nLab
parallelogram identity

The Parallelogram Identity

Idea

The parallelogram identity is an identity which characterises those norms which are the norms associated with inner products. An inner product can be considered as being the structure required to define the angle between two vectors and a norm can be considered as being the structure required to define the length of a vector. From standard Euclidean geometry, lengths and angles (almost) determine each other so knowing one, we should be able to define the other.

If length is known, angle can be defined by the cosine law:

c 2=a 2+b 22abcosC c^2 = a^2 + b^2 - 2 a b \cos C

Although this formula could be used to define “angle” for any length (that is, norm), not every length admits a sensible notion of an angle. There are several ways to describe what the fundamental properties of angles should be. One is to say that angles should add:

θ(u,v)+θ(v,w)=θ(u,w) \theta(u,v) + \theta(v,w) = \theta(u,w)

This needs careful interpretation since the angle between two vectors is slightly ambiguous: one can choose the internal angle or the external angle and it is not possible to make a consistent choice. However, modulo that uncertainty, the above is a reasonable property to insist on.

A special case of this is when w=uw = -u. This leads to the following diagram.

Layer 1 u -u u u v v ϕ \phi ψ \psi

The cosine law for this special case leads to two formulae:

u+v 2 =u 2+v 22uvcos(ψ) uv 2 =u 2+v 22uvcos(ϕ) \begin{aligned} {\|u + v\|^2} &= {\|u\|^2} + {\|v\|^2} - 2{\|u\|}{\|v\|} \cos(\psi) \\ {\|u - v\|^2} &= {\|u\|^2} + {\|v\|^2} - 2{\|u\|}{\|v\|} \cos(\phi) \end{aligned}

By imposing the assumption that angles should add correctly, we deduce that ψ=πϕ\psi = \pi - \phi and thus cos(ψ)=cos(ϕ)\cos (\psi) = - \cos(\phi). Summing the two lines above leads to the parallelogram identity:

u+v 2+uv 2=2u 2+2v 2 {\|u + v\|^2} + {\|u - v\|^2} = 2 {\|u\|^2} + 2 {\|v\|^2}

If a norm satisfies this identity, then the definition of angle (using the cosine identity) satisfies all the basic properties of angles that one can consider.

Inner Products

A norm which satisfies the parallelogram identity is the norm associated with an inner product. Using the parallelogram identity, there are three commonly stated equivalent forumlae for the inner product; these are called the polarization identities. (The notation can vary a little, but it is usually some form of round or angled brackets.) These formulae hold for vector spaces over \mathbb{R}; there are similar formulae for \mathbb{C}, but they have more terms.

u,v =12(u+v 2u 2v 2) =12(u 2+v 2uv 2) =14(u+v 2uv 2) \begin{aligned} \langle u, v \rangle &= \frac{1}{2} \left({\|u + v\|^2} - {\|u\|^2} - {\|v\|^2} \right) \\ &= \frac{1}{2} \left({\|u\|^2} + {\|v\|^2} - {\|u - v\|^2}\right) \\ &= \frac{1}{4} \left({\|u + v\|^2} - {\|u - v\|^2}\right) \end{aligned}

It is possible to show directly that this is an inner product. Certain properties are easy to deduce directly from the formulae.

Lemma

The function (u,v)u,v(u,v) \to \langle u, v\rangle satisfies the following properties:

  1. u,u0\langle u, u \rangle \ge 0
  2. u,u=0\langle u, u \rangle = 0 if and only if u=0u = 0
  3. u,0=0\langle u, 0 \rangle = 0
  4. u,v=v,u\langle u, v \rangle = \langle v, u \rangle
  5. λu,λv=λ 2u,v\langle \lambda u, \lambda v \rangle = \lambda^2 \langle u, v \rangle
  6. u,v+w=u,v+u,w\langle u, v + w \rangle = \langle u, v \rangle + \langle u, w\rangle
Proof

All but the last of these is a simple deduction from the formulae. The last is as well, but takes a little work to get it in the correct form. We multiply by 44 to simplify the notation.

