Inlinear algebra, a branch ofmathematics, thepolarization identity is any one of a family of formulas that express theinner product of twovectors in terms of thenorm of anormed vector space. If anorm 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 anyinner product space satisfies theparallelogram law:${\displaystyle \Vert x+y{\Vert }^2+\Vert x-y{\Vert }^2=2\Vert x{\Vert }^2+2\Vert y{\Vert }^2.}$ In fact, as observed byJohn von Neumann,^[1] the parallelogram law characterizes those norms that arise from inner products. Given anormed space${\displaystyle (H,\Vert \cdot \Vert )}$, the parallelogram law holds for${\displaystyle \Vert \cdot \Vert }$ if and only if there exists an inner product${\displaystyle ⟨\cdot ,\cdot ⟩}$ on${\displaystyle H}$ such that${\displaystyle \Vert x{\Vert }^2=⟨x,x⟩}$ for all${\displaystyle x\in H,}$ in which case this inner product is uniquely determined by the norm via the polarization identity.^[2][3]

Anyinner product on a vector space induces a norm by the equation$${\displaystyle \Vert x\Vert =\sqrt{⟨x,x⟩}.}$$ The polarization identities reverse this relationship, recovering the inner product from the norm.Every inner product satisfies:$${\displaystyle \Vert x+y{\Vert }^2=\Vert x{\Vert }^2+\Vert y{\Vert }^2+2\mathrm{Re}⟨x,y⟩\text{ for all vectors }x,y.}$$ 

Solving for${\displaystyle \mathrm{Re}⟨x,y⟩}$  gives the formula${\displaystyle \mathrm{Re}⟨x,y⟩=\frac{1}{2}\left(\Vert x+y{\Vert }^2-\Vert x{\Vert }^2-\Vert y{\Vert }^2\right).}$  If the inner product is real then${\displaystyle \mathrm{Re}⟨x,y⟩=⟨x,y⟩}$  and this formula becomes a polarization identity for real inner products.

If the vector space is over thereal numbers then the polarization identities are:^[4] 

These various forms are all equivalent by theparallelogram law:^{[proof 1]}$${\displaystyle 2\Vert x{\Vert }^2+2\Vert y{\Vert }^2=\Vert x+y{\Vert }^2+\Vert x-y{\Vert }^2.}$$ 

This further implies that${\displaystyle L^p}$  class is not aHilbert space whenever⁠${\displaystyle p\ne 2}$ ⁠, as the parallelogram law is not satisfied. For the sake of counterexample, consider${\displaystyle x=1_A}$  and${\displaystyle y=1_B}$  for any two disjoint subsets${\displaystyle A,B}$  of general domain${\displaystyle \mathrm{\Omega }\subset {\mathbb{R}}^n}$  and compute the measure of both sets under parallelogram law.

For vector spaces over thecomplex numbers, the above formulas are not quite correct because they do not describe theimaginary 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 isantilinear in the first or the second argument. The notation${\displaystyle ⟨x|y⟩,}$  which is commonly used in physics will be assumed to beantilinear in thefirst argument while${\displaystyle ⟨x,y⟩,}$  which is commonly used in mathematics, will be assumed to be antilinear in itssecond argument. They are related by the formula:$${\displaystyle ⟨x,y⟩=⟨y|x⟩\text{ for all }x,y\in H.}$$ 

Thereal 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${\displaystyle x,y\in H}$  is always equal to:^{[4][proof 1]} 

It is always asymmetric map, meaning that^{[proof 1]}$${\displaystyle R(x,y)=R(y,x)\text{ for all }x,y\in H,}$$ and it also satisfies:^{[proof 1]}$${\displaystyle R(ix,y)=-R(x,iy)\text{ for all }x,y\in H,}$$ which in plain English says that to move a factor of${\displaystyle i}$  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.

Unlike its real part, theimaginary part of a complex inner product depends on which argument is antilinear.

