View source for Polarization identity

{{short description|Formula relating the norm and the inner product in a inner product space}}
{{About|quadratic forms|formulas for higher-degree polynomials|Polarization of an algebraic form}}

[[File:Parallelogram law.svg|thumb|Vectors involved in the polarization identity <math>2\|x\|^2 + 2\|y\|^2 = \|x+y\|^2 + \|x-y\|^2.</math> ]]
In [[linear algebra]], a branch of [[mathematics]], the '''polarization identity''' is any one of a family of formulas that express the [[inner product]] of two [[Vector (geometry)|vectors]] in terms of the [[Norm (mathematics)|norm]] of a [[normed vector space]]. 
If a [[Norm (mathematics)|norm]] arises from an inner product then the polarization identity can be used to express this inner product entirely in terms of the norm. The polarization identity shows that a norm can arise from at most one inner product; however, there exist norms that do not arise from any inner product. 

The norm associated with any [[inner product space]] satisfies the [[parallelogram law]]: <math>\|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2.</math> 
In fact, as observed by [[John von Neumann]],{{sfn|Lax|2002|p=53}} the parallelogram law characterizes those norms that arise from inner products. 
Given a [[Normed vector space|normed space]] <math>(H, \|\cdot\|)</math>, the parallelogram law holds for <math>\|\cdot\|</math> if and only if there exists an inner product <math>\langle \cdot, \cdot \rangle</math> on <math>H</math> such that <math>\|x\|^2 = \langle x,\ x\rangle</math> for all <math>x \in H,</math> in which case this inner product is uniquely determined by the norm via the polarization identity.<ref name=Blanchard>{{cite book|author=[[Philippe Blanchard]], Erwin Brüning|chapter=Proposition 14.1.2 (Fréchet–von Neumann–Jordan)|chapter-url=https://books.google.com/books?id=1g2rikccHcgC&pg=PA192|page=192|title=Mathematical methods in physics: distributions, Hilbert space operators, and variational methods|year=2003|publisher=Birkhäuser|isbn=0817642285}}</ref><ref name=Teschl>{{cite book|author=[[Gerald Teschl]]|title=Mathematical methods in quantum mechanics: with applications to Schrödinger operators|chapter=Theorem 0.19 (Jordan–von Neumann)|page=19|url=https://www.mat.univie.ac.at/~gerald/ftp/book-schroe/|isbn=978-0-8218-4660-5|year=2009|publisher=American Mathematical Society Bookstore}}</ref>

== Polarization identities ==

Any [[Inner product space|inner product]] on a vector space induces a norm by the equation
<math display=block>\|x\| = \sqrt{\langle x, x \rangle}.</math>
The polarization identities reverse this relationship, recovering the inner product from the norm.
Every inner product satisfies:
<math display=block>\|x + y\|^2 = \|x\|^2 + \|y\|^2 + 2\operatorname{Re}\langle x, y \rangle \qquad \text{ for all vectors } x, y.</math>

Solving for <math>\operatorname{Re}\langle x, y \rangle</math> gives the formula <math>\operatorname{Re}\langle x, y \rangle = \frac{1}{2} \left(\|x+y\|^2 - \|x\|^2 - \|y\|^2\right).</math> If the inner product is real then <math>\operatorname{Re}\langle x, y \rangle = \langle x, y \rangle</math> and this formula becomes a polarization identity for real inner products.

=== Real vector spaces ===

If the vector space is over the [[real number]]s then the polarization identities are:{{sfn|Schechter|1996|pp=601-603}}
<math display="block">\begin{alignat}{4}
\langle x, y \rangle
&= \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2\right) \\[3pt]
&= \frac{1}{2} \left(\|x+y\|^2 - \|x\|^2 - \|y\|^2\right) \\[3pt]
&= \frac{1}{2} \left(\|x\|^2 + \|y\|^2 - \|x-y\|^2\right). \\[3pt]
\end{alignat}</math>

These various forms are all equivalent by the [[parallelogram law]]:<ref group="proof" name="RealPartEquivalentFormsAndFormulasProof" />
<math display="block">2\|x\|^2 + 2\|y\|^2 = \|x+y\|^2 + \|x-y\|^2.</math>

This further implies that <math>L^p</math> class is not a [[Hilbert space]] whenever {{tmath|1= p\neq 2 }}, as the parallelogram law is not satisfied. For the sake of counterexample, consider <math>x=1_A</math> and <math>y=1_B</math> for any two disjoint subsets <math>A,B</math> of general domain <math>\Omega\subset\mathbb{R}^n</math> and compute the measure of both sets under parallelogram law.

