Polarization identity

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

${\displaystyle {\begin{aligned}\langle u,v\rangle &={\frac {1}{2}}\left(\|u+v\|^{2}-\|u\|^{2}-\|v\|^{2}\right)\\[3pt]&={\frac {1}{2}}\left(\|u\|^{2}+\|v\|^{2}-\|u-v\|^{2}\right)\\[3pt]&={\frac {1}{4}}\left(\|u+v\|^{2}-\|u-v\|^{2}\right).\end{aligned}}}$

These various forms are all equivalent by the parallelogram law:

${\displaystyle 2\|{\textbf {u}}\|^{2}+2\|{\textbf {v}}\|^{2}=\|{\textbf {u}}+{\textbf {v}}\|^{2}+\|{\textbf {u}}-{\textbf {v}}\|^{2}.}$

For vector spaces over the complex numbers, the above formulas are not quite correct. They assume ${\displaystyle \langle x,y\rangle +\langle y,x\rangle =2\langle x,y\rangle ,}$ but for a complex inner product, this sum instead cancels out the imaginary part. However, an analogous expression does ensure that both real and imaginary parts are retained. The real part of any inner product (no matter which coordinate is antilinear and no matter if it is real or complex) is a symmetric bilinear map that is always equal to:

${\displaystyle {\begin{alignedat}{4}R(x,y):&=\operatorname {Re} \langle x\mid y\rangle =\operatorname {Re} \langle x,y\rangle \\&={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right)\\\end{alignedat}}}$

The complex part of the inner product depends on whether it is antilinear in the first or the second coordinate.

If the inner product ${\displaystyle \langle x\mid y\rangle }$ is antilinear in the first coordinate, then for all ${\displaystyle x,y\in V,}$

${\displaystyle {\begin{alignedat}{4}\langle x\mid y\rangle &={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}-i\|x+iy\|^{2}+i\|x-iy\|^{2}\right)\\&=R(x,y)-iR(x,iy).\\\end{alignedat}}}$

The last equality is similar to the formula expressing a linear functional in terms of its real part.

If the inner product ${\displaystyle \langle x,\ y\rangle }$ is antilinear in the second coordinate then for all ${\displaystyle x,y\in V,}$

${\displaystyle {\begin{alignedat}{4}\langle x,\ y\rangle &={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}+i\|x+iy\|^{2}-i\|x-iy\|^{2}\right)\\&=R(x,y)+iR(x,iy).\\\end{alignedat}}}$

This expression can be phrased symmetrically:

${\displaystyle \langle x,y\rangle ={\frac {1}{4}}\sum _{k=0}^{3}i^{k}\left\|x+i^{k}y\right\|^{2}.}$[3]

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

${\displaystyle \langle x,\ y\rangle ={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right)}$ for all ${\displaystyle x,y\in V.}$

We need prove that this formula defines an inner product which induces the norm ${\displaystyle \|\cdot \|}$. That is, we must show:

1. ${\displaystyle \langle x,x\rangle =\|x\|^{2},\quad x\in V}$
2. ${\displaystyle \langle x,y\rangle =\langle y,x\rangle ,\quad x,y\in V}$
3. ${\displaystyle \langle x+z,y\rangle =\langle x,y\rangle +\langle z,y\rangle \quad }$ for all ${\displaystyle x,y,z\in V,}$
4. ${\displaystyle \langle \alpha x,y\rangle =\alpha \langle x,y\rangle \quad }$ for all ${\displaystyle x,y\in V,}$ and all ${\displaystyle \alpha \in \mathbb {R} }$

(This axiomatization omits positivity, which is implied by (1) and the fact that ||·|| is a norm.)

For properties (1) and (2), we simply substitute: ${\displaystyle \langle x,x\rangle ={\frac {1}{4}}\left(\|x+x\|^{2}-\|x-x\|^{2}\right)=\|x\|^{2}}$, and ${\displaystyle \|x-y\|^{2}=\|y-x\|^{2}}$.