Antilinear in first argument

The polarization identities for the inner product${\displaystyle ⟨x|y⟩,}$  which isantilinear in thefirst argument, are

 

where${\displaystyle x,y\in H.}$  The second to last equality is similar to the formulaexpressing a linear functional${\displaystyle \phi }$  in terms of its real part:${\displaystyle \phi (y)=\mathrm{Re}\phi (y)-i(\mathrm{Re}\phi )(iy).}$ 

Antilinear in second argument

The polarization identities for the inner product${\displaystyle ⟨x,y⟩,}$  which isantilinear in thesecond argument, follows from that of${\displaystyle ⟨x|y⟩}$  by the relationship:${\displaystyle ⟨x,y⟩:=⟨y|x⟩=\overline{⟨x|y⟩}\text{ for all }x,y\in H.}$  So for any${\displaystyle x,y\in H,}$ ^[4]

 

This expression can be phrased symmetrically as:^[5]$${\displaystyle ⟨x,y⟩=\frac{1}{4}\sum _{k=0}^3{i}^k{\Vert x+i^{k}y\Vert }^2.}$$ 

Summary of both cases

Thus if${\displaystyle R(x,y)+iI(x,y)}$  denotes the real and imaginary parts of some inner product's value at the point${\displaystyle (x,y)\in H\times H}$  of its domain, then its imaginary part will be: where the scalar${\displaystyle i}$  is always located in the same argument that the inner product is antilinear in.

Using⁠${\displaystyle R(ix,y)=-R(x,iy)}$ ⁠, the above formula for the imaginary part becomes: 

In a normed space${\displaystyle (H,\Vert \cdot \Vert ),}$  if theparallelogram law$${\displaystyle \Vert x+y{\Vert }^2+\Vert x-y{\Vert }^2=2\Vert x{\Vert }^2+2\Vert y{\Vert }^2}$$ holds, then there exists a uniqueinner product${\displaystyle ⟨\cdot ,\cdot ⟩}$  on${\displaystyle H}$  such that${\displaystyle \Vert x{\Vert }^2=⟨x,x⟩}$  for all${\displaystyle x\in H.}$ ^[4][1]

Another necessary and sufficient condition for there to exist an inner product that induces a given norm${\displaystyle \Vert \cdot \Vert }$  is for the norm to satisfyPtolemy's inequality, which is:^[6]$${\displaystyle \Vert x-y\Vert \Vert z\Vert +\Vert y-z\Vert \Vert x\Vert \ge \Vert x-z\Vert \Vert y\Vert \text{ for all vectors }x,y,z.}$$ 

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

If${\displaystyle A:H\to Z}$  is alinearisometry between two Hilbert spaces (so${\displaystyle \Vert Ah\Vert =\Vert h\Vert }$  for all${\displaystyle h\in H}$ ) then$${\displaystyle ⟨Ah,Ak{⟩}_Z=⟨h,k{⟩}_H\text{ for all }h,k\in H;}$$ that is, linear isometries preserve inner products.

If${\displaystyle A:H\to Z}$  is instead anantilinear isometry then$${\displaystyle ⟨Ah,Ak{⟩}_Z=\overline{⟨h,k{⟩}_H}=⟨k,h{⟩}_H\text{ for all }h,k\in H.}$$ 

The second form of the polarization identity can be written as$${\displaystyle \Vert \mathbf{\text{u}}-\mathbf{\text{v}}{\Vert }^2=\Vert \mathbf{\text{u}}{\Vert }^2+\Vert \mathbf{\text{v}}{\Vert }^2-2(\mathbf{\text{u}}\cdot \mathbf{\text{v}}).}$$ 

This is essentially a vector form of thelaw of cosines for thetriangle formed by the vectors⁠${\displaystyle \mathbf{\text{u}}}$ ⁠,⁠${\displaystyle \mathbf{\text{v}}}$ ⁠, and⁠${\displaystyle \mathbf{\text{u}}-\mathbf{\text{v}}}$ ⁠. In particular,$${\displaystyle \mathbf{\text{u}}\cdot \mathbf{\text{v}}=\Vert \mathbf{\text{u}}\Vert \Vert \mathbf{\text{v}}\Vert \cos \theta ,}$$ where${\displaystyle \theta }$  is the angle between the vectors${\displaystyle \mathbf{\text{u}}}$  and⁠${\displaystyle \mathbf{\text{v}}}$ ⁠.

