}\qquad {\text{ or }}\qquad {\frac {\|x\|^{2}+\|y\|^{2}-\|x-y\|^{2}}{2}}.}$$

In the complex case it is given by:

$${\displaystyle \langle x,y\rangle ={\frac {\|x+y\|^{2}-\|x-y\|^{2}}{4}}+i{\frac {\|ix-y\|^{2}-\|ix+y\|^{2}}{4}}.}$$

For example, using the ${\displaystyle p}$-norm with ${\displaystyle p=2}$ and real vectors ${\displaystyle x}$ and ${\displaystyle y,}$ the evaluation of the inner product proceeds as follows:

$${\displaystyle {\begin{aligned}\langle x,y\rangle &={\frac {\|x+y\|^{2}-\|x-y\|^{2}}{4}}\\[4mu]&={\tfrac {1}{4}}\left(\sum _{i}|x_{i}+y_{i}|^{2}-\sum _{i}|x_{i}-y_{i}|^{2}\right)\\[2mu]&={\tfrac {1}{4}}\left(4\sum _{i}x_{i}y_{i}\right)\\&=x\cdot y,\\\end{aligned}}}$$

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 ${\displaystyle \|\cdot \|}$ is for the norm to satisfy Ptolemy's inequality:[3]

$${\displaystyle \|x-y\|\,\|z\|~+~\|y-z\|\,\|x\|~\geq ~\|x-z\|\,\|y\|\qquad {\text{ for all vectors }}x,y,z.}$$

• Commutative property – Property of some mathematical operations
• François Daviet – Italian mathematician and military officer (1734–1798)
• Inner product space – Generalization of the dot product; used to define Hilbert spaces
• Minkowski distance – Mathematical metric in normed vector space
• Normed vector space – Vector space on which a distance is defined
• Polarization identity – Formula relating the norm and the inner product in a inner product space
• Ptolemy's inequality

• Weisstein, Eric W. "Parallelogram Law". MathWorld.
• The Parallelogram Law Proven Simply at Dreamshire blog
• The Parallelogram Law: A Proof Without Words at cut-the-knot