For property (3), it is convenient to work in reverse. We seek to show that

${\displaystyle \|x+z+y\|^{2}-\|x+z-y\|^{2}{\overset {?}{=}}\|x+y\|^{2}-\|x-y\|^{2}+\|z+y\|^{2}-\|z-y\|^{2}}$

Equivalently,

${\displaystyle 2(\|x+z+y\|^{2}+\|x-y\|^{2})-2(\|x+z-y\|^{2}+\|x+y\|^{2}){\overset {?}{=}}2\|z+y\|^{2}-2\|z-y\|^{2}}$

Now we apply the parallelogram identity:

${\displaystyle 2\|x+z+y\|^{2}+2\|x-y\|^{2}=\|2x+z\|^{2}+\|2y+z\|^{2}}$

${\displaystyle 2\|x+z-y\|^{2}+2\|x+y\|^{2}=\|2x+z\|^{2}+\|z-2y\|^{2}}$

Thus the claim we seek is

${\displaystyle {\cancel {\|2x+z\|^{2}}}+\|2y+z\|^{2}-({\cancel {\|2x+z\|^{2}}}+\|z-2y\|^{2}){\overset {?}{=}}2\|z+y\|^{2}-2\|z-y\|^{2}}$

${\displaystyle \|2y+z\|^{2}-\|z-2y\|^{2}{\overset {?}{=}}2\|z+y\|^{2}-2\|z-y\|^{2}}$

But the latter claim can be verified by subtracting the following two further applications of the parallelogram identity:

${\displaystyle \|2y+z\|^{2}+\|z\|^{2}=2\|z+y\|^{2}+2\|y\|^{2}}$

${\displaystyle \|z-2y\|^{2}+\|z\|^{2}=2\|z-y\|^{2}+2\|y\|^{2}}$

Thus (3) holds.

It is straightforward to verify by induction that (3) implies (4), as long as we restrict to α∈ℤ. But "(4) when α∈ℤ" implies "(4) when α∈ℚ". And any positive-definite, real-valued, ℚ-bilinear form satisfies the Cauchy–Schwarz inequality, so that ⟨·,·⟩ is continuous. Thus ⟨·,·⟩ must be ℝ-linear as well.

The second form of the polarization identity can be written as

${\displaystyle \|{\textbf {u}}-{\textbf {v}}\|^{2}=\|{\textbf {u}}\|^{2}+\|{\textbf {v}}\|^{2}-2({\textbf {u}}\cdot {\textbf {v}}).}$

This is essentially a vector form of the law of cosines for the triangle formed by the vectors ${\displaystyle {\textbf {u}}}$, ${\displaystyle {\textbf {v}}}$, and ${\displaystyle {\textbf {u}}-{\textbf {v}}}$. In particular,

${\displaystyle {\textbf {u}}\cdot {\textbf {v}}=\|{\textbf {u}}\|\,\|{\textbf {v}}\|\cos \theta ,}$

where ${\displaystyle \theta }$ is the angle between the vectors ${\displaystyle {\textbf {u}}}$ and ${\displaystyle {\textbf {v}}}$.

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

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

Then

${\displaystyle {\begin{aligned}\|{\textbf {u}}+{\textbf {v}}\|^{2}&=({\textbf {u}}+{\textbf {v}})\cdot ({\textbf {u}}+{\textbf {v}})\\[3pt]&=({\textbf {u}}\cdot {\textbf {u}})+({\textbf {u}}\cdot {\textbf {v}})+({\textbf {v}}\cdot {\textbf {u}})+({\textbf {v}}\cdot {\textbf {v}})\\[3pt]&=\|{\textbf {u}}\|^{2}+\|{\textbf {v}}\|^{2}+2({\textbf {u}}\cdot {\textbf {v}}),\end{aligned}}}$

and similarly

${\displaystyle \|{\textbf {u}}-{\textbf {v}}\|^{2}=\|{\textbf {u}}\|^{2}+\|{\textbf {v}}\|^{2}-2({\textbf {u}}\cdot {\textbf {v}}).}$

Forms (1) and (2) of the polarization identity now follow by solving these equations for u · v, while form (3) follows from subtracting these two equations. (Adding these two equations together gives the parallelogram law.)

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

${\displaystyle Q(v)=B(v,v),}$

then

${\displaystyle {\begin{aligned}2B(u,v)&=Q(u+v)-Q(u)-Q(v),\\2B(u,v)&=Q(u)+Q(v)-Q(u-v),\\4B(u,v)&=Q(u+v)-Q(u-v).\end{aligned}}}$

The so-called symmetrization map generalizes the latter formula, replacing Q by a homogeneous polynomial of degree k defined by Q(v) = B(v, ..., v), where B is a symmetric k-linear map.[4]

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 ε-quadratic forms and ε-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.

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.