# Polarization identity

https://en.wikipedia.org/wiki/Polarization_identity

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 vectors in terms of the norm of a normed vector space. Equivalently, the polarization identity describes when a norm can be assumed to arise from an inner product. In that terminology:  

The norm associated with any inner product space satisfies the parallelogram law. In fact, the parallelogram law characterizes those normed spaces that arise from inner product spaces. Explicitly, this means that a normed space $(H,\|\,\cdot \,\|),$ the parallelogram law holds if and only if there exists an inner product $\langle \,\cdot \,,\,\cdot \,\rangle$ on $H$ such that $\|x\|^{2}=\langle x,\ x\rangle$ for all $x\in H.$ ## Formulas

Any inner product on a vector space induces a norm by the equation

$\|x\|={\sqrt {\langle x,x\rangle }}.$ The polarization identities reverse this relationship, recovering the inner product from the norm.

### Real vector spaces

If the vector space is over the real numbers, then expanding out the squares of binomials reveals

{\begin{alignedat}{4}\langle x,y\rangle &={\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]&={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right).\end{alignedat}} These various forms are all equivalent by the parallelogram law:

$2\|x\|^{2}+2\|y\|^{2}=\|x+y\|^{2}+\|x-y\|^{2}.$ ### Complex vector spaces

For vector spaces over the complex numbers, 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 in the first or the second argument. The notation $\langle x|y\rangle ,$ which is commonly used in physics will be assumed to be antilinear in the first argument while $\langle x,\,y\rangle ,$ which is commonly used in mathematics, will be assumed to be antilinear its the second argument. They are related by the formula:

$\langle x,\,y\rangle =\langle y\,|\,x\rangle \quad {\text{ for all }}x,y\in H.$ 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 is always equal to:

{\begin{alignedat}{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{alignedat}} The equalities $R(x,y)=R(y,x)$ and $R(y,ix)=-R(x,iy)$ hold for all vectors $x,y\in H.$ Antilinear in first argument

For the inner product $\langle x\,|\,y\rangle ,$ which is antilinear in the first argument, for any $x,y\in H,$ {\begin{alignedat}{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)-iR(x,iy)\\&=R(x,y)+iR(ix,y).\\\end{alignedat}} The second to last equality is similar to the formula expressing a linear functional $\varphi$ in terms of its real part: $\varphi (y)=\operatorname {Re} \varphi (y)-i(\operatorname {Re} \varphi )(iy).$ Antilinear in second argument

The formula for the inner product $\langle x,\ y\rangle$ which is antilinear in the second argument, follows from that of $\langle x\,|\,y\rangle$ by the relationship: $\langle x,\ y\rangle :=\langle y\,|\,x\rangle ={\overline {\langle x\,|\,y\rangle }}.$ So for any $x,y\in H,$ {\begin{alignedat}{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)+iR(x,iy)\\&=R(x,y)-iR(ix,y).\\\end{alignedat}} This expression can be phrased symmetrically as: 

$\langle x,y\rangle ={\frac {1}{4}}\sum _{k=0}^{3}i^{k}\left\|x+i^{k}y\right\|^{2}.$ ## Reconstructing the inner product

In a normed space $(H,\|\,\cdot \,\|),$ if the parallelogram law

$\|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^{2}+2\|y\|^{2}$ holds, then there is an inner product on $H$ such that $\|x\|^{2}=\langle x,\ x\rangle$ for all $x\in H.$ 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

$\langle x,\ y\rangle ={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right)\quad {\text{ for all }}x,y\in H.$ We need prove that this formula defines an inner product which induces the norm $\|\,\cdot \,\|.$ That is, we must show:

1. $\langle x,x\rangle =\|x\|^{2},\quad x\in H$ 2. $\langle x,y\rangle =\langle y,x\rangle ,\quad x,y\in H$ 3. $\langle x+z,y\rangle =\langle x,y\rangle +\langle z,y\rangle \quad {\text{ for all }}x,y,z\in H,$ 4. $\langle \alpha x,y\rangle =\alpha \langle x,y\rangle \quad {\text{ for all }}x,y\in H{\text{ and all }}\alpha \in \mathbb {R}$ (This axiomatization omits positivity, which is implied by (1) and the fact that $\|\,\cdot \,\|$ is a norm.)