=== Complex vector spaces ===

For vector spaces over the [[complex number]]s, the above formulas are not quite correct because they do not describe the [[imaginary part]] of the (complex) inner product. 
However, an analogous expression does ensure that both real and imaginary parts are retained. 
The complex part of the inner product depends on whether it is [[Antilinear map|antilinear]] in the first or the second argument. 
The notation <math>\langle x | y \rangle,</math> which is commonly used in physics will be assumed to be [[Antilinear map|antilinear]] in the {{em|first}} argument while <math>\langle x,\, y \rangle,</math> which is commonly used in mathematics, will be assumed to be antilinear in its {{em|second}} argument. 
They are related by the formula:
<math display=block>\langle x,\, y \rangle = \langle y \,|\, x \rangle \quad \text{ for all } x, y \in H.</math>

The [[real part]] of any inner product (no matter which argument is antilinear and no matter if it is real or complex) is a symmetric bilinear map that for any <math>x, y \in H</math> is always equal to:{{sfn|Schechter|1996|pp=601-603}}<ref group=proof name=RealPartEquivalentFormsAndFormulasProof />
<math display=block>\begin{alignat}{4}
R(x, y) 
:&= \operatorname{Re} \langle x \mid y \rangle = \operatorname{Re} \langle x, y \rangle \\
&= \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2\right) \\
&= \frac{1}{2} \left(\|x+y\|^2 - \|x\|^2 - \|y\|^2\right) \\[3pt]
&= \frac{1}{2} \left(\|x\|^2 + \|y\|^2 - \|x-y\|^2\right). \\[3pt]
\end{alignat}</math>

It is always a [[symmetric map]], meaning that<ref group=proof name=RealPartEquivalentFormsAndFormulasProof />
<math display=block>R(x, y) = R(y, x) \quad \text{ for all } x, y \in H,</math>
and it also satisfies:<ref group=proof name=RealPartEquivalentFormsAndFormulasProof /> 
<math display=block>R(ix, y) = - R(x, iy) \quad \text{ for all } x, y \in H,</math>
which in plain English says that to move a factor of <math>i</math> to the other argument, introduce a negative sign. These properties can be proven either from the properties of inner products directly or from properties of norms by using the polarization identity.

{{collapse top|title={{anchor|Proof of formulas and equivalent forms}}Proof of properties of <math>R</math> using the polarization identity|left=true}}
Let 
<math display=block>R(x, y) := \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2\right).</math> 
Then <math>R(y, x)=\frac{1}{4} \left(\|y+x\|^2 - \|y-x\|^2\right)=\frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2\right)=R(x, y),</math>

which proves that {{tmath|1= R(x, y) = R(y, x) }}.

Additionally,

<math>R(ix, y)=\frac{1}{4} \left(\|ix+y\|^2 - \|ix-y\|^2\right)</math>

<math>=\frac{1}{4} \left(\|x+(1/i)y\|^2 - \|x-(1/i)y\|^2\right)</math>


<math>=\frac{1}{4} \left(\|x-iy\|^2 - \|x+iy\|^2\right)</math>

<math>=-\frac{1}{4} \left( \|x+iy\|^2 - \|x-iy\|^2\right)</math>

<math>=-R(x, iy),</math>

which proves that <math>R(ix, y) = - R(x, iy).</math> 
<math>\blacksquare</math>
<!--Proof:
<math display=block>\begin{alignat}{4}
R(y,ix)&=\frac{1}{2}\left(\langle y,ix \rangle+\langle ix,y \rangle\right) \\
&=\frac{1}{2i}\left(\langle iy,ix \rangle-\langle ix,iy \rangle\right) \\
&=\frac{1}{2i}\left(-i\langle iy,x \rangle-i\langle x,iy \rangle\right) \\
&=-\frac{1}{2}\left(\langle x,iy \rangle+\langle iy,x \rangle\right) \\
&=-R(x,iy)
\end{alignat}</math>-->

{{collapse bottom}}

Unlike its real part, the [[imaginary part]] of a complex inner product depends on which argument is antilinear. 