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

The basic relation between the norm and thedot product is given by the equation$${\displaystyle \Vert \mathbf{\text{v}}{\Vert }^2=\mathbf{\text{v}}\cdot \mathbf{\text{v}}.}$$ 

Then and similarly$${\displaystyle \Vert \mathbf{\text{u}}-\mathbf{\text{v}}{\Vert }^2=\Vert \mathbf{\text{u}}{\Vert }^2+\Vert \mathbf{\text{v}}{\Vert }^2-2(\mathbf{\text{u}}\cdot \mathbf{\text{v}}).}$$ 

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

The standardJordan–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.^[7]

Define various senses of orthogonality:

• isosceles:$\Vert x+y\Vert =\Vert x-y\Vert$ 
• Roberts’:$\Vert x+ty\Vert =\Vert x-ty\Vert$  for all scalar$t$ .
• Pythagorean:${\Vert x+y\Vert }^2=\Vert x{\Vert }^2+{\Vert y\Vert }^2$ 
• Birkhoff–James:$\Vert x\Vert \le \Vert x+ty\Vert$  for all scalar$t$ .

Let$V$  be a vector space over the real or complex numbers. Let$\Vert \cdot \Vert$  be a norm over$V$ . 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$V$  is over the complex numbers.

• (von Neumann–Jordan condition) The norm satisfies the parallelogram identity.
• (weakened von Neumann–Jordan condition)$\Vert x+y{\Vert }^2+\Vert x-y{\Vert }^2=4$  for all unit vectors$x,y$ . That is, the norm satisfies the parallelogram identity for unit vectors.
• For any$x,y\in V$ , the set of points equidistant to$x,y$  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$W\subset V$ , for every$x\in W$ , there exists$y\in W$  that is Roberts’ orthogonal to$x$ .
• Isosceles orthogonality implies Pythagorean orthogonality.
• Pythagorean orthogonality implies isosceles orthogonality.
• If$x,y$  are Pythagorean orthogonal, then so are$x,-y$ .
• Birkhoff–James orthogonality is symmetric.
• If$\Vert x\Vert =\Vert y\Vert$  and$t,s$  are real, then$\Vert tx+sy\Vert =\Vert sx+ty\Vert$ .

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$\{x\cos \theta +y\sin \theta :\theta \in [0,2\pi ]\}$ , for some unit vectors$x,y$ .

The Banach-Mazur rotation problem: Given aseparableBanach space$V$  such that for any two unit vectors$x,y,$  there exists a linear surjective isometry$T$  such that$T(x)=y$  or$T(y)=x$ , is$V$  isometrically isomorphic to a Hilbert space?

The general case of the problem isopen. 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.^[8]

The polarization identities are not restricted to inner products. If${\displaystyle B}$  is anysymmetric bilinear form on a vector space, and${\displaystyle Q}$  is thequadratic form defined by$${\displaystyle Q(v)=B(v,v),}$$ then 

The so-calledsymmetrization map generalizes the latter formula, replacing${\displaystyle Q}$  by a homogeneous polynomial of degree${\displaystyle k}$  defined by${\displaystyle Q(v)=B(v,\dots ,v),}$  where${\displaystyle B}$  is a symmetric${\displaystyle k}$ -linear map.^[9]

The formulas above even apply in the case where thefield ofscalars hascharacteristic 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 inL-theory; for brevity, in this context "symmetric bilinear forms" are often referred to as "symmetric forms".

These formulas also apply to bilinear forms onmodules over acommutative ring, though again one can only solve for${\displaystyle B(u,v)}$  if 2 is invertible in the ring, and otherwise these are distinct notions. For example, over the integers, one distinguishesintegral quadratic forms from integralsymmetric forms, which are a narrower notion.

More generally, in the presence of a ring involution or where 2 is not invertible, one distinguishes${\displaystyle \epsilon }$ -quadratic forms and${\displaystyle \epsilon }$ -symmetric forms; 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" (integralquadratic form) and "twos in" (integralsymmetric form) was understood – see discussion atintegral quadratic form; and in thealgebraization ofsurgery theory, Mishchenko originally usedsymmetricL-groups, rather than the correctquadraticL-groups (as in Wall and Ranicki) – see discussion atL-theory.

Finally, in any of these contexts these identities may be extended tohomogeneous polynomials (that is,algebraic forms) of arbitrarydegree, where it is known as thepolarization formula, and is reviewed in greater detail in the article on thepolarization of an algebraic form.

• Inner product space – Vector space with generalized dot product
• Law of cosines – Generalization of Pythagorean theorem
• Mazur–Ulam theorem – Surjective isometries are affine mappings
• Minkowski distance – Vector distance using pth powers
• Parallelogram law – Sides and diagonals have equal sums of squares
• Ptolemy's inequality – Relation between distances of four points

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