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

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 any of the expressions: $${\displaystyle {\begin{aligned}\langle x,y\rangle &={\tfrac {1}{4}}{\bigl (}\|x+y\|^{2}-\|x-y\|^{2}{\bigr )}\\[5mu]&={\tfrac {1}{2}}{\bigl (}\|x+y\|^{2}-\|x\|^{2}-\|y\|^{2}{\bigr )}\\[5mu]&={\tfrac {1}{2}}{\bigl (}\|x\|^{2}+\|y\|^{2}-\|x-y\|^{2}{\bigr )}.\end{aligned}}}$$

In the complex case it is given by: $${\displaystyle \langle x,y\rangle ={\tfrac {1}{4}}{\bigl (}\|x+y\|^{2}-\|x-y\|^{2}{\bigr )}+{\tfrac {1}{4}}i{\bigl (}\|ix-y\|^{2}-\|ix+y\|^{2}{\bigr )}.}$$

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: $${\displaystyle {\begin{aligned}\langle x,y\rangle &={\tfrac {1}{4}}{\bigl (}\|x+y\|^{2}-\|x-y\|^{2}{\bigr )}\\[8mu]&={\tfrac {1}{4}}{\Bigl (}\sum _{i}|x_{i}+y_{i}|^{2}-\sum _{i}|x_{i}-y_{i}|^{2}{\Bigr )}\\[2mu]&={\tfrac {1}{4}}{\Bigl (}4\sum _{i}x_{i}y_{i}{\Bigr )}\\&=x\cdot y,\\\end{aligned}}}$$ 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: For any three vectors ⁠${\displaystyle x}$⁠, ⁠${\displaystyle y}$⁠, and ⁠${\displaystyle z}$⁠,^[3] $${\displaystyle \|x-y\|\,\|z\|~+~\|y-z\|\,\|x\|~\geq ~\|x-z\|\,\|y\|.}$$

• François Daviet – Royal Sardinian Army officer and mathematician (1734–1798)
• Inner product space – Vector space with generalized dot product
• Minkowski distance – Vector distance using pth powers
• 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 – Relation between distances of four points