'''Antilinear in first argument'''

The polarization identities for the inner product <math>\langle x \,|\, y \rangle,</math> which is [[Antilinear map|antilinear]] in the {{em|first}} argument, are
:<math>\begin{alignat}{4}
\langle x \,|\, y \rangle 
&= \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2 - i\|x + iy\|^2 + i\|x - iy\|^2\right) \\
&= \frac{1}{4} \sum_{k=0}^3 i^k\|x+(-i)^ky\|^2 \\
&= R(x, y) - i R(x, iy) \\
&= R(x, y) + i R(ix, y) \\
\end{alignat}</math>
where <math>x, y \in H.</math> 
The second to last equality is similar to the formula [[Real and imaginary parts of a linear functional|expressing a linear functional]] <math>\varphi</math> in terms of its real part: <math>\varphi(y) = \operatorname{Re} \varphi(y) - i (\operatorname{Re} \varphi)(i y).</math> 

'''Antilinear in second argument'''

The polarization identities for the inner product <math>\langle x, \ y \rangle,</math> which is [[Antilinear map|antilinear]] in the {{em|second}} argument, follows from that of <math>\langle x \,|\, y \rangle</math> by the relationship: 
<math>\langle x, \ y \rangle := \langle y \,|\, x \rangle = \overline{\langle x \,|\, y \rangle} \quad \text{ for all } x, y \in H.</math> 
So for any <math>x, y \in H,</math>{{sfn|Schechter|1996|pp=601-603}}
: <math>\begin{alignat}{4}
\langle x,\, y \rangle 
&= \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2 + i\|x + iy\|^2 - i\|x - iy\|^2\right) \\
&= R(x, y) + i R(x, iy) \\
&= R(x, y) - i R(ix, y). \\
\end{alignat}</math>

This expression can be phrased symmetrically as:<ref>{{Cite web|last=Butler|first=Jon|date=20 June 2013|title=norm - Derivation of the polarization identities?|url=https://math.stackexchange.com/questions/425173/derivation-of-the-polarization-identities|url-status=live|archive-url=https://archive.today/20201014185358/https://math.stackexchange.com/questions/425173/derivation-of-the-polarization-identities|archive-date=14 October 2020|access-date=2020-10-14|website=Mathematics Stack Exchange}} See Harald Hanche-Olson's answer.</ref>
<math display=block>\langle x, y \rangle = \frac{1}{4} \sum_{k=0}^3 i^k \left\|x + i^k y\right\|^2.</math>

'''Summary of both cases'''

Thus if <math>R(x, y) + i I(x, y)</math> denotes the real and imaginary parts of some inner product's value at the point <math>(x, y) \in H \times H</math> of its domain, then its imaginary part will be:
<math display=block>I(x, y) ~=~ 
\begin{cases}
~R({\color{red}i} x, y)  & \qquad \text{ if antilinear in the } {\color{red}1} \text{st argument} \\
~R(x, {\color{blue}i} y) & \qquad \text{ if antilinear in the } {\color{blue}2} \text{nd argument} \\
\end{cases}</math>
where the scalar <math>i</math> is always located in the same argument that the inner product is antilinear in.

Using {{tmath|1= R(ix, y) = - R(x, iy) }}, the above formula for the imaginary part becomes:
<math display=block>I(x, y) ~=~ 
\begin{cases}
-R(x, {\color{black}i} y) & \qquad \text{ if antilinear in the } {\color{black}1} \text{st argument} \\
-R({\color{black}i} x, y) & \qquad \text{ if antilinear in the } {\color{black}2} \text{nd argument} \\
\end{cases}</math>

== Reconstructing the inner product ==

In a normed space <math>(H, \|\cdot\|),</math> if the [[parallelogram law]]
<math display=block>\|x+y\|^2 ~+~ \|x-y\|^2 ~=~ 2\|x\|^2+2\|y\|^2</math>
holds, then there exists a unique [[inner product]] <math>\langle \cdot,\ \cdot\rangle</math> on <math>H</math> such that <math>\|x\|^2 = \langle x,\ x\rangle</math> for all <math>x \in H.</math>{{sfn|Schechter|1996|pp=601-603}}{{sfn|Lax|2002|p=53}}

{{math proof|proof=
We will only give the real case here; the proof for complex vector spaces is analogous.

By the above formulas, if the norm is described by an inner product (as we hope), then it must satisfy
<math display=block>\langle x, \ y \rangle = \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2\right) \quad \text{ for all } x, y \in H,</math>
which may serve as a definition of the unique candidate <math>\langle \cdot, \cdot \rangle  </math> for the role of a suitable inner product. Thus, the uniqueness is guaranteed.

It remains to prove that this formula indeed defines an inner product and that this inner product induces the norm <math>\|\cdot\|.</math> 
Explicitly, the following will be shown:
# <math>\langle x, x \rangle = \|x\|^2, \quad x \in H</math>
# <math>\langle x, y \rangle = \langle y, x \rangle, \quad x, y \in H</math>
# <math>\langle x+z, y\rangle = \langle x, y\rangle + \langle z, y\rangle \quad \text{ for all } x, y, z \in H,</math> 
# <math>\langle \alpha x, y \rangle = \alpha\langle x, y \rangle \quad \text{ for all } x, y \in H \text{ and all } \alpha \in \R</math>
(This axiomatization omits [[Positive-definite bilinear form|positivity]], which is implied by (1) and the fact that <math>\|\cdot\|</math> is a norm.)

For properties (1) and (2), substitute: <math display=inline>\langle x, x \rangle = \frac{1}{4} \left(\|x+x\|^2 - \|x-x\|^2\right) = \|x\|^2,</math> and <math>\|x-y\|^2 = \|y-x\|^2.</math>

For property (3), it is convenient to work in reverse. 
It remains to show that 
<math display=block>\|x+z+y\|^2 - \|x+z-y\|^2 \overset{?}{=} \|x+y\|^2 - \|x-y\|^2 + \|z+y\|^2 - \|z-y\|^2</math>
or equivalently,
<math display=block>2\left(\|x+z+y\|^2 + \|x-y\|^2\right) - 2\left(\|x+z-y\|^2 + \|x+y\|^2\right) \overset{?}{=} 2\|z+y\|^2 - 2\|z-y\|^2.</math>

Now apply the parallelogram identity:
<math display=block>2\|x+z+y\|^2 + 2\|x-y\|^2 = \|2x+z\|^2 + \|2y+z\|^2</math>
<math display=block>2\|x+z-y\|^2 + 2\|x+y\|^2 = \|2x+z\|^2 + \|z-2y\|^2</math>
Thus it remains to verify:
<math display=block>\cancel{\|2x+z\|^2} + \|2y+z\|^2 - (\cancel{\|2x+z\|^2} + \|z-2y\|^2) \overset{?}{{}={}} 2\|z+y\|^2 - 2\|z-y\|^2</math>
<math display=block>\|2y+z\|^2 - \|z-2y\|^2 \overset{?}{=} 2\|z+y\|^2 - 2\|z-y\|^2</math>

But the latter claim can be verified by subtracting the following two further applications of the parallelogram identity:
<math display=block>\|2y+z\|^2 + \|z\|^2 = 2\|z+y\|^2 + 2\|y\|^2</math>
<math display=block>\|z-2y\|^2 + \|z\|^2 = 2\|z-y\|^2 + 2\|y\|^2</math>

Thus (3) holds.

It can be verified by induction that (3) implies (4), as long as <math>\alpha \in \Z.</math> 
But "(4) when <math>\alpha \in \Z</math>" implies "(4) when <math>\alpha \in \Q</math>". 
And any positive-definite, [[Characteristic zero|real-valued]], <math>\Q</math>-bilinear form satisfies the [[Cauchy–Schwarz inequality]], so that <math>\langle \sdot,\sdot \rangle</math> is continuous. 
Thus <math>\langle \sdot,\sdot \rangle</math> must be <math>\R</math>-linear as well. 
}}

Another necessary and sufficient condition for there to exist an inner product that induces a given norm <math>\|\cdot\|</math> is for the norm to satisfy [[Ptolemy's inequality]], which is:<ref>{{Cite journal|last=Apostol|first=Tom M.|date=1967|title=Ptolemy's Inequality and the Chordal Metric|url=https://www.tandfonline.com/doi/pdf/10.1080/0025570X.1967.11975804|journal=Mathematics Magazine|volume=40|issue=5|pages=233–235| language=en| doi=10.2307/2688275|jstor=2688275|url-access=subscription}}</ref>
<math display=block>\|x - y\| \, \|z\| ~+~ \|y - z\| \, \|x\| ~\geq~ \|x - z\| \, \|y\| \qquad \text{ for all vectors } x, y, z.</math>

== Applications and consequences ==

If <math>H</math> is a complex Hilbert space then <math>\langle x \mid y \rangle</math> is real if and only if its imaginary part is {{tmath|1= 0 = R(x, iy) = \frac{1}{4} \left(\Vert x+iy \Vert^2 - \Vert x-iy \Vert^2\right) }}, which happens if and only if {{tmath|1= \Vert x+iy \Vert = \Vert x-iy \Vert }}. 
Similarly, <math>\langle x \mid y \rangle</math> is (purely) imaginary if and only if {{tmath|1= \Vert x+y \Vert = \Vert x-y \Vert }}. 
For example, from <math>\|x+ix\| = |1+i| \|x\| = \sqrt{2} \|x\| = |1-i| \|x\| = \|x-ix\|</math> it can be concluded that <math>\langle x | x \rangle</math> is real and that <math>\langle x | ix \rangle</math> is purely imaginary. 

=== Isometries ===

If <math>A : H \to Z</math> is a [[Linear map|linear]] [[isometry]] between two Hilbert spaces (so <math>\|A h\| = \|h\|</math> for all <math>h \in H</math>) then 
<math display=block>\langle A h, A k \rangle_Z = \langle h, k \rangle_H \quad \text{ for all } h, k \in H;</math>
that is, linear isometries preserve inner products. 

If <math>A : H \to Z</math> is instead an [[Antilinear map|antilinear]] isometry then 
<math display=block>\langle A h, A k \rangle_Z = \overline{\langle h, k \rangle_H} = \langle k, h \rangle_H \quad \text{ for all } h, k \in H.</math> 

=== Relation to the law of cosines ===

The second form of the polarization identity can be written as
<math display=block>\|\textbf{u}-\textbf{v}\|^2 = \|\textbf{u}\|^2 + \|\textbf{v}\|^2 - 2(\textbf{u} \cdot \textbf{v}).</math>

This is essentially a vector form of the [[law of cosines]] for the [[triangle]] formed by the vectors {{tmath|1= \textbf{u} }}, {{tmath|1= \textbf{v} }}, and {{tmath|1= \textbf{u}-\textbf{v} }}. 
In particular,
<math display=block>\textbf{u}\cdot\textbf{v} = \|\textbf{u}\|\,\|\textbf{v}\| \cos\theta,</math>
where <math>\theta</math> is the angle between the vectors <math>\textbf{u}</math> and {{tmath|1= \textbf{v} }}.

The equation is numerically unstable if u and v are similar because of [[catastrophic cancellation]] and should be avoided for numeric computation.

=== Derivation ===

The basic relation between the norm and the [[dot product]] is given by the equation
<math display=block>\|\textbf{v}\|^2 = \textbf{v} \cdot \textbf{v}.</math>

Then
<math display=block>\begin{align}
\|\textbf{u} + \textbf{v}\|^2
&= (\textbf{u} + \textbf{v}) \cdot (\textbf{u} + \textbf{v}) \\[3pt]
&= (\textbf{u} \cdot \textbf{u}) + (\textbf{u} \cdot \textbf{v}) + (\textbf{v} \cdot \textbf{u}) + (\textbf{v} \cdot \textbf{v}) \\[3pt]
&= \|\textbf{u}\|^2 + \|\textbf{v}\|^2 + 2(\textbf{u} \cdot \textbf{v}),
\end{align}</math>
and similarly
<math display=block>\|\textbf{u} - \textbf{v}\|^2 = \|\textbf{u}\|^2 + \|\textbf{v}\|^2 - 2(\textbf{u} \cdot \textbf{v}).</math>

Forms (1) and (2) of the polarization identity now follow by solving these equations for {{tmath|1= \textbf{u} \cdot \textbf{v} }}, while form (3) follows from subtracting these two equations. 
(Adding these two equations together gives the parallelogram law.)

== Generalizations ==

=== Jordan–von Neumann theorems ===
The standard [[Pascual Jordan|Jordan]]–[[John von Neumann|von Neumann]] theorem, as stated previously, is that the if a norm satisfies the parallelogram law, then it can be induced by an inner product defined by the polarization identity. There are variants of the theorem.<ref>{{Cite journal |last=Day |first=Mahlon M. |date=1947 |title=Some Characterizations of Inner-Product Spaces |url=https://www.jstor.org/stable/1990458 |journal=Transactions of the American Mathematical Society |volume=62 |issue=2 |pages=320–337 |doi=10.2307/1990458 |issn=0002-9947}}</ref>

Define various senses of orthogonality:

* isosceles: <math display="inline">\|x+y \| =\|x-y \|</math>
* Roberts’: <math display="inline">\left\|x+ty\right\|=\left\|x-ty\right\|</math> for all scalar <math display="inline">t</math>.
* Pythagorean: <math display="inline">\left\|x+y\right\|^2=\|x\|^2+\left\|y\right\|^2</math>
* Birkhoff–James: <math display="inline">\|x\| \leq \|x + ty \|</math> for all scalar <math display="inline">t</math>.

Let <math display="inline">V</math> be a vector space over the real or complex numbers. Let <math display="inline">\|\cdot\|</math> be a norm over <math display="inline">V</math>. We consider conditions for which the norm is induced by an inner product. In the following statements, whenever a scalar appears, the scalar may be restricted to be merely real, even when <math display="inline">V</math> is over the complex numbers.

* (von Neumann–Jordan condition) The norm satisfies the parallelogram identity.
* (weakened von Neumann–Jordan condition) <math display="inline">\|x + y\|^2 + \|x - y\|^2 = 4</math> for all unit vectors <math display="inline">x,y</math>. That is, the norm satisfies the parallelogram identity for unit vectors.
* For any <math display="inline">x, y \in V</math>, the set of points equidistant to <math display="inline">x, y</math> is flat, that is, an affine subspace.
* Orthogonality in either isosceles or Roberts’ sense is either additive or homogeneous on one variable.
* For every two-dimensional subspace <math display="inline">W \subset V</math>, for every <math display="inline">x \in W</math>, there exists <math display="inline">y \in W</math> that is Roberts’ orthogonal to <math display="inline">x</math>.
* Isosceles orthogonality implies Pythagorean orthogonality.
* Pythagorean orthogonality implies isosceles orthogonality.
* If <math display="inline">x, y</math> are Pythagorean orthogonal, then so are <math display="inline">x, -y</math>.
* Birkhoff–James orthogonality is symmetric.
* If <math display="inline">\|x\|=\|y\|</math> and <math display="inline">t, s</math> are real, then <math display="inline">\|t x+s y\|=\|s x+t y\|</math>.

For the real vector space, there is also the condition:

* Any two-dimensional slice of the unit sphere is an ellipse, that is, parameterizable as <math display="inline">\{x \cos\theta + y \sin\theta : \theta \in [0, 2\pi]\}</math>, for some unit vectors <math display="inline">x, y</math>.
{{Math proof|title=Proof|proof=Since the internal [[John ellipsoid|John ellipse]] <math display="inline">E</math> is unique, for any bijective linear map <math display="inline">T</math> that preserves the unit circle, it must have <math display="inline">TE = E</math>. Since <math display="inline">E</math> must touch the circle at some point <math display="inline">x</math>, we may map <math display="inline">x</math> to any other point <math display="inline">y</math> on the circle, thus every point <math display="inline">y</math> touches the ellipse <math display="inline">E</math>. Thus the circle is the ellipse.}}

The Banach-Mazur rotation problem: Given a [[Separable space|separable]] [[Banach space]] <math display="inline">V</math> such that for any two unit vectors <math display="inline">x, y,</math> there exists a linear surjective isometry <math display="inline">T</math> such that <math display="inline">T(x) = y</math> or <math display="inline">T(y) = x</math>, is <math display="inline">V</math> isometrically isomorphic to a Hilbert space? 

The general case of the problem is [[Open problem|open]]. When the space is parable finite-dimensional, the answer is yes. In other words, given a finite-dimensional normed vector space over the real or complex numbers, if any point on the unit sphere can be mapped (rotated) to any other point by a linear isometry, then the norm is induced by an inner product.<ref>{{cite journal
 | last1 = Becerra Guerrero | first1 = Julio
 | last2 = Rodríguez-Palacios | first2 = A.
 | hdl = 10662/18957
 | issue = 1
 | journal = Extracta Mathematicae
 | mr = 1914238
 | pages = 1–58
 | title = Transitivity of the norm on Banach spaces
 | volume = 17
 | year = 2002}}</ref>