For properties (1) and (2), substitute: ${\textstyle \langle x,x\rangle ={\frac {1}{4}}\left(\|x+x\|^{2}-\|x-x\|^{2}\right)=\|x\|^{2},}$ and $\|x-y\|^{2}=\|y-x\|^{2}.$ For property (3), it is convenient to work in reverse. It remains to show that

$\|x+z+y\|^{2}-\|x+z-y\|^{2}{\overset {?}{=}}\|x+y\|^{2}-\|x-y\|^{2}+\|z+y\|^{2}-\|z-y\|^{2}$ or equivalently,
$2(\|x+z+y\|^{2}+\|x-y\|^{2})-2(\|x+z-y\|^{2}+\|x+y\|^{2}){\overset {?}{=}}2\|z+y\|^{2}-2\|z-y\|^{2}.$ Now apply the parallelogram identity:

$2\|x+z+y\|^{2}+2\|x-y\|^{2}=\|2x+z\|^{2}+\|2y+z\|^{2}$ $2\|x+z-y\|^{2}+2\|x+y\|^{2}=\|2x+z\|^{2}+\|z-2y\|^{2}$ Thus it remains to verify:

${\cancel {\|2x+z\|^{2}}}+\|2y+z\|^{2}-({\cancel {\|2x+z\|^{2}}}+\|z-2y\|^{2}){\overset {?}{{}={}}}2\|z+y\|^{2}-2\|z-y\|^{2}$ $\|2y+z\|^{2}-\|z-2y\|^{2}{\overset {?}{=}}2\|z+y\|^{2}-2\|z-y\|^{2}$ But the latter claim can be verified by subtracting the following two further applications of the parallelogram identity:

$\|2y+z\|^{2}+\|z\|^{2}=2\|z+y\|^{2}+2\|y\|^{2}$ $\|z-2y\|^{2}+\|z\|^{2}=2\|z-y\|^{2}+2\|y\|^{2}$ Thus (3) holds.

It can be verified by induction that (3) implies (4), as long as $\alpha \in \mathbb {Z} .$ But "(4) when $\alpha \in \mathbb {Z}$ " implies "(4) when $\alpha \in \mathbb {Q}$ ". And any positive-definite, real-valued, $\mathbb {Q}$ -bilinear form satisfies the Cauchy–Schwarz inequality, so that $\langle \cdot ,\cdot \rangle$ is continuous. Thus $\langle \cdot ,\cdot \rangle$ must be $\mathbb {R}$ -linear as well.

## Applications and consequences

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

### Isometries

If $A:H\to Z$ is a linear isometry between two Hilbert spaces (so $\|Ah\|=\|h\|$ for all $h\in H$ ) then

$\left\langle Ah,Ak\right\rangle _{Z}=\left\langle h,k\right\rangle _{H}\quad {\text{ for all }}h,k\in H;$ that is, linear isometries preserve inner products.

If $A:H\to Z$ is instead an antilinear isometry then

$\left\langle Ah,Ak\right\rangle _{Z}={\overline {\left\langle h,k\right\rangle _{H}}}=\left\langle k,h\right\rangle _{H}\quad {\text{ for all }}h,k\in H.$ ### Relation to the law of cosines

The second form of the polarization identity can be written as

$\|{\textbf {u}}-{\textbf {v}}\|^{2}=\|{\textbf {u}}\|^{2}+\|{\textbf {v}}\|^{2}-2({\textbf {u}}\cdot {\textbf {v}}).$ This is essentially a vector form of the law of cosines for the triangle formed by the vectors ${\textbf {u}},{\textbf {v}},$ and ${\textbf {u}}-{\textbf {v}}.$ In particular,

${\textbf {u}}\cdot {\textbf {v}}=\|{\textbf {u}}\|\,\|{\textbf {v}}\|\cos \theta ,$ where $\theta$ is the angle between the vectors ${\textbf {u}}$ and ${\textbf {v}}.$ ### Derivation

The basic relation between the norm and the dot product is given by the equation

$\|{\textbf {v}}\|^{2}={\textbf {v}}\cdot {\textbf {v}}.$ Then

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

## Generalizations

### Symmetric bilinear forms

The polarization identities are not restricted to inner products. If $B$ is any symmetric bilinear form on a vector space, and $Q$ is the quadratic form defined by

$Q(v)=B(v,v),$ then
{\begin{aligned}2B(u,v)&=Q(u+v)-Q(u)-Q(v),\\2B(u,v)&=Q(u)+Q(v)-Q(u-v),\\4B(u,v)&=Q(u+v)-Q(u-v).\end{aligned}} The so-called symmetrization map generalizes the latter formula, replacing $Q$ by a homogeneous polynomial of degree $k$ defined by $Q(v)=B(v,\ldots ,v),$ where $B$ is a symmetric k-linear map. 

The formulas above even apply in the case where the field of scalars has 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 modules over a commutative ring, though again one can only solve for $B(u,v)$ if 2 is invertible in the ring, and otherwise these are distinct notions. For example, over the integers, one distinguishes integral quadratic forms from integral symmetric forms, which are a narrower notion.

More generally, in the presence of a ring involution or where 2 is not invertible, one distinguishes $\varepsilon$ -quadratic forms and $\varepsilon$ -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" (integral quadratic form) and "twos in" (integral symmetric form) was understood – see discussion at integral quadratic form; and in the algebraization of surgery theory, Mishchenko originally used symmetric L-groups, rather than the correct 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 polynomials (that is, algebraic forms) of arbitrary 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.