4u,v+w =u+v+w 2uvw 2 =(12u+v)+(12u+w) 2(12uv)+(12uw) 2 =(12u+v)+(12u+w) 2+(12u+v)(12u+w) 2 (12uv)(12uw) 2(12uv)+(12uw) 2 =212u+v 2+212u+w 2212uv 2212uw 2 =212u+v 2212uv 2 +212u+w 2212uw 2 =812u,v+812u,w \begin{aligned} 4\langle u, v + w \rangle &= {\|u + v + w\|^2} - {\|u - v - w\|^2} \\ &={\left\|\left(\frac{1}{2}u + v\right) + \left(\frac{1}{2}u + w\right)\right\|^2} - {\left\|\left(\frac{1}{2}u - v\right) + \left(\frac{1}{2}u - w \right)\right\|^2} \\ &={\left\|\left(\frac{1}{2}u + v\right) + \left(\frac{1}{2}u + w\right)\right\|^2} + {\left\|\left(\frac{1}{2}u + v\right) - \left(\frac{1}{2}u + w\right)\right\|^2} \\ &\qquad - {\left\|\left(\frac{1}{2}u - v\right) - \left(\frac{1}{2}u - w\right)\right\|^2} - {\left\|\left(\frac{1}{2}u - v\right) + \left(\frac{1}{2}u - w \right)\right\|^2} \\ &=2{\left\|\frac{1}{2}u + v\right\|^2} + 2{\left\|\frac{1}{2}u + w\right\|^2} - 2{\left\|\frac{1}{2}u - v\right\|^2} - 2{\left\|\frac{1}{2}u - w\right\|^2} \\ &= 2{\left\|\frac{1}{2}u + v\right\|^2} - 2{\left\|\frac{1}{2}u - v\right\|^2} \\ &\quad+ 2{\left\|\frac{1}{2}u + w\right\|^2} - 2{\left\|\frac{1}{2}u - w\right\|^2}\\ &= 8\langle \frac{1}{2}u,v \rangle + 8\langle \frac{1}{2}u,w\rangle \end{aligned}

To get it as stated, we then apply this in the special case of w=0w = 0 to deduce that u,v=212u,v\langle u, v \rangle = 2\langle \frac{1}{2} u, v\rangle. Substituting this back in to the above, we obtain the form in the statement.

The properties above were all reasonably straightforward deductions from the definition. There is one more property that is needed which is a little more complicated. First, we note a useful result about the continuity of the supposed inner product.

Lemma

The map (u,v)12(u+v 2u 2v 2)(u,v) \mapsto \frac{1}{2}({\|u + v\|^2} - {\|u\|^2} - {\|v\|^2}) is continuous.

Proof

This is a simple consequence of composition of continuous maps.

Using that, we can prove the following property of the proposal for an inner product.

Proposition

The map u,v\langle u,v\rangle satisfies the identity

u,λv=λu,v \langle u,\lambda v \rangle = \lambda \langle u, v \rangle

for all vectors uu, vv, and real numbers λ\lambda.

Proof

We first use the formula u,v+w=u,v+u,w\langle u, v + w \rangle = \langle u, v \rangle + \langle u, w\rangle to prove that for any nn \in \mathbb{N}

u,nv=nu,v. \langle u, n v \rangle = n \langle u, v\rangle.

We prove this by induction. It is clearly true for the case n=1n = 1. Assume that it holds for nn, then

u,(n+1)v=u,nv+v=u,nv+u,v=nu,v+u,v=(n+1)u,v \langle u, (n+1)v \rangle = \langle u, n v + v \rangle = \langle u, n v\rangle + \langle u, v \rangle = n \langle u, v \rangle + \langle u, v \rangle = (n + 1)\langle u, v \rangle

whence, by induction, it holds for all nn \in \mathbb{N}.

Next we observe that it holds for all nn \in \mathbb{Z}. It clearly holds for n=0n = 0, and for nn \in \mathbb{N} then

0=u,0=u,vv=u,v+u,v 0 = \langle u, 0 \rangle = \langle u, v - v \rangle = \langle u, v \rangle + \langle u, - v \rangle

whence

u,v=u,v. \langle u, -v \rangle = - \langle u, v \rangle.