=== Symmetric bilinear forms ===

The polarization identities are not restricted to inner products. 
If <math>B</math> is any [[symmetric bilinear form]] on a vector space, and <math>Q</math> is the [[quadratic form]] defined by
<math display=block>Q(v) = B(v, v),</math>
then
<math display=block>\begin{align}
2 B(u, v) &= Q(u + v) - Q(u) - Q(v), \\
2 B(u, v) &= Q(u)     + Q(v) - Q(u - v), \\
4 B(u, v) &= Q(u + v)        - Q(u - v).
\end{align}</math>

The so-called [[Homogeneous polynomial|symmetrization map]] generalizes the latter formula, replacing <math>Q</math> by a homogeneous polynomial of degree <math>k</math> defined by <math>Q(v) = B(v, \ldots, v),</math> where <math>B</math> is a symmetric <math>k</math>-linear map.<ref>{{harvnb|Butler|2013}}. See Keith Conrad (KCd)'s answer.</ref>

The formulas above even apply in the case where the [[Field (mathematics)|field]] of [[Scalar (mathematics)|scalars]] has [[Characteristic (algebra)|characteristic]] two, though the left-hand sides are all zero in this case. 
Consequently, in characteristic two there is no formula for a symmetric bilinear form in terms of a quadratic form, and they are in fact distinct notions, a fact which has important consequences in [[L-theory]]; for brevity, in this context "symmetric bilinear forms" are often referred to as "symmetric forms".

These formulas also apply to bilinear forms on [[Module (mathematics)|modules]] over a [[commutative ring]], though again one can only solve for <math>B(u, v)</math> if 2 is invertible in the ring, and otherwise these are distinct notions. For example, over the integers, one distinguishes [[integral quadratic form]]s from integral {{em|symmetric}} forms, which are a narrower notion.

More generally, in the presence of a ring involution or where 2 is not invertible, one distinguishes [[ε-quadratic form|<math>\varepsilon</math>-quadratic form]]s and [[ε-symmetric form|<math>\varepsilon</math>-symmetric form]]s; a symmetric form defines a quadratic form, and the polarization identity (without a factor of 2) from a quadratic form to a symmetric form is called the "[[symmetrization]] map", and is not in general an isomorphism. This has historically been a subtle distinction: over the integers it was not until the 1950s that relation between "twos out" (integral {{em|quadratic}} form) and "twos in" (integral {{em|symmetric}} form) was understood – see discussion at [[integral quadratic form]]; and in the [[algebraization]] of [[surgery theory]], Mishchenko originally used {{em|symmetric}} ''L''-groups, rather than the correct {{em|quadratic}} ''L''-groups (as in Wall and Ranicki) – see discussion at [[L-theory]].

=== Homogeneous polynomials of higher degree ===

Finally, in any of these contexts these identities may be extended to [[homogeneous polynomial]]s (that is, [[algebraic form]]s) of arbitrary [[Degree of a polynomial|degree]], where it is known as the [[polarization formula]], and is reviewed in greater detail in the article on the [[polarization of an algebraic form]].

== See also ==

* {{annotated link|Inner product space}}
* {{annotated link|Law of cosines}}
* {{annotated link|Mazur–Ulam theorem}}
* {{annotated link|Minkowski distance}}
* {{annotated link|Parallelogram law}}
* {{annotated link|Ptolemy's inequality}}

== Notes and references ==

{{reflist}}
{{reflist|group=note}}
{{reflist|group=proof|refs=
<ref group=proof name=RealPartEquivalentFormsAndFormulasProof>A proof can be [[#Proof of formulas and equivalent forms|found here]].</ref>}}

== Bibliography ==
{{sfn whitelist|CITEREFLax2002|CITEREFRudin1991|CITEREFSchechter1996}}
* {{Lax Functional Analysis}} <!-- {{sfn|Lax|2002|p=}} -->
* {{Rudin Walter Functional Analysis|edition=2}} <!-- {{sfn|Rudin|1991|p=}} -->
* {{Schechter Handbook of Analysis and Its Foundations}} <!-- {{sfn|Schechter|1996|p=}} -->

{{Hilbert space}}
{{Lp spaces}}
{{Banach spaces}}
{{Functional Analysis}}

[[Category:Abstract algebra]]
[[Category:Linear algebra]]
[[Category:Functional analysis]]
[[Category:Vectors (mathematics and physics)]]
[[Category:Norms (mathematics)]]
[[Category:Algebraic identities]]