Next, we prove it for pq\frac{p}{q} \in \mathbb{Q}. In fact, it is sufficient to prove it for 1n\frac{1}{n} with nn \in \mathbb{N}. For that we observe that

u,1nv=nnu,1nv=1nu,nnv=1nu,v. \langle u, \frac{1}{n} v \rangle = \frac{n}{n} \langle u, \frac{1}{n} v \rangle = \frac{1}{n} \langle u, \frac{n}{n} v \rangle = \frac{1}{n} \langle u, v \rangle.

To get to λ\lambda \in \mathbb{R} we need to appeal to continuity. We find a sequence (q n)(q_n) \in \mathbb{Q} such that (q n)λ(q_n) \to \lambda and see that by continuity

u,λv=u,(limq n)v=limu,q nv=limq nu,v=λu,v. \langle u, \lambda v \rangle = \langle u, (\lim q_n) v \rangle = \lim \langle u, q_n v \rangle = \lim q_n \langle u, v \rangle = \lambda \langle u, v \rangle.

If {\|{-}\|} is a norm on a vector space satisfying the parallelogram law then the function

(u,v)12(u+v 2u 2v 2) (u,v) \mapsto \frac{1}{2}\left({\|u + v\|^2} - {\|u\|^2} - {\|v\|^2}\right)

is an inner product.

Isometries on inner product spaces

Because lengths determine inner products, and because inner products respect linear structure, isometries must also respect linear (or affine) structure:

Theorem

Let VV be a vector space with a norm satisfying the parallelogram law. Then an isometry (distance-preserving function) f:VVf \colon V \to V is an injective linear affine map.

Proof

Put v=f(0)v = f(0). Then the map xf(x)vx \mapsto f(x) - v is also an isometry, so that without loss of generality we may assume f(0)=0f(0) = 0, and prove that ff is a linear isomorphism.

Letting d(x,y)d(x, y) denote the distance between xx and yy, we have

xy=d(x,y)=d(f(x),f(y))=f(x)f(y){\|x-y\|} = d(x, y) = d(f(x), f(y)) = {\|f(x)-f(y)\|}

and then, putting y=0y=0, we have x=f(x){\|x\|} = {\|f(x)\|}. Hence

x,y = 12(x 2+y 2xy 2) = 12(f(x) 2+f(y) 2f(x)f(y) 2) = f(x),f(y)\array{ \langle x, y\rangle & = & \frac1{2}\left({\|x\|^2} + {\|y\|^2} - {\|x-y\|^2}\right) \\ & = & \frac1{2}\left({\|f(x)\|^2} + {\|f(y)\|^2} - {\|f(x)-f(y)\|^2}\right) \\ & = & \langle f(x), f(y) \rangle }

i.e., ff preserves the inner product.

Let WW be the linear subspace spanned by the image f(V)f(V). Then WW inherits an inner product, and if {e i}\{e_i\} is an orthonormal basis of VV (without restriction on cardinality), then {u i=f(e i)}\{u_i = f(e_i)\} is also an orthonormal basis of WW, since ff preserves the inner product. Next, for any linear combination ax+bya x + b y,

f(ax+by),u i W = ax+by,e i V = ax,e i V+by,e i V = af(x),u i W+bf(y),u i W = af(x)+bf(y),u i W\array{ \langle f(a x + b y), u_i \rangle_W & = & \langle a x + b y, e_i \rangle_V \\ & = & a \langle x, e_i\rangle_V + b\langle y, e_i\rangle_V \\ & = & a \langle f(x), u_i \rangle_W + b \langle f(y), u_i \rangle_W \\ & = & \langle a f(x) + b f(y), u_i \rangle_W }

for each u iu_i, whence f(ax+by)=af(x)+bf(y)f(a x + b y) = a f(x) + b f(y). Thus ff is a linear map, and a linear isomorphism onto the subspace WW because it carries a basis e ie_i of VV to a basis u iu_i of WW.

Summarizing, any isometry f:VVf \colon V \to V is of the form

f(x)=Mx+bf(x) = M x + b

where MM is a linear injection and bb is a vector.

For information on related results, see isometry and Mazur-Ulam theorem.

Revised on April 16, 2014 19:13:52 by Toby Bartels (98.16.